复分析手写笔记整理
复分析手写笔记整理
说明：以下内容按原图
IMG_2596.JPG至IMG_2639.JPG的顺序识别整理，并按抽代笔记示例的 Markdown 风格排版。手写页中划掉内容以~~删除线~~保留；OCR 难以确认的笔迹已尽量按上下文保留。
1. 复数、序与基本拓扑
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Formal def: \mathbb{P}[x] is the ring of polynomials in x with coefficient in \mathbb{R}, let (x^2+1) be the maximal ideal generated by x^2+1 in \mathbb{P}[x]. then the ce the ring (\mathbb{C}; t, x) \cong \mathbb{P}[x]/x^2+1.
\mathbb{Z} is represented by a point or a vector on plan. |z| := \text{modulus of } z = \sqrt{x^2+y^2}. \theta = \arg(z).
operations: z_1 = x_1 + iy_1, z_2 = x_2 + iy_2, z_3 + z_4 = (x_1+x_2) + (y_1+y_2), z_1 \cdot z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)
\bar{z}_1 = x_1 - iy_1, called complex conjugate of z_1. \bar{z}_1 \cdot z_1 = |z_1|^2 \equiv x_1^2 + y_1^2.
We can also represent z in Euler’s formula e^{i\theta} = \cos\theta + i\sin\theta
proof: convergent power series: e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \dots + \frac{(i\theta)^{2n}}{2^{2n}} + \dots + i\theta \cos\theta + i\sin\theta. if clearly converges absolutely. \Rightarrow e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \dots + \frac{(i\theta)^{2n}}{2^{2n}} + \dots + i\theta \cos\theta + i\sin\theta.
We can also represent z in polar form z = r e^{i\theta} where r = |z|. \theta = \arg(z).
De Moivre theorem: z^n = r^n e^{in\theta} (it’s easy to prove use plastopolar form).
col: Double angle formula suppose z = e^{i\theta}, z^2 = (\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\sin\theta.
by De Moivre theorem z^2 = \cos 2\theta + i\sin 2\theta \Rightarrow \cos 2\theta = \cos^2\theta - \sin^2\theta, \sin 2\theta = 2i\sin\theta.
Rationalize denominator \frac{1}{1+zi} \times \frac{1-zi}{1-zi} = \frac{1-zi}{z}.
Roots of unity: z^n = 1 has n roots in complex number. z^n narrely, e^{i\theta}, e^{\frac{2\pi i}{n}}, e^{\frac{in-yki}{n}}, all of satisfy z^n = e^{2\pi ii} = 1, n \in \mathbb{Z}. they distribute around unit cycle.
Suppose we have a polynomial P(z) with real coefficients. we claim the roots of P(z) sum to a real number.
proof: P(z) = az^n + a(-\frac{1}{z})z^{n-1} + \dots + a(z-an)(z-a{n-1})\dots(z-a_1), a \in \mathbb{R}, P(z) = az^n + a(-\frac{1}{z})z^{n-1} + \dots + a(z-a_n), we want to show the coefficient of z^n is all real.
Total ordering^{v}: (using \leq) is a binary relation satisfying (1) any two elements are comparable i.e. Ha, b \in \mathbb{F}, other a \leq b or a \leq a (2) a \leq a and a \leq a (3) a \leq b, b \leq c \Rightarrow a \leq c (transitivity). (4) a \leq b and a \leq a (conjugate).
\Leftrightarrow a = b.
Def: field (\mathbb{F}, x, t, \dots) together with a total order \leq on \mathbb{F} is an ordered field if (1) a \leq b \Rightarrow a+c \leq b \not\leq c. (2) 0 \leq a and 0 \leq b
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\Rightarrow 0 \leq a \cdot b
Proof: \mathbb{C} is not an ordered field,
Either z \leq 0 or z > 0 if z \leq 0 \leq z \Rightarrow 0 \leq z < 0 \Rightarrow 0 \leq z < 0, so for any 0 \leq a in \mathbb{C} we have a \leq -a then a + 0 \leq -a + a \Rightarrow 0 \leq a contradiction
if z \leq 0, so 0 \leq z, x \Rightarrow 0 \leq (-z)^2 = -1, the same as above
solve z^2 = a + zb: suppose z^2 = a + zb, z = x + iy, (x + iy)^2 = x^2 - y^2 + 2xy \Rightarrow x^2 - y^2 = a, 2xy = b \Rightarrow x = \frac{b}{2y} \Rightarrow \frac{b^2}{4y^2} - y^2 = a^4 \Rightarrow b^2 - 4y^4 = a^4 y^2 \Rightarrow y^2 = \frac{b^2}{4}
since, e^{i\theta} = \cos \theta + i \sin \theta, \cos \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}, \cos \sqrt{\theta} = \frac{e^{i\theta} + e^{-i\theta}}{2}, e^{\frac{i\theta}{2}} = \frac{e^{i\theta} + e^{-i\theta}}{2}, e^{\frac{i\theta}{2} + \pi in} = \frac{e^{i\theta} + e^{-i\theta}}{2}, e^{\frac{i\theta}{2} - \pi in} = \frac{e^{i\theta} - e^{-i\theta}}{2}
= \frac{e^{i\theta}}{2} \left( \cos \frac{1}{2} + i \sin \frac{1}{2} \right) + \frac{e^{i\theta}}{2} \left( \cos \frac{1}{2} - i \sin \frac{1}{2} \right)
1 + \cos \theta + \cos 2\theta + \dots + \cos \theta = \text{Re} \left( e^0 + e^{i\theta} + e^{i2\theta} + \dots + e^{in\theta} \right), suppose y = e^0 + e^{i\theta} + \dots
+ e^{i\theta}, e^{i\theta} y - y = e^{i\theta} e^{in\theta} - 1 \Rightarrow y = \frac{e^{in\theta} - 1}{e^{i\theta} - 1}
Def: A sequence z_1, z_2, \dots converges to w \in \mathbb{C} if \lim_{n \rightarrow \infty} |z_n - w| = 0.
likely, the sequence is cauchy if \lim_{n \rightarrow \infty} |z_n - w| = 0.
A set is open if every point in a interior point. A set is closed if its complement is open.
Disk \overline{Dr(z_0)} := {z \in \mathbb{C} \mid |z - z_0| \leq r}, Dr(z_0) := {z \in \mathbb{C} \mid |z - z_0| < r}, \overline{Dr(z_0)} := {z \in \mathbb{C} \mid |z - z_0| \leq r}, Dr(z_0): the open unit centered on z_0.
Let \mathbb{R} \subset \mathbb{C} be a set \text{diam}(\mathbb{R}) := \sup_{z \in \mathbb{R}} |z - w|.
compactness (\frac{|z| - |z_0|}{2}) \in \mathbb{C}. closed with boundary diameter. (All are the same in \mathbb{R})
Verify the identity \sin \theta + \sin \phi = 2 \sin \frac{\theta + \phi}{2} \cos \frac{\theta - \phi}{2}, using Euler’s formula e^{i\theta} = \cos \frac{\theta}{2} + i \sin \frac{\theta}{2}, e^{i\phi} = \cos \frac{\phi}{2} - i \sin \frac{\phi}{2}, e^{i\theta} + e^{i\phi} = e^{i(\theta + \phi)}
PHS = 2 \left( \frac{e^{i\theta} - e^{i\phi}}{2i} \right) \left( \frac{e^{i\theta} + e^{i\phi}}{2} \right) = \frac{e^{i\theta} + e^{i\phi} - e^{i(\theta + \phi)}}{2i}
e^{i\theta} = \frac{1}{2} = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2}
eg. (b) \frac{1}{2} = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2} \Rightarrow z = \frac{1}{2}
Defn: a connected open set is called a region. A set is connected if there are the open sets U \cap V = S, U \cap V = S, U \cap V = \emptyset. Path connected set is path connected if for every z_1, z_2 \in S, z_1 a path \gamma \in S \Rightarrow z_2 a path \gamma’ \in S
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z_1 to z_2.
f(z) (a single value function on \Omega \subset \mathbb{C}) is cartinous if for every z_1 > 0, z_2 > 0, |z_1 - z_2| < \delta \Rightarrow |f(z_1) - f(z_2)| < \varepsilon, z_1, z_2 \in \Omega.
f(z) is differentiable at z_0 \in \Omega if \lim_{h \rightarrow 0} \frac{f(z_0 + h) - f(z_0)}{h} exists. this def means h limit should exist whereaver h \rightarrow 0 at any path.
Examples that not differentiable. f(z) = \bar{z}, approach 0 along real axis, the limit is \infty, approach 0 along imaginary axis the limit is -\infty.
Differentiable function polynomial proof: P(z) = an z^n + a{n-1} z^{n-1} + \dots + a_0
P(z0 + h) - P(z_0) = a_n [(z_0 + h) - z_0] + a{n-1} [(z_0 + h) - z_0^{n-1}] + \dots + a_0 - a_0 = a_n h^{n-1} f(z_0).
+ a_{n-1} h^{n-1} f(z_0) + \dots + a_1 h^{n-1} f(z_0) = a_n h^{n-1} f(z_0) = h a_n z_0^{n-1} f(z_0) = h a_n z_0^{n-1} f(z_0) + a_0 z_0
Holomorphic Holomorphic function: Function are single valued and defined on an open set \Omega with a exact limit of \lim_{h \rightarrow 0} \frac{f(z_0 + h) - f(z_0)}{h}. h \in \Omega.
Recall f(z) is holomorphic C complex differentiable / regular / analytic at z_0 \in \mathbb{C} if the limit f’(z0) := \lim{h \rightarrow 0} \frac{f(z_0 + h) - f(z_0)}{h} exist for h \in \mathbb{C}.
If f is holomorphic everywhere on complex plane, it is called entire
eg. play horizontals, e^z, triangular functions
examples of functions that not holomorphic: \bar{z}, arg z, |z|.
let: f : u + iv \rightarrow u(x,y) if h is real f’(z) = \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}; if h is imaginary,
f’(z) = \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}. A necessary condition of f is holomorphic both real and imaginary part at the equal \Rightarrow \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} and \frac{\partial v}{\partial x} = -\frac{\partial v}{\partial y}.
complex - Riemann equations (equations)
Jacobian \left| \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right| then the Jacobian \left| \frac{\partial u, v}{\partial x, y} \right| = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial v}{\partial y} \frac{\partial u}{\partial x}; if a function is holomorphic then the Jacobian \left| \frac{\partial u, v}{\partial x, y} \right| = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial v}{\partial y} \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial v}{\partial y} \frac{\partial u}{\partial x} = 0
\left(\frac{\partial v}{\partial x}\right)^2 = |f’(z)| preserve angles?
Assume enough differentiable. \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial x \partial y} like wise \frac{\partial f}{\partial y} = -\frac{\partial f}{\partial x f}. u is homogeneous likely, vis harmonic.
\Rightarrow \frac{\partial f}{\partial y} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 0, u is homogeneous.
find v that makes f = u + iv holomorphic.
Assume u = x^2 - y^2 find v that makes f = u + iv holomorphic.
\frac{\partial v}{\partial x} = 2x = \frac{\partial v}{\partial y}, \Rightarrow v = 2xy.
\Rightarrow v = 2xy.
Def: \frac{\partial f}{\partial z} = \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} and \frac{\partial f}{\partial z} = f’(z).
cR equation \Rightarrow \frac{\partial f}{\partial z} = 0 and \frac{\partial f}{\partial z} = f’(z).
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Theorem: If u and v are continuously differentiable and satisfy the CR equation on \Lambda, then f is holomorphic (Note: \Lambda is open, because \mathbb{Z}^2 satisfy the crossing CR equations at 0, but is not holomorphic.
proof: f(x+h) - f(x) = \frac{d^n}{dx^n} h_1 + \frac{d^n}{dx^n} h_2 + i \left( \frac{d^n}{dx^n} h_1 + \frac{d^n}{dx^n} h_2 \right) + o(h). (h = h_1 + ih_2)
= \frac{d^n}{dx} i \left( \frac{d^n}{dx} h_1 \right) + \frac{d^n}{dx} h_2 + i \frac{d^n}{dx} h_2 = \left( \frac{d^n}{dx} - i \frac{d^n}{dx} \right) h_1 + \left( \frac{d^n}{dx} + i \frac{d^n}{dx} \right) h_2 = \left( \frac{d^n}{dx} - i \frac{d^n}{dx} \right) h_1 + i \left( \frac{d^n}{dx} + i \frac{d^n}{dx} \right) h_2
= \left( \frac{d^n}{dx} - i \frac{d^n}{dx} \right) h_1. So f(z) = \lim_{h \rightarrow 0} \frac{f(z+h) - f(z)}{h} = \frac{d^n}{dx} - i \frac{d^n}{dx} = 0.
Example: f_1 and f_2 are simultaneously differentiable; the CR equation set f(z) = u + i v, f(z) = i \cdot u(x, y) - i \cdot v(x, y). the CR equation of \alpha f(z) is x \frac{d^n}{dx^n} = y \frac{d^n}{dy^n}, \frac{d^n}{dy^n} = - \frac{d^n}{dx}, the CR equation is \frac{d^n}{dx} = \frac{d^{n-1}}{dy}.
Exercise: A harmonic function satisfies \frac{d^n}{dx^n} = 0, \frac{d^n}{dz^n} = \frac{1}{4} \left( \frac{d^2}{dx^2} + \frac{1}{i} \frac{d^2}{dy^2} \right).
= \frac{i}{4} \left( \frac{d^2}{dx^2} + \frac{1}{i} \frac{d^2}{dy^2} \right) - \frac{1}{i} \frac{d^2}{dx^2} + \frac{d^2}{dx^2} \frac{d^2}{dy^2} = 0.
A rational function is the ratio to of two polynomials. like \frac{P_1(z)}{P_2(z)}.
hr. A rational function is holomorphic among from the zeros of derivation of: A consequence of prop 2.2 of:
prop 2.2 of tb: If \alpha and g are holomorphic in \Lambda, then (1) f_{\alpha g} is holomorphic. (f{\alpha g})’ = f’{\alpha} g’(z) f_g is holomorphic (f_g)’ = f_g’ + f’_g(z) g \neq 0 \frac{f’_g}{g} = \frac{f’_g}{g’} is holomorphic, (\frac{f’_g}{g})’ = \frac{f’_g - g’}{g’^2} (4) g of is holomorphic, (g(f(z)))’ = (f(z))’ f’(z).
All the proof are the same as real value function
Exercise: if a holomorphic function has |f| = C C \neq 0, then f is a constant proof: suppose f = u + iv is not constant. u^2 + v^2 = C^2 \Rightarrow x \frac{d^n}{dx^n} + y \frac{d^n}{dy^n} = 0.
\Rightarrow 2u \frac{d^n}{dx} + 2v \frac{d^n}{dy} = -v \frac{d^n}{dx}, also u \frac{d^n}{dy} = -v \frac{d^n}{dy}. \Rightarrow x \frac{d^n}{dx} = -v \frac{d^n}{dy} and \frac{d^n}{dy} = -v \frac{d^n}{dx} \Rightarrow \frac{d^n}{dx} / \frac{d^n}{dy} = \frac{x}{u} = -v = u \Rightarrow u^2 + v^2 = 0. contradiction.
Def: A complete metric space is a metric space on which every cauchy sequence {a_n} is a complex metric space and {a_n} is not complete metric. {a_n} is a complex metric space.
Power series \sum_{n=0}^{\infty} a_n z^n, a_n \in \mathbb{C}. the power series converges absolutely if \lim_{n \rightarrow \infty} |a_n| = 1.
Geometric series \sum_{n=0}^{\infty} z^n \leq \frac{1}{1-z} for |z| R the de te the series diverges |z| don’t know. where R = \lim_{n \rightarrow \infty} \sup |a_n|^{1/n} (Haddard) (\frac{1}{R}) = \infty and
2. 全纯函数、Cauchy-Riemann 方程与调和函数
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\frac{1}{r_0} = 0). call R the radius of convergence.
poit: suppose R is given by Hadamard formula. when |z| r^n for r \geq |z|. so this sequence larger than a divergence sequence \sum_{n=1}^{\infty} |z|, it diverges.
(1) \sum_{n=0}^{\infty} n \cdot z^n does not converge anywhere on unit circle. Let SN = \sum{n=0}^N n z^n if z is convergent, S_N whould be cauchy so \lim{N \rightarrow \infty} S{N+1} - S_N = 0 i.e. \lim_{N \rightarrow \infty} \frac{1}{N} z^N = 0, contradiction.
(2) start \sum_{n=1}^{\infty} \frac{z^n}{h^n} converge everywhere on unit cycle take \sum_{n=1}^{\infty} \frac{|z|^n}{h^n} = \frac{1}{h^{n+1}} it converges, so \sum_{n=1}^{\infty} \frac{z^n}{h^n} converges absolutely.
(3) \sum_{n=1}^{\infty} \frac{z^n}{h^n} converges everywhere on unit cycle except z \geq |z|. let C_n = \frac{1}{z}, honoritionally as n \rightarrow \infty. \sum_{n=0}^{\infty} z^n is bounded as long as z \neq |z| independently of the series converge by something like Dirichlet criticon.
If \arg z is equal to a rational multiple of 2\pi, this follows. say \arg z = 2\pi so z^n + z^{n+1} + \dots + z^{n+k-1} = 0. because it distribute uniformly around circle.
If \arg z is not a rational multiple of 2\pi, we say \arg z = \pi - 2\pi, we can approximate z by multiple rational numbers. arbitrarily well. Hence the \sum sum is still independently of n/V.
Then the power series f(z) = \sum_{n=0}^{\infty} a_n z^n is a holomorphic function for |z| < R.
For |z| < R, we also have f(z) = \sum_{n=0}^{\infty} n a_n z^{n-1} with the same radius as f.
pt: \lim{n \rightarrow \infty} \sup |a_n|^{\frac{1}{n}} = \lim{n \rightarrow \infty} \frac{1}{n^{\frac{1}{n}}} \lim{n \rightarrow \infty} \sup |a_n|^{\frac{1}{n}} = \lim{n \rightarrow \infty} \frac{1}{n^{\frac{1}{n}}} \sup |a_n|^{\frac{1}{n}} = \frac{1}{R} so f(z) and f’(z) have similar convergent radius.
let fN(z) = \sum{n=1}^N \frac{z^n}{h^n} an z^{n-1} + \sum{n=N+1}^{\infty} \frac{z^n}{h^n} a_n z^{n-1} = S_N(z) + T_N(z). define g(z) = \sum{n=0}^{\infty} h a_n z^{n-1} - g(z_0) = \sum{n=N+1}^{\infty} \frac{z^n}{h^n} an |z|^{n-1} + \sum{n=N+1}^{\infty} \frac{z^n}{h^n} an |z|^{n-1} - g(z_0) = \sum{n=N+1}^{\infty} \frac{z^n}{h^n} an |z|^{n-1} + \sum{n=N+1}^{\infty} \frac{z^n}{h^n} a_n |z|^{n-1} - g(z_0) = 0 as N \rightarrow \infty.
so fN(z) = \sum{n=N+1}^{\infty} \frac{z^n}{h^n} an |z|^{n-1} + \sum{n=N+1}^{\infty} \frac{z^n}{h^n} a_n |z|^{n-1} - g(z_0) = 0 as N \rightarrow \infty.
col. A power series is infinitely complex differentiable within the radius of convergence by theorem above. f is called analytic at z_0 if \exists a power series \sum_{n=0}^{\infty} a_n z^n with positive radius.
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convergence st f(z) = \sum_{h=0}^{\infty} a_h (z-z_0)^h on an absolute neighbourhood of z_0. If z_0 be a point inside the radius of convergence of function f, then f has a convergent power series expansion around z_0. Say f(z) = a0 + \sum{h=0}^{\infty} a_h z^h is the power series around 0, so we have f(z) = a0 + \sum{h=0}^{\infty} ah (z-z_0)^h = f(z_0) + \sum{h=0}^{\infty} ah (z-z_0)^h = f(z_0) + f(z_0) t(z-z_0) + \dots = \sum{h=0}^{\infty} f(z_0) (z-z_0)^h.
Show for |z| < 1 \sum_{h=1}^{\infty} \frac{z^h}{h!} = \frac{1}{1-z}.
claim - within the radius of absolute convergence order of f, not important.
|z| < 1 is \sum_{h=0}^{\infty} z^h = \frac{1}{1-z} (z \neq 0). |z| < 1 is \sum_{h=0}^{\infty} z^h = \frac{1}{1-z}. |z| < 1 is \sum_{h=0}^{\infty} z^h = \frac{1}{1-z}.
Find the radius of convergence of (1/z)^n z^n. \sum_{h=0}^{\infty} h! z^{n+h} = h! \frac{1}{1-z} z^n. |z| < 1 is \sum_{h=0}^{\infty} h! z^{n+h} = h! \frac{1}{1-z} z^n = \frac{1}{1-z} z^n. |z| < 1 is \sum_{h=0}^{\infty} h! z^n = \frac{1}{1-z} z^n.
|z| < 1 is a subsequence of {z^n}, the radius is still 1. obviously
intergraphing integrals. z(t): [a, b] \rightarrow \mathbb{C} is a parametrised path / a map of real interval [a, b] into \mathbb{C}. f is a closed curve if z(a) = z(b) / is simple if f(z) = z(t_2), t_1 \in t_2 \Rightarrow t_1 = a, t_2 \geq b or t_1 \geq b, t_2 \geq a. Is smooth if z’(t) is at every point and the left-sided limit of derivatives at the endpoint / is piecewise smooth if z is continuous on [a, b] and \exists finitely many points a_1, \dots, a_n in [a, b]. St z(t) is smooth in intervals [a, a_1], \dots, [a_n, b].
Two parametrisations z: [a, b] \rightarrow \mathbb{C} and \hat{z}: [a, b] \rightarrow \mathbb{C} are equivalent if \exists a continuously differentiable bijection f: [a, b] \rightarrow \mathbb{C} from [a, b] to [a, b] with f(s) \geq 0 and \hat{z}(s) = z(f(s)).
Take counter clockwise the orientation let \gamma be the smooth path with parametrisation z: [a, b] \rightarrow \mathbb{C}. Def f(z) dz = \int_a^b f(z(t)) z(t) dt. It’s well-defined since chain rule tells us it won’t be affected by parametrisation / similarly defined for \gamma is clockwise a piecewise smooth.
A primitive for function f on \mathbb{N} is c \notin \mathbb{F} such that F is holomorphic on \mathbb{N} and F’(z) = f(z). If a continuous function f has a primitive F in \mathbb{N}, and \gamma is a path with image within \mathbb{N} then f(z) dz = F(w_1) - F(w_2) it’s only dependent on the endpoint and \gamma.
If f has a primitive on \mathbb{N} \Leftrightarrow f(z) dz = 0. F is a closed periodic in \mathbb{N}. If f has a primitive it’s clear from (i) if f has a destroyed define
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f(z) = \int_C f(z) dz \gamma is a path in \Lambda from fixed z_0 to z, we check it’s a primitive.
Demarck: when does a primitive exist?
Green’s theorem: Let C be a positively oriented piecewise smooth simple closed curve in \mathbb{R}^2, let D be the region bounded by C if L and M are functions of x and y defined on an open set containing D and having continuous partial derivatives then \int_C L dx + M dy = \int_D (\frac{\partial L}{\partial x} + \frac{\partial L}{\partial y}) dx dy if \frac{\partial L}{\partial x} = \frac{\partial L}{\partial y}. Apply Green theorem to real part and imaginary part 0 = \int_C f dz = \int_C u dx - \text{Vol} + i \int_C u dy + \text{Vol} x \Leftrightarrow \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} and \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} CR-equation !!! i.e. f has a primitive \Leftrightarrow f is holomorphic.
example: C is the unit cyclicle with two orientations z(\theta) = e^{i\theta} dz = i e^{i\theta} d\theta \int_C \int_C e^{i\theta} d\theta = i \int_0^{2\pi} e^{i(\pi+i\theta)} d\theta = \int_0^0, \pi+i\theta \neq 0 if C is any truely oriented circle not containing 0, z^n is always holomorphic \Rightarrow if f has primitive \Rightarrow \int_C z^n dz = 0.
Explicit: if |a| < r < |b| then \int_C \frac{1}{(z-a)(z-b)} dz = \frac{2\pi i}{a-b}, C with radius r. \int_C \frac{1}{(z-a)(z-b)} dz = \frac{1}{a-b} \left[ \int_C \frac{1}{z-a} dz - \int_C \frac{1}{z-b} dz \right]. \int_C \frac{1}{z-b} dz = 0 because Region doesn’t contain b thus \frac{1}{z-b} is holomorphic \int_C \frac{1}{z-a} dz = \int_C \frac{\frac{dz}{z}}{z} C’ is unit cyclicle around a.
Example: f(z) is analytic on a neighbourhood of a closed curve C share that \int_C f(z) f(z) dz is purely imaginary.
Let f = u + iv. then we derive \int_C f df = \int_C (u - iv) (du + idv) = \int_C u du + i \int_C v dv then we derive u du + v dv = d(\frac{u^2+v^2}{2}) has primitive, so real part is 0 \Rightarrow give a parameterisation for f. For f, z = t + ib \sqrt{1-a^2}, f’’_{zz} = t - ib \sqrt{1-a^2}.
Jordan curve theorem let C be a simple closed curve in \mathbb{R}^2 then the outcomplex of C in \mathbb{R}^2 consist of two disjoint open sets, call inside and outside. proof:
housat’s theorem: If \Lambda is open set in C, and T (T is a 1-dimensional contour, whose interior is also contained in \Lambda, then \int_T f(z) dz = 0 whenever f is holomorphic in \Lambda. proof: f is the subdivision \Delta then then \int_T f dz = \int_T f dz. subdivide again T_2: Let T_1^* be the triangle having the largest subdivide again T_2:
IMG_2603
\int_0^{\pi/2} f(z) dz \leq 4 \int_0^{\pi/4} f(z) dz \leq -4 \int_0^{\pi/2} f(z) dz \quad \text{①}, we have a point z_0 in the interior of T_1^*; since f is holomorphic in \sqrt{z}, it convergent power series around z_0 in T_1^*; so \int_{z_0}^{z_0+1} a_1 (z-z_0)^n dz = a_0 + a_1 (z-z_0) + o(z-z_0) + o(z-z_0), as z-z_0 \rightarrow 0. a_0 + a_1 (z-z_0) of course get 0 when get integral on T_1^*. So \int_{T_1^*} f(z) dz = 0.
\int{T_1^*} a_1 z + a_1 (z-z_0) + o(z-z_0) dz \leq \frac{a_1}{2} \int{T_1^*} (z-z_0) dz. Define f(z) = \int_{T_1^*} a_1 z dz, where dz is diameter of T_1^* (h is perimeter of T_1^*) = \frac{2\pi}{T1^*} \int{T_1^*} a_1 z dz, in O \int{T_1^*} f(z) dz \leq \frac{a_1}{2} \int{T_1^*} (z-z_0) dz. a0 + a_1 (z-z_0) + o(z-z_0) dz = \frac{a_0}{2} \int{T_1^*} (z-z_0) dz + o(z-z_0), so \int_{T_1^*} f(z) dz = 0. (col If f is holomorphic in an open \Omega that contains a polygon P in its interior, then \int_{P} f(z) dz = 0. proof: we can divide the play gon to 4. Thus, if f is holomorphic in a disk then \int_{D} f(z) dz = 0 for every closed curve in the disk. f has a primitive in the disk. proof: assume the disk has centre on O def: F(z) = \int_{T_2} f(w) dw, F_{xz} is \int_{T_2} (z-z_0) dz like that we want to show F(z) is holomorphic on the disk and f(z) = f(z). let calculate to derivative of F. F(z+h) - F(z).
\int{T_2} f(w) dw = \int{T_2} f(w) dz if T_1^* = T_2. \int{T_2} f(w) dw = \int{T_2} f(z) dz if T_1^* = T_2. \int{T_2} f(w) dw = \int{T_2} f(z) dz.
we can have f(w) = f(z) + \epsilon(w), where |\epsilon(w)| \rightarrow 0 as w \rightarrow z. so it is \int{T_2} f(z) dw + \int{T_2} \epsilon(w) dw = f(z) + \epsilon(z). so \lim_{w \rightarrow z} \frac{f(z) - f(w)}{w-z} = f(z).
comment: the theorem is not just true for disk, like f(z) we have same will want to find the winding number of \Gamma around P P \in P contain in \Gamma. so its winding number for non transpose intersection is where the tangent to the ray and to \Gamma are linearly dependent: don’t count non-transverses.
(i): suppose a piecewise smooth closed curve \Gamma does not pass through a thin circle C. (the winding number of \Gamma around C is |C|). (winding number can be the thickness, the line between C and P \in P rotates as P go over C). (pf: Recall for a circle of radius \epsilon centred on C \int_{C} \frac{dz}{z-a} = 2\pi i, we can the curve to the same thickness of circle around a, and as the area between the circle and C is 0 because \int_{C} \frac{dz}{z-a} is holomorphic, so we only compute h-times circle derive 2\pi i h. Ex: show that for x \in \Gamma \int{C} \frac{dz}{z-a} = \int{C} e^{-ix} e^{-ix^2} e^{-2\pi i x} dx. Don’t count \int_{C} e^{-ix} e^{-ix^2} dx integrate around C. C is \Gamma. \int_{C} e^{-ix} e^{-ix^2} dx = 0.
\int_{T_1^*} f(z) dz = 0 or \int_{T_1^*} f(z) dz = 0
f(z) = \int_{T_1^*} a_1 z dz + a_1 (z-z_0) + o(z-z_0)
\int{T_1^*} f(z) dz = \int{T_1^*} a_1 z dz + a_1 (z-z_0) + o(z-z_0)
\int{T_1^*} f(z) dz = \int{T_1^*} a_1 z dz + a_1 (z-z_0) + o(z-z_0)
\int_{T_1^*} f(z) dz = 0
IMG_2604
\int_1 e^{-\pi i (R + i y)} dy \rightarrow 0 as R \rightarrow 0, because the modulo goes to 0. and…
\int1 e^{-\pi i z} = \lim{z \rightarrow \infty} \int{-\infty}^{\infty} e^{-\pi z^2} dz = 1, \lim{R \rightarrow \infty} \int_{-\infty}^{\infty} e^{-\pi i z^2} dz = -\infty \int_{-\infty}^{\infty} e^{\pi i(x+iy)} dx since the whole \int_{-\infty}^{\infty} e^{\pi i x+i y} dx > 0 (holo). so m| = e^{\pi y^2} \int_{-\infty}^{\infty} e^{-\pi i x^2 - 2\pi i y x} dx. D.
Ex: show that \int_0^{\infty} \sin x^2 dx = \int_0^{\infty} \cos x^2 dx = \sqrt{\frac{\pi}{2}} (Frenel Integrals).
f = e^{-x^2} is holomorphic everywhere. i. integrate around that area
\int{t_1}^{t_2} f dz = \lim{R \rightarrow \infty} \int_{t_1}^{t_2} \frac{1}{z} |f(z)| dz = \int_0^{t_2} e^{-x^2} dx = \frac{1}{\sqrt{\pi}}, \lim{R \rightarrow \infty} \int{t_2}^{t_1} f dz = \frac{1}{\sqrt{\pi}}
\lim{R \rightarrow \infty} \int{t_1}^{t_2} e^{-t^2} e^{i t^2 \cos \theta} d\theta = 0. i. calculate modulos, t^2 \leq \int_0^{t_2} |e^{-t^2 \cos \theta}| d\theta = \int_0^{t_2} |e^{-t^2 \cos \theta}| R |d\theta|
= \int_0^{t_2} |e^{-t^2 \cos \theta}| R |d\theta| = \int_0^{t_2} |e^{-t^2 \cos \theta}| R |d\theta| = \int_0^{t_2} |e^{-t^2 \cos \theta}| R |d\theta|
\int_1 e^{-z^2} dz = \int_0^1 e^{-i z^2 \cos \theta} d\theta dr. \int_0^1 e^{-z^2 \cos \theta} d\theta = \int_0^1 e^{-z^2 \cos \theta} dz = \int_0^1 e^{-z^2 \cos \theta} dz = \int_0^1 e^{-z^2 \cos \theta} dz
so \int_0^{\infty} \int_0^{\pi/2} (\cos^2 r) dr + \int_0^{\infty} \sin^2 r dr = \int_0^{\pi/2} \int_0^{\pi/2} \cos^2 r dr - \int_0^{\pi/2} \int_0^{\pi/2} \sin^2 r dr = \pi \int_1 f dz = -\frac{1}{\sqrt{\pi}} \Rightarrow \int_0^{\pi/2} \int_0^{\pi/2} \cos^2 r dr = \int_0^{\pi/2} \sin^2 r dr = \frac{1}{\sqrt{\pi}}
Enter spiral (x(t), y(t)) = (\int_0^t \cos x^2 dt, \int_0^t \sin x^2 dt). may like (1)
Cauchy’s integral formula: suppose f is holomorphic on open set containing the closing of a disk D. If C is the tvely oriented boundary of D, then f(z) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-\epsilon} dz.
proof: consider (2) C = [z_0, z_1]. Argue that \frac{1}{2\pi i} \int{z_0}^{z_1} \frac{f(z)}{z-\epsilon} dz + \frac{1}{2\pi i} \int{z_1}^{\infty} \frac{f(z)}{z-\epsilon} dz = 0, I_z is the tag-vely oriented circle with radius \epsilon. \partial_z f(z) = \frac{f(z)}{z-\epsilon} - \frac{f(z)}{z-z_1} = \frac{f(z)}{z-z_1} - \frac{f(z)}{z-z_0}.
\int_{z_0}^{z_1} \frac{f(z)}{z-\epsilon} dz = 2\pi \epsilon / M \rightarrow 0. so \int0^{\pi/2} \frac{f(z)}{z-\epsilon} dz = \int{z_0}^{\pi/2} \frac{f(z)}{z-\epsilon} dz = -2\pi i f(z_0)
complete the proof.
\int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz is the assumption of previous thr.
\int_0^{\pi/2} \frac{f(z)}{z-\epsilon} dz is the assumption of previous thr. f^{(1)}(z) = \frac{1}{2\pi i} \int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz
so \int_0^{\pi/2} \frac{f(z)}{z-\epsilon} dz = 2\pi \epsilon / M \rightarrow 0. so \int0^{\pi/2} \frac{f(z)}{z-\epsilon} dz = \int{z_0}^{\pi/2} \frac{f(z)}{z-\epsilon} dz = -2\pi i f(z_0)
complete the proof.
Induction: Cauchy’s formula is the assumption of case h=0. Assume f^{(h-1)}(z) = \frac{1}{(h-1)!} \int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz, we need to calculate \int{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz = \int{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz.
using A^{(h-1)} = (A-1) \left( A^{(h-1)} + A^{(h-2)} + \dots + A^{(h-1)} \right), with A = \frac{1}{(h-1)!} \int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz.
when h \rightarrow 0, f^{(h)}(z) = \frac{1}{2\pi i} \int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz.
where \epsilon is tvely oriented circle, \intC \int{z0}^{\infty} \frac{f(z)}{z-\epsilon} dz = \int{z0}^{\infty} \int{z_0}^{\pi/2} \frac{f(z)}{z-\epsilon} dz = -2\pi i f(z_0)
take z \rightarrow 0, so \frac{1}{2\pi i} \int_{z_0}^{\infty} \frac{f(z)}{z-\epsilon} dz = 0
IMG_2605
|f^{(n)}(z0)| \leq n! \sup{\frac{2\pi i}{2n}} |f(z)|, where C is the circle radius R centred at z_0.
translate z_0 to 0, plura metrized integral using z = Re^{i\theta}.
Suppose how that C is not simple, but has a minding number n.
Then f(z) = \frac{1}{2\pi i n C} \int_C \frac{f(z)}{z^{n-1}} dz.
f(z) is then translated to 0 to yield \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
Hence evaluate for x \in [0, 1].
\int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
to evaluate the integral \frac{1}{z^{n-1}} \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \frac{1}{z^{n-1}} \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
\int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
\int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
Thm: Suppose f is holo on an open set \Lambda. If D is a disk center at z_0 whose closure is in \Lambda, then f has a power series expansion at z_0.
f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n for all z \in D, with a_n = f^{(n)}(z_0).
p_f: recall that for os(RK), \frac{1}{1+z} = \frac{1}{1+z} + \frac{z}{(1+z)^2} + \dots, converge. start with f(z) = \sum_{n=0}^{\infty} \frac{f(z)}{z^{n-1}} dz.
note that \frac{1}{1+z} = \frac{1}{1+z} + \frac{z}{(1+z)^2} + \dots so \frac{1}{1+z} \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz = \int_0^{2\pi} \frac{f(z)}{z^{n-1}} dz.
note that the power series is convergent and become geometric series is absolutely convergent.
Thm: Suppose f is holo in a connected \Lambda and f is equal to 0 on a sequence of points in \Lambda with a limit point in \Lambda, then f \equiv 0.
p_f take z_0 be the limit point, then f(z_0) = 0 by continuity. Since f is holo it has a convergent power series centred on z_0. f(z) = a_n (z-z_0)^n.
[1+(z-z_0)] is nonzero. \sum [(z-z_0)] \rightarrow 0 as z \rightarrow z_0, on a neighbourhood of z_0 [1+(z-z_0)] is nonzero. \sum a_n \rightarrow 0 on a neighbourhood on an open set contain z_0.
but the set of points on which f is zero is also closed, then the open set is closed so it’s \Lambda or \emptyset.
Def: Suppose f is analytic on a connected \Lambda, and g is analytic on a connected n with n 1 and C_n = \frac{1}{n-1} + \frac{1}{n-2}. show
3. Cauchy 定理、同伦与积分公式
IMG_2606
that C_n are the coefficient of the power series of rational function and use this to derive formula for C_n
for a sequence {a_n} a generating function is an analytic function power series expansion is given by \sum_{n=0}^{\infty} a_n z^n. Suppose R(z) is a generation for the fibonacci sequence the R(z) = z f(z) + z^2 f(z) + z \Rightarrow f(z) = \frac{z}{1-z^2}.
Cn = \frac{1}{\sum{k=1}^{\infty} \frac{R(z)}{f(z)}} ds, R(z) = \frac{1}{f(z)} \left( \frac{1}{1-A_1 z} - \frac{1}{1-A_2 z} \right), \lambda_1 = \frac{1+is}{2}, \lambda_2 = \frac{1-is}{2}. and
so far, suppose u and v functions is a function defined on \mathbb{R} (open), if u and v are continuously differentiable, and satisfy the CR egs. \frac{du}{dx} = \frac{dv}{dy}, \frac{dv}{dz} = \frac{du}{dx} then f is holo. The power series f(z) = \sum_{n=0}^{\infty} a_n z^n define a holo function with its disk of convergence. If f is holo in a disk, then \int_{D} f(z) dz = 0, f is any close loop in disk. Suppose f holo in an open set C that contain closure of disk D. If C denotes the boundary of D with the ariation then f(z) = \frac{1}{2\pi i} \int_{D} f(z) dz, for any z in D, also f^{(n)}(z) = \frac{1}{2\pi i} \int_{D} \frac{f(z)}{(z-z_0)^n} dz. |f^{(n)}(z0)| \leq n! \sup{R^n} \frac{|f(z)|}{|z-z_0|^n}, C is the boundary of disk with radius R centred on z_0.
Thm. suppose f is holo on an open set \mathbb{R}. If D is a disk controlled at z_0 whose closure is in \mathbb{R} then f has a convergent power series expansion on neighbourhood of z_0 with a_n = \frac{f^{(n)}(z_0)}{|z-z_0|^n}.
(ol if f is entire and bounded then f is constant. i.e. |f| \leq b \Rightarrow f = c M p.f. for any z_0 \in \mathbb{C}, the count in equality give that |f(z_0)| \leq \frac{b}{R} since |z| is the radius of a circle controlled on z_0. Since f is entire we can choose |z| \rightarrow +\infty. If then f’(z_0) = 0 so f is a constant.
Thm. Every polynomial P(z) = a_n z^n + \dots + a_1 z + a_0 of degree n \geq 1 has a roots in \mathbb{C}, count with multiplicity p_i. If polynomial has no roots claim \overline{p(z)} is bounded and entire entire is obvious. to show the claim, when |z| > R for sufficiently large R |\overline{p(z)}| \sim \frac{1}{p_i} so it’s bounded, when |z| \leq R is a compact set s.e. with out zero point, so bounded in it, thus we prove the claim. by the \overline{p(z)} is constant contradiction so p(z) has a root. r. we write P(z) = (z-r_1) P_1(z), repeat the anywhere for P_i(z) so we get idea.
IMG_2607
constructing holomorphic functions
Moreover there: suppose f is continuous in an open disk D and for any triangle T in D, then f is holo in D (convert Gaussian theorem)
proof: \int_{T} f(z) dz > 0 previous argument says f has a primitive F on D, so F|_T = f.
F is holo on D \Rightarrow F is C^\infty on D so f is holo.
Then {fn}{n=1}^{\infty} is a holo functions that converges uniformly to f in every compact subset of \Omega then f is holo in \Omega.
pt: let D be a disk D \times C \setminus \Omega and T \subset D then \int_T f_n(z) dz = 0 (by Gaussian) since f_n \rightarrow f uniformly \int_T df_n(z) dz = 0, so f is holo in \Omega \setminus T by Morera.
Then: In the previous theorem, not only do the function converge, but the derivative of all orders converge uniformly to f on the compact subsets.
pt: for s > 0, \text{def } \Lambda_s = {z \in \Omega \mid D_s(z) \subset \Omega} for z_0 \in \Omega, f^{(k)}(z_0) = f^{(k)}(z_0) take \delta as small as we like.
Then: let T(z, s) be def for (z, s) \in \Omega \times [0, 1], suppose f satisfy (1) F(z, s) is holo for any fixed s \in [0, 1] (2) F is continuous in \Omega \times [0, 1]. then the function on \Omega def by f^{(k)}(z, s) ds is holo
pt: let D is a disk contained in \Omega. Define fn(z) = \frac{1}{n} \sum{k=1}^n F(z, \frac{k}{n}) since [0, 1] is compact. f is uniformly continuous in s, so for any s \in [0, 1], |fn(z) - f(z)| \leq \frac{1}{n} \sum{k=1}^n \frac{k}{n} = \frac{1}{n} \frac{n}{n} = \frac{1}{n} = 1 since |s - z| \delta.
|fn(z) - F(z, s)| \leq \frac{1}{n} \sum{k=1}^n \frac{k}{n} |F(z, \frac{k}{n}) - F(z, s)| \leq \frac{1}{n} \sum_{k=1}^n \frac{k}{n} = \frac{1}{n} \frac{n}{n} = 1 so f_n \rightarrow f uniformly, so f is holo.
Example: f(s) = \int_0^\infty e^{-t} t^{s-1} dt for \Re(s) > 0 when n \in \mathbb{N}^* |f(n)| = n! Recall: suppose f and g are holo in a connect extending holo functions f, g on sequence of distinct points with a compact point in \Omega then f = g in \Omega.
pt: easy col from 4.8. then f = g in \Omega. Reflection around the real axis then f_+ and f_- agree on I and f+ = f- is holo in \Omega_+, f- = f+ on \Omega_-.
proof: Morera theorem any triangle we take contain \Omega_+ and \Omega_- we divide the part in \Omega_+ and \Omega_- all part has its intergul O, then f_+ and f_- are holo in O.
IMG_2608
Thm (Schwarz reflective principle) suppose f is holo on \Lambda^+ that extends continuously to \bar{I}, where f|_{\bar{I}} = f|_I n is real then f can be extended to a holo function f on \Lambda^- = \Lambda^+ \cup \bar{\Lambda}
pt: take z \in \Lambda^- def df = f(\bar{z}) f is holo (check by yourself) f on a neighbourhood of z_0 \in \Lambda^+ we have f(z) = \bar{z} a_n(z-z_0) so f(\bar{z}) = \bar{z} a_n(\bar{z}-z_0)^n = \bar{z} \bar{a}_n(z-z_0)^n, since f on \bar{I} is real so a_n is real \hookrightarrow \bar{a}_n = a_n so f(\bar{z}) = \bar{z} a_n(z-z_0)^n
Thm: (Runge approximation) let k be a compact set, and suppose f is holo function on open set \Lambda where k \subset \Lambda, then f can be approximated uniformly on K by rational function whose singularities are contained in \cup K.
pt: let z \in K f(z) = \frac{1}{2\pi i} \int_{z_0}^z \frac{f(y)}{y-z} dy r is a piecewise line outside K, in \Lambda and is straight path segment r_1, \dots, r_n. For any [ri, r{i+1}] \rightarrow C, \int_{z_0}^z \frac{f(y)}{y-z} dy = \int_0^1 \frac{f(r_i(t))}{r_i(t)-z} dr_i(t) dt. let f(z, t) = \frac{f(r_i(t))}{r_i(t)-z} hence by compactness, f(z, t) is uniformly continuous on K \times [ri, r{i+1}] by private privous the Riemann sums (each \sigma of which of rational functions with singular outside k) covarge to \int_0^1 F(z, t) dt uniformly since f(z) = \frac{1}{2\pi i} \int_{z_0}^z \frac{f(y)}{y-z} dy
Thm In the previous theorem, when C \setminus K is path connected, f can be approximated uniformly by polynomials pt: z_i in C \setminus K s.t. |z_i| < 1 then \frac{1}{z_i-z_1} = \frac{1}{\bar{z}_1} \cdot \left( \frac{1}{1-\frac{z_i}{\bar{z}_1}} \right) = \frac{1}{z_i-z_1} covarges it follows that powers of \frac{1}{z_i-z_1} and powers of \frac{1}{z_i-z_1} \times \text{poly} in \bar{z} can be uniformly approximated by polynomials
let \rho = d(z_0, 1)/20, 20 \in \Lambda \setminus K. there is a path from z_0 to \bar{z}_1, cut with z_0, w_1, \dots, w_n = \bar{z}_1 d(wi, w{i+1}) < \rho next step - show it that the works with w_i in place of z_i, it works with wide place of w_i \bar{z}i = \bar{z}{i+1} |wi - w{i+1}| < \rho, then |\frac{wi - w{i+1}}{zi - w{i+1}}| < 1, so it can occur to a controlling power set ies. i.e. be approximated by poly of in \bar{z} finish proof by induction
Example: If f(z) is holo real on the real axis, and imaginary on the imaginary axis show that is odd i.e. f(-z) = -f(z) If f is real on the axis, f(z) = \bar{z} a_n(z-z_0)^n since f is real on real axis, then a_n are
IMG_2609
all real. If imaginary on imaginary axis then f(z) = \frac{e^{iz}}{z} is odd. Take z_0=0 an i^n y^n, so a_n=0 for all n is even. \Rightarrow the x f is odd. (take z_0=0 so it’s odd around z_0, we can extend to a whole odd function)
f(z) has an isolated singularities at z_0 if \exists an open set U containing z_0 s.t. f(z) is holo on U \setminus {z_0}
Types of isolated singularities = 0 removable singularities \oplus poles \oplus essential singularities. \bar{x} of 0: \frac{\sin z}{z}, define \frac{\sin 0}{0} = 1, so its a holo function
\bar{x} of \oplus \frac{1}{z^3} + \frac{1}{z^2} + \frac{1}{z} + z. at z=0 \bar{x} of 0 e^z, z \rightarrow 0 has no limit as z \rightarrow 0
An isolated singularity is a removable singularity at z_0 if \lim_{z \rightarrow z_0} (z-z_0)/(z-z_0) = 0 why is lim z or z^* not a removable singularity at z=0? It’s not isolated.
Thm: If f has a removable singularities at z_0, \exists! holo function h on U, that agrees with f on U \setminus {z_0}.
pf: def h(z) = \frac{f(z)}{z-z_0}, C is contain in U. f(z) at z_0 f(z) at z_0 f(z) at z_0
f(z) at z_0 has has radius \delta. d the def of removable singularity. \lim_{z \rightarrow z_0} f(z) = 0, si since f is holo in the interior of C. \int_C \frac{f(z)}{z-z_0} dz
= \int_{z_0}^z \frac{f(z)}{z-z_0} dz. and h(z) is holo h(z) is given by the cauchy integral formula for the so is first derivative.
Def: An isolated singularity at z_0 is a pole if it is not removable but \exists a natural number m s.t. \lim_{z \rightarrow z_0} (z-z_0)^{m+1} f(z) = 0. The smallest m is possible nis called the order of the pole if m \geq 1 it is a simple pole.
Def: all other isolated singularities are called essential singularities. The point z_0 is a zero for f if f(z_0) = 0. Recall that for a non-constant function, the zero is isolated.
Thm: Suppose f is holo on an open set \Omega, with a zero at z_0. suppose also that f is not constant. Then \exists a neighborhood U \subset \Omega of z_0 a hon-vanishing holox function g on U and a unique h \in U s.t. f(z) = g(z) where h is the multiplicity of zero or order of z_0
pf: take the power series expansion around z=z_0. D. A punctured disk (or deleted neighborhood) {z | 0 < |z-z_0| < r} is a punctured disk for deleted neighborhood.
r > 0. If f has a pole at z_0 \in \Omega, then in a neighborhood of z_0 \exists a non-constant function h s.t. f(z) = g(z) where g is holo on U and h is a simple pole. Thm: If f has a pole at z_0 \in \Omega, then in a neighborhood of z_0 \exists a non-constant function h s.t. f(z) = g(z) where g is holo on U and h is a simple pole.
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holo function h and a unique n f \mathbb{V} s.t. f(z) = h(z)(z-z_0)^{-h}.
pf. by def of pole, If f has a pole of order n, then (z-z_0)^h f(z) has removable singularities at z=z_0. h is then the holo function from the thin about removable singularities.
is called principal part and b_1 is the residue.
If P(z) is the principal part of f, then \frac{1}{2\pi i} \int{z_0}^1 P(z) dz = b{n-1}.
thm: If f has a pole of order n at z_0, then \text{res}{z_0} f = \lim{z \rightarrow z_0} \frac{(d z)^{n-1}}{(z-z_0)^n}.
pf. recause an ance of Tuler expansion (0! = 1)
Kiernan st. sphere
Hemisphere of the Kiernan sphere is the unit circle. take north shere to the unit circle. north pole one of its
stereographic projection give out a formula. If is holo assure any from a few poles is not considered to be a function from the Kiernan sphere to itself. For ex: z \mapsto \bar{z}.
we can classify \infty as being either .. Degenerate.
② removable sig singularities.
③ pole
④ essential singularities sin z
has an essential singularities at \infty of a function f by cond consider f(\frac{1}{z}) at z=0. (claim: an entire function has infinite radius of convergence. \Rightarrow It is a function with a non essential singularity at \infty is polynomial.)
in this case suppose f is holomorphic on an open set containing a circle in its interior, with the exception of a pole at z_0 inside the circle. \int_{z_0}^1 f(z) dz = 2\pi i res of f.
for suppose now that f is holo except for pole at z_1, z_2. and f is not necessarily sin pole then \int{z_1}^1 f(z) dz = \sum{z=z_1}^{z_2} 2\pi i W(f, z_k) res of f. \Rightarrow z \in e^{i\pi} poles of f(z) at z_1.
show that \int_{z_1}^1 \frac{dx}{f(x)} = \frac{\pi}{\text{sin } a} so we have only one pole calculated. \Rightarrow z = i\pi + 2\pi i \sin \theta.
IMG_2611
*
Chapter 2 第7題 (第1問)
Q. f(x) = \frac{e^{(x+2)} - e^{2x}}{e^{(x+1)} - e^x}, \Rightarrow \psi(z) = \psi(z)/\psi(0) = \psi(0) = 1, 以下第1問解說 \psi(z) = 0, \psi(0) = 1, \quad \sum \psi_k = \varphi \circ e^k = \varphi \circ \dots \circ e^k
若 \varphi = \psi(z) = z + a_n z^n + O(z^{n+1}), \psi_k(z) = z + ka_n z^n + O(z^{n+1}) 且 \psi(0) = 0, \psi(0) \neq 0
由 Cauchy 解法得 \psi_k(z) = \frac{1}{z^{n+1}} \frac{f(z)}{f(0)} = \frac{1}{z^{n+1}} \frac{e^{(z+2)} - e^{2z}}{e^{(z+1)} - e^z} = \frac{e^{(z+2)} - e^z}{e^{(z+1)} - e^z}
\psi = 1 \rightarrow 1, 故得 \psi_k(0) = \frac{1}{z^{n+1}} \frac{e^{(z+2)} - e^{2z}}{e^{(z+1)} - e^z}
\psi_k(z) \rightarrow z \rightarrow 0, 矛盾, 故 \psi_k(z) \rightarrow 0, z \rightarrow 2, a_n = 0. □
- 令 f(x) = \frac{f(x) - f(0)}{f(0)} 得 f’(x) = 0, f(0) = 1, 是要证 f 是 \mathbb{R} 向下的函数.
由 第1問解說得 f(x) = \frac{e^x - 1}{1 - e^x}, 由 f’(x) = \frac{e^x - 1}{1 - e^x}, 由 f’(0) = 1, 由 f’(x) = \frac{e^x - 1}{1 - e^x}, 由 f’(x) = \frac{e^x - 1}{1 - e^x} 得 f’(x) = \frac{e^x - 1}{1 - e^x}
由 f(x) = \frac{e^x - 1}{1 - e^x} 得 f’(x) = \frac{e^x - 1}{1 - e^x}
由 f(x) = \frac{e^x - 1}{1 - e^x} 得 f’(x) = \frac{e^x - 1}{1 - e^x}
\pi(0, 1) has simple poles at interger with residue 1
Example: show that \sum_{n=0}^{\infty} \frac{x^n}{(x+1)^{n+1}} = \frac{1}{1+x} (or f(x) = \frac{1}{1+x}) where any simple poles of residue 0 are poles of \pi(0, 1). Also poles of \pi(0, 2) at -1/2.
4. 幂级数、Laurent 展开、留数与奇点
IMG_2612
tes u f(z) = \lim_{z \rightarrow u} \frac{d}{dz} (u f(z)) \frac{\pi \cot(\pi z)}{\cot(\pi z)} = \lim_{z \rightarrow u} \frac{d}{dz} \pi \cot(\pi z) = - \frac{\pi^2}{(\sin(\pi u))^2}, take u a large circle \sim e^{i\pi z} radius R is bounded so |f(z)| \sim \frac{\pi^2}{(\sin(\pi u))^2}, |\cot(\pi z)| \sim \frac{1}{\pi^2} R \rightarrow \infty.
\int{R \rightarrow \infty} \int{z \in R} f(z) dz = 0, so \sum_{n=-\infty}^{\infty} \frac{\pi^2}{(\sin(\pi u))^2} = 0.
similarly \sum_{n=-\infty}^{\infty} \frac{\pi^2}{(\sin(\pi u))^2} = \frac{\pi}{a} \cot(\pi a), a > 0, the take f(z) = \frac{\pi \cot(\pi z)}{z} like wise poles of f occur at the integral line \int{R \rightarrow \infty} \int{z \in R} f(z) dz = 0 \Rightarrow z residue of f \rightarrow 0.
res u f(z) = \frac{1}{\pi \cot(\pi a)}, we have simple pole at z = \pm ia. res ia f(z) = \lim_{z \rightarrow ia} \frac{1}{(z-ia)} \cot(\pi z), \lim_{z \rightarrow -ia} \frac{1}{(z-ia)} \cot(\pi z) = \frac{\pi \cot(\pi a)}{\pi \cot(\pi a)} = \frac{\pi}{\pi} = 1.
\text{res-in } f(z) = \lim_{z \rightarrow -ia} f(z) = \frac{\pi \cot(\pi a)}{\pi \cot(\pi a)} = 1.
Recall 3 types of isolate singularities (1) we show that if z_0 is a removable singularity of f, \exists an open set U contain z_0 and a holo function h on U such that s.t. -f = \partial h on U \setminus {z_0} and f is bounded.
(2) z_0 is a pole of order n of f iff \bar{f} has a zero of order n at z_0 i.e. \lim_{z \rightarrow z_0} (z-z_0)^n f(z) exist and is not 0.
Col. suppose f has an isolated singularities at z_0, then z_0 is a pole of f iff H(z) \rightarrow \infty as z \rightarrow z_0.
pf: only need to show “it” suppose |f(z)| \rightarrow \infty as z \rightarrow z_0 then \bar{f} \rightarrow 0 as z \rightarrow z_0, z_0 is a pole of f.
these contrast with e^{\frac{1}{z}} near 0. approach along the real axis \rightarrow +\infty along the real axis \rightarrow 0, along imaginary axis \rightarrow -\infty. bounded with lent oscillate.
Thm (Cassorati verierstrass) f is holo in the practiced pick \partial h(z_0)/z_0 has an essential singularity the image of Dr(z_0) \setminus {z_0} under f is dense in complex plane i.e. every in \mathbb{C} is a limit point of the image under f of a sequence in Dr(z_0) \setminus {z_0}.
pf: Assume the image of f is not dense, \exists w and g > 0, s.t. |f(z) - w| > \delta.
\forall z \in Dr(z_0) \setminus {z_0} def g(z) = \sqrt{(z-w)^2}, then |g(z)| < \frac{1}{\delta} bounded on Dr(z_0) \setminus {z_0}.
g(z) is bounded \Rightarrow to a holo function on Dr(z_0) if \bar{g}(z_0) = 0 \Rightarrow z_0 is a pole if \bar{g}(z_0) \neq 0, so g(z_0) is removable, contradict.
def: a function f on an open set U is nero morphic if \exists a sequence of points {z_0, z_1, \dots} with no limit point in U, s.t. |f| is
IMG_2613
holo on \sqrt{{z_0, z_1, \dots}} (z) The points {z_0, z_1, \dots} are all poles of f.
The function that is nero in the extended complex plane is rational pf: At each of f is bounded as z \rightarrow z_0. By assumption f has a pole of a removable signality at z_0. Either way there can be only finitely many sig. singularities the (first there is no singularities near north pole the complement is compact so only finite singularities let {z_1, \dots, z_n} be the pole of f at each singularity, subtract from f its principal part (z_1, \dots, z_n) what left all the principle parts are subtracted is a bounded entire function \Rightarrow is a constant by Louville the since the principal part are all rational functions f is a finite sum of rational function.
A rational function is determined up to multiplicative constant by the location and multiplicities of its zeros and poles.
Thm: (Argument principle). Sup f is nero in an open set containing a circle C and disk D bounded by circle C suppose f has no poles or zeros on C then \frac{1}{2\pi i} \int_{C} \frac{f(z)}{z} dz = # {\text{zeros of } f \text{ in } D} - # {\text{poles of } f \text{ in } D} poles and zeros are counted with multiplicity.
pf: If f is holo with a zero of order n at z_0, f(z) = (z-z_0)^n g(z), g(z_0) \neq 0 and g(z_0) |g(z)| < \infty, \frac{f(z)}{z} = n(z-z_0)^{n-1} \frac{g(z)}{g(z_0)}. g(z) is holo so \int_{C} \frac{f(z)}{z} dz = n \cdot 2\pi i n = 2\pi i n since large calculation show that a pole of order n contribute -n in the contain integral.
Thm f and g are holo in an open set containing a circle C and the disk D that it bounds. If |H(z)| > |g(z)| for all z \in C then f and g have the same number (counted with multiplicity) of zeros in D.
pf: let f_+(z) = f(z) + g(z) in D, h+ = \text{number of zeros of } f+ \text{ in } D, h- = \text{number of zeros of } g- \text{ in } D, so h+ = \frac{1}{2\pi i} \int{C} \frac{f_+(z)}{z} dz by the argument principle. h_- is \frac{1}{2\pi i} \int{C} \frac{f-(z)}{z} dz continuous for z \in C and z \in C. so h_+ is a continuous function of t, but h_- is \in AV^* so h_+ is independent of t. D.
To show that all 5 sol to equations x_1 x_2 x_3 + 8z - 1 = 0 are containing the dole |z| \leq 2. take f(z) = z^5, g(z) = 8z - 1. If |z| \leq 2 and |g| \leq 1, so f has 5 zeros in |z| \leq 2. hence f_+ has 5 not zeros.
Thm and h = \frac{1}{2\pi i} \int_{C} \frac{f(z)}{z} dz are counted with multiplicity.
Thm f and g are holo in an open set containing a circle C that it bounds. If |H(z)| > |g(z)| for all z \in C then f_+ and f_- have the same number (counted with multiplicity) of zeros in D.
IMG_2614
show that only one of these 5 roots is contained in the disk z < |z| take f = 8z-1, g = 2z^5.
f has exactly 1 zeros in |z| < |z| so |f| \leq |g|. obviously
Ex: f and g are holo in a connected open set contain the disk |z| < |z|
f has a simple zero at z=0, and vanishes nowhere else on |z| \leq |z| def.
f_2(z) = f(z) + \varepsilon g(z). show that \varepsilon small sufficiently small, then (1) f_2(z) is
a unique zero in |z| \leq |z| z_0 is the zero of f_2(z), show that the path z_0 is continuous.
pf: (1) since pf, g banded |z| \leq |z|, for small enough \varepsilon, |f| \leq |\varepsilon g| satisfy
the assumption of Rouché thm. so zero is unique and simple. (z) z_0 = z_0
\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z) + \varepsilon g(z)}{f(z) + \varepsilon g(z)} z dz by argument thm. on |z|=1, the assumption implys z_0
writen like this is continuous function of \varepsilon,
Thm suppose f is holo on punctured neighbourhood Dr(z_0) \setminus {z_0} and \lim_{z \rightarrow z_0} f(z) = 0 then there is a holo function h on Dr(z_0) that
h = f on Dr(z_0) \setminus {z_0} and g(z) = \begin{cases} (z-z_0)^2 f(z) & \text{on } Dr(z_0) \setminus {z_0} \ 0 & \text{on } z=z_0 \end{cases}
pf: def g(z) = (z-z_0)^2 f(z) g(z) = \begin{cases} (z-z_0)^2 f(z) & \text{on } Dr(z_0) \setminus {z_0} \ 0 & \text{on } z=z_0 \end{cases} g(z) is holo in Dr(z_0) \setminus {z_0} \Rightarrow g(z) is holo
in Dr(z_0) with g’(z_0) = 0 so g(z) = g(z_0) + g’(z)(z-z_0) + \frac{g’’(z_0)}{2}(z-z_0)^2 + \dots \Rightarrow g(z) is holo in Dr(z_0) \setminus {z_0} and g = f on Dr(z_0) \setminus {z_0} g is holo in Dr(z_0) \setminus {z_0}
Recall Argument principle: Suppose f is nero in an open set containing circle C and its interior. If f has no poles or zeros on C then \frac{1}{2\pi i} \int_C \frac{f(z)}{f(z)} dz = #{\text{zeros of } f \text{ inside } C} - #{\text{poles of } f \text{ inside } C}
C (count with multiplicities)
Rouché thm. Suppose f and g are holo in an open set containing a circle C and its interior. If |f(z)| > |g(z)| f is holo in C
then f and fg have same number of zeros inside C.
def: a map is open if it maps open set to open set. Thm: (open mapping thm) If f is holo and nonconstant in a connected open set \Lambda, then f is open.
pf: \text{let } w_0 = f(z_0). want to show that for some \varepsilon > 0, all points in the interior of C we have w_0 = f(z_0) - w in the interior of C and w is a neighbourhood of z_0 in the interior of C.
IMG_2615
= f(z) - w_0 + (w_0 - w). Let check a circle of radius \delta centered on z_0 where \delta is small enough s.t. C \subseteq \mathbb{R}. Then C is a lower bound on F(z) - w_0, we can choose w sufficiently close to w_0 s.t. F on C satisfying |F| > (w - w_0) so g must have a zero. Since F has no zero on C. Then Maximum modulus principle: If f is a nonconstant holo function on an open set \mathbb{R}, then f does not reach a maximum on \mathbb{R}.
pf: suppose |f| reach a maximum at point z_0 \in \mathbb{R}, then must be in U = f(z_0) and a_1 z_0 \in \mathbb{R} s.t. f(z_1) = U so must be z \in U, |z| > |f(z_0)| nontradition.
Col. suppose \mathbb{R} is a connected open set with compact closure \mathbb{R}, If f is holo on \mathbb{R} and continues on \mathbb{R}, then \sup{f \in \mathbb{R}} |f(z)| \leq \sup{z \in \mathbb{R}} |f(z)|.
pf: If f is nonconstant, \exists max f on \mathbb{R}, due to compactness and continuity of f, and due to the Maximum modulus then, the max must in \mathbb{R} if f is a constant, obviously.
Fourier series
Suppose f is holo in Rdisk D_r(z_0) around z_0 and f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n in \mathbb{R}[z_0] then the coefficient of power series expand of f are given by a_n = \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + r e^{i\theta}) e^{-in\theta} d\theta and \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + r e^{i\theta}) e^{in\theta} d\theta = 0 for n < 0. pf: a_n = \frac{f(z_0)}{n!} \pm \text{by canchy’s integral formula we get } a_n = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z_0 + r e^{i\theta})}{(r e^{i\theta} + r)^{n+1}} d\theta = \frac{1}{2\pi i} \int_0^{2\pi} (z + r e^{i\theta}) e^{-in\theta} d\theta, when n < 0 \frac{f(z_0)}{(z - z_0)^{n+1}} is holo so integral is 0.
Col. (main value property) If f is holo in a disk D_r(z_0), then f(z) = \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + r e^{i\theta}) d\theta.
pf: a_0 = f(z_0) \neq 0 by previous then a_0 = \frac{1}{2\pi i} \int_0^{2\pi} f(z_0 + r e^{i\theta}) d\theta = 0. Col. If f is holo in D_r(z_0) and u = Re(f), then u(z_0) = \frac{1}{2\pi i} \int_0^{2\pi} u(z_0 + r e^{i\theta}) d\theta = HK och KR just need to take real part of f.
Multivalued functions
Suppose f(z) is a complex multivalued function. A point a \in C is called
IMG_2616
a branch point of f if (1) f(a) has only one value or might not be defined (2) for small \varepsilon and a circle C of radius C extended on a, a choice f(z) at one point on C cannot continuously extend over all of C.
Ex: \log z / \log z = \log e^{iz} = \log r + iz so \log |z| = \log z = \log |z| + 2\pi i \arg(z) + 2\pi i \operatorname{inh} h \in \mathbb{Z}.
for the function f(z) = z^{-1}, h \in \mathbb{N}. 0 is called a branch point of order h. 0 is a logarithmic branch point of \log z for more general functions. h is a branch point of order h if h \in \mathbb{N} is the smallest number of times it is necessary to travel around C to get back to the original value of f. If the finite n doesn’t exist, it’s a logarithm branch point of f. f is not holo at a branch point.
To def = “branches” of multivalue f functions we need the notion of homotopy.
def: let \Lambda be an open set in complex plane \mathbb{C}, and let [0, 1] \rightarrow \Lambda be arcs sharing the same endpoints [0, 1] is homotopic to [0, 1] if f: [0, 1] \times [0, 1] \rightarrow \Lambda (continuous in s and t that s \cdot t = t \cdot s, H(0, t) = \gamma_0(t), H(1, t) = \gamma_1(t), H(s, 0) = \gamma_1(0) = \gamma_0(0), H(s, 1) = \gamma_0(1) = \gamma_1(1)).
thm. Suppose \gamma_0 and \gamma_1 are homo in \Lambda and f is holo in \Lambda then \int_0^1 f dz = \int_1^\infty f dz.
def: A connected set \Lambda is called simply connected if any two arcs with common endpoint are homo.
thm. A holo function on a simply connected domain \Lambda has a primitive this is the restatement of thm s \cdot t.
pf: choose a base point w_0 \in \Lambda and f(z) = \int_{\Gamma} f(w) dw where \Gamma is any path from w_0 to z. so \lim_{z \rightarrow w_0} \frac{f(z+th) - f(z)}{th} = f(z).
Branch of a multivalue function: choose an open, simply connected set \Lambda not containing any singularities of the multivalue function f. Choose a base point z_0 is \Lambda, f has k values at z_0, {f_1(z_0), \dots, f_k(z_0)} choose one of value f_i f_j(z_0) on \Lambda define a function F(z) = fi(z_0) + \int{\Gamma} f_j(z) dz (where \Gamma is an arc from z_0 to z). this function F is a branch of f.
IMG_2617
Ex: f(z) = \log(z) that take \lambda = 0 & we vertically real axis. choose f_1(1) = 0. then the function f(z) = \text{of } \int_z \frac{dz}{z} is a branch of \log z is called ‘principle branch’ of \log z: f(z) = \log|z| + i \arg(z) -\pi < \arg z < \pi
let f(z) be a complex multivalued function. A point is in a branch point if (1) f(a) has only one value (may not be defined) for small \epsilon and a circle C_\epsilon(a) of radius \epsilon around a, a choice of f(z) at a point on the circle can’t be contained continuously extend over circle.
consider the possibility of branch point at w in the Riemann sphere. Eg. The function k^z, k \in \mathbb{C} has no branch point k^z = e^{z \log k} is a multivalue function.
k is a constant when we go around a circle, the choice of \log k is not varying. z is single value, so it’s an expentension function.
A Riemann surface is informally a surface obtained by clearing together (copies of) branch cuts.
Recall Branches of multivalued function choose an open, simply connected set \Lambda not captaining any branch points or singularities of the multivalue function f choose a base point z_0 in \Lambda and a value f_1(z_0) of f at z_0. On \Lambda we have function f_1(z_0) | \int_z f(w) dw where \int_z is a path in \Lambda from z_0 to z, the function is called a branch/branch of the multivalue function.
Eg: f(z) = \sqrt{1+z^2} = \sqrt{(1+z)(1-z)} \sqrt{z(z-1)} take w = z - i for z = i at 0 for w = i = re^{i\theta}. On C_\epsilon(i) f(z) \sim \sqrt{z} i w \sqrt{z} has a branch point w = re^{i\theta} for w = re^{i\theta}. \sqrt{w} = \int_{z_0}^w e^{i\theta} d\theta so as \theta varies from 0 to \pi/2 similarly there is a branch at z = i but \sqrt{w} can be a branch like of f at 0 and is not a branch point.
Take a power series expansion of \int_{z_0}^w \sqrt{z} d\theta around z = 0 we get \sqrt{z} = \int_{z_0}^w e^{i\theta} d\theta show that it is a function which has radius of convergence!
IMG_2618
claim: the radius of convergence is the distance from the nearest branch point or some problem point.
Ig: \log z and \log -z are never equal, it has the differential in z.
In fact Bernoulli states that \log(-z)^2 = \log z^2 \Rightarrow z(\log(-z)) = z \log z?
Thm: If f is nowhere vanishing function holo function in simply connected region \Omega, then there exist a holo function g on \Omega satisfy
proof: choose a point z_0 in \Omega define g(z) = \int_{f(z)} \frac{f’(z)}{f’(z)} dw + C_0, where C_0 = f(z_0). f is a path in \Omega from z_0 to z. g’(z) = \frac{f’(z)}{f(z)}, so \frac{d}{dz}(f(z)) = \frac{d}{dz}(f(z)).
\Rightarrow f’(z) e^{-g(z)} - f(z) \frac{f’(z)}{f(z)} e^{-g(z)} = 0. \Rightarrow f(z) e^{-g(z)} is a contour. since f(z_0) = e^{g(z_0)}, f(z) = e^{g(z)}.
Ig: describe the branch point of f(z) = \sqrt{1-z^2}. what is the smallest number of branch points cuts headed to obtain a region on which f is single valued
Branch point z=0, z=-1. Checks the branch point of square root function x is not a branch point. it can be x = \pm \sqrt{-1}, so the
answer is x.
Ig: show that all the values of i are real. After there are any other points such that z^i is real.
i^2 = e^{i \log i} = e^{\frac{i\pi}{2}} e^{\frac{i\pi}{2}} e^{\frac{i\pi}{2}} = e^{-i\pi} e^{-i\pi} e^{-i\pi} = e^{-2i\pi} e^{-2i\pi} = e^{-4i\pi}. so \log z = 4\pi + i\theta. \theta \in \mathbb{R}, so |z| = e^{i\pi}, n \in \mathbb{N}.
Ig: what are the branch points of \cos^{-1} z. \cos^{-1} z = \frac{e^{iW} + e^{-iW}}{2} \Rightarrow z = \frac{e^{iW} + e^{-iW}}{2}. \cos^{-1} z = \frac{z + \sqrt{z^2 - 1}}{2} and \sin^{-1} z = \frac{z - \sqrt{z^2 - 1}}{2}. so W = \frac{1}{2} \log(z + \sqrt{z^2 - 1}). Branch point of z + \sqrt{z^2 - 1} is z + \sqrt{z^2 - 1}. so branch point z + \sqrt{z^2 - 1} has no branch point since z + \sqrt{z^2 - 1} \neq 0. so branch point is z.
where are the branch points and branch points cut of \log \sqrt{-z}? so is \log \sqrt{-z} branch point z=0, z=1, so z is a branch point z_1. so branch point cut can be x axis x > z_1. z_1 is anti-x. x is integer co. z_1 is integer. x is integer.
IMG_2619
z=0 is a pole not branch point a is rational a = \frac{p}{q}. z is z=0 is a branch point because z=0 is a branch point of z since p and q relatively prime, we need to go around the circle q times to get back to a rational value since every circle z^a changes z if a is irrational, as \theta varies over (0, 2\pi) on C_z(0) argument of z^a varies over 2\pi a, then there is no h s.t. 2\pi a h is a multiple of 2\pi, so it’s logarithm branch point.
when a = a_1 + i a_2 on C_z(0) z^a = e^{a_1 + i a_2} e^{i \theta(a_1 + i a_2)} = e^{a_1 + i a_2 - \theta a_2} e^{i \theta a_1} a_2 \neq 0 \Rightarrow |z^a| always change \Rightarrow \theta is a logarithm branch point to by. \log \frac{z-1}{z+1} branch point z=\pm 1 \log \frac{z-1}{z+1} = \log(z-1) - \log(z+1) ensily see z \neq \pm 1 are all logarithm branch point.
From the garden of delights: problem we residue then to compute \int_{a^{\pm 1}}^{a^{\pm 1}} e^{i \theta} d\theta def: say f(z) is a - periodic if f(z+a) = f(z) for all z, a \neq 0.
Ex: \cos z, sin z is 2\pi periodic e^z is 2\pi i periodic (e^{az} = e^a e^b and e^{2\pi i}) def say f(z) is a - quasi-periodic if \frac{f(z+a)}{f(z)} is a - periodic \frac{f(z+a)}{f(z)} = \frac{f(z)}{f(z+a)} \Rightarrow f(z+a) = f^2(z)
Fundamental lemma: if f: U \rightarrow \mathbb{C} is holo a - quasiperiodic then \intC f(z) dz = \int_C \frac{f(z)}{f(z+a)} dz = \int{a+C} \frac{f(z)}{f(z+a)} dz C can be any curve z \rightarrow a+z
pf: RHS = \intC \frac{f(z)}{f(z+a)} dz = \int{a+C} \frac{f(z)}{f(z+a)} dz = \dots = \int_C \frac{f(z+a)}{f(z+a)} dz = \dots
\int_C \frac{f(z) - f(z+a)}{f(z)} dz = \int_C f(z) dz we want to use lemma take f(z) = e^{-z^2} f(z+a) = e^{-2az-a^2}
\frac{f(z)}{f(z+a)} = e^{-2a(z-a)} - e^{-2az+a^2} claim f is - quasi-periodic we want to show e^{-a^2} = e^{a^2} \Rightarrow e^{2a^2} = e^{2az} \Rightarrow a^2 = 2\pi i \Rightarrow a = \sqrt{2\pi i} \cdot \frac{e^{iz}}{e^{iz+a}}
we need to find poles z = 2\pi i - 2\pi i n so \int{\mathbb{C}} e^{-z^2} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz we need to find poles z = 2\pi i - 2\pi i n so \int{\mathbb{C}} e^{-z^2} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz
\int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz we need to find poles z = 2\pi i - 2\pi i n so \int{\mathbb{C}} e^{-z^2} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz
\int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz = \int_{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz we need to find poles z = 2\pi i - 2\pi i n so \int{\mathbb{C}} e^{-z^2} dz = \int{\mathbb{C}} \frac{e^{-z^2}}{1+e^{2\pi i z}} dz
Extra credit: function \phi(x) which is i-quasi-periodic
IMG_2620
forget about function f(z) > 0. so our function can only have isolate zeros. Dr = disk of radius r, Cr = circle of radius r (around O). The Jarsen’s formula let \lambda be an open set that contains D_R suppose f is holo on \lambda. f(z) \neq 0, and vanishes nowhere on C_R let {z_1, \dots, z_k} be the zeros of f in D_R, counted with multiplicity, then \log |f(z)| = \sum_{k=1}^k \log \left| \frac{z_k}{R} \right| + \frac{1}{2\pi} \int_0^{2\pi} \log |f(Re^{i\theta})| d\theta
pt: for real number \log f_1, f_2 = \log f_3 + \log f_4 always hold. so if f_1, f_2 satisfy the thm, then so does f_3, f_4. {z_1, f_2} = {z_1, z_2} \cup {z_3, f_4} z_3 g(z) holo on \lambda s.t. f(z) = (z-z_1) \cdots (z-z_k) g(z) where g(z) \neq 0 on \lambda. It suffices to prove the thm for g(z) and z-z_1 for g(z) to want to show \log |g(0)| = \frac{1}{2\pi} \int_0^{2\pi} \log |g(Re^{i\theta})| d\theta.
let h = \log g so ① follows from the mean value property of fleal part. Remark: f holomorph and non-vanishing, \log f is holo and \log |f| is hormonic since \log |f| = Re(\log f). for z-z_1 want to show \log |z_1| = \log \frac{z_1}{R}.
later put is 0. \int_0^{2\pi} \log |Re^{i\theta} - z_1| d\theta = \log |z_1| + \frac{1}{2\pi} \int_0^{2\pi} \log \left| e^{z_1\theta} - \frac{z_1}{R} \right| d\theta we want to show \int_0^{2\pi} \log |e^{z_1\theta} - \frac{z_1}{R}| d\theta = \int_0^{2\pi} \log \left| 1 - \frac{z_1}{Re^{i\theta}} \right| d\theta let \theta = -i\theta then it is -\int_0^{2\pi} \log \left| 1 - \frac{z_1}{Re^{i\theta}} \right| d\theta since \frac{z_1}{Re^{i\theta}} < 1. \Rightarrow 1 - \frac{z_1}{Re^{i\theta}} z_1 is holo
and non-vanishing so \log f so z_1 g(z) = \log \left( 1 - \frac{z_1}{Re^{i\theta}} z_1 \right) from mean value then \int_0^{2\pi} \log \left( 1 - \frac{z_1}{Re^{i\theta}} z_1 \right) d\theta = \log \left( 1 - \frac{z_1}{Re^{i\theta}} 0 \right) = 0. so satisfy.
suppose f is holo on D_R pf(r) = \prod f zeros of f, counted with multiplicity in D_R hf(r) is a static function.
Then suppose f(0) \neq 0 and f is non-vanish on C_R then \int_0^{2\pi} f(r) dr = \frac{1}{2\pi} \int_0^{2\pi} \log |f(Re^{i\theta})| d\theta - \log |f(0)|.
pt: let R = \frac{1}{2\pi} \int_0^{2\pi} f(r) dr be the zeros of f inside D_R then \int_0^{2\pi} f(r) \frac{dr}{r} plug in the Jarsen formula to get the result. for further the theorem then follows by def: Order of function let f be an entire function if \exists p > 0 and constants A, \beta > 0. s.t. |f(z)| \leq A \frac{|z|^{\beta}}{|z|^p} then f has order of growth \leq p. the order of growth is p. pf = \frac{1}{p} f.
IMG_2621
pple: the order of growth at (1), z^n (2), \sin z (3), \sin z e^z (1)
thm: If f is an entire function with order of growth \leq f, then (1)
h(r) \leq Cr^p \quad \exists C > 0 and H > 0 all sufficient by large r. (ii) If z_1, z_2, \dots
note the ion zeros of f, then for all s \geq 1 we have \sum_{k=1}^{\infty} \frac{1}{(z_k)^s} < \infty.
f: \text{Infinite products} \quad \int_0^\infty h(x) \frac{dx}{x} = \frac{1}{\pi} \int_0^{2\pi} \log |f(x)| dx = \int_0^\infty \log |f(x)| dx let f(x)
the \int_{-\infty}^{\infty} \frac{1}{x} \int_0^{2\pi} \log |f(x)| dx = \log |f(z)| (1) since h(r) is knotonic
increasing with r \quad \int_0^{2\pi} h(x) \frac{dx}{x} \geq h(r) \int_r^{\infty} \frac{1}{x} dx = h(r) \log(2r - h(r) \log r) = h(r) \log 2 \cdot (1)
and we have \int_0^{2\pi} \log |f(x)| dx \leq \int_0^{2\pi} \log |f(x)| dx \leq C_1 r^p. (3) apply
(2) we get (1) \sum{k=1}^{\infty} |z_k|^{-s} = \sum{j=0}^{\infty} \sum{k=1}^{p-1} |z_k|^{-s} \leq \sum{j=0}^{\infty} 2^{-sj} h(j^{s+1}) \leq C \sum_{j=0}^{\infty} |z_j|^{s+1}
Infinite products. let {an}{n=1}^{\infty} be a sequence of complex numbers, then
\frac{1}{n} (1 + a_n) converges if the limit \lim_{n \rightarrow \infty} \frac{1}{n} (1 + a_n) exist and no zeros
prop If \sum_{n=1}^{\infty} |a_n| < \infty, the product \frac{1}{n} (1 + a_n) converge if converge to zero
iff one of factors is zero
pt: If \sum_{n=1}^{\infty} |a_n| converge with out logs of generality suppose |a_n| is
If |a_n| < \epsilon, |ta_n| is with in the radius of convergence of \log |ta_n| in suppose
z \rightarrow 1. power series expansion \Rightarrow |\log |ta_n|| < 2\epsilon |ta_n| when z < 0 so \frac{1}{n} |ta_n|
= \frac{1}{n} \log |ta_n|. \log |tan| < 2a_n \Rightarrow \sum{n=1}^{\infty} \log |ta_n| converges so \sum_{n=1}^{\infty} \log |ta_n| converges
since \exp is continuous so \exp \sum_{n=1}^{\infty} \log |ta_n| converges so \sum_{n=1}^{\infty} \log |ta_n| converges
if |ta_n| \neq 0, \forall n, the product is of the form e^{\sum_{n=1}^{\infty} \log |ta_n|} and is knotonic
Recall: If \sum_{n=1}^{\infty} |a_n| < \infty, the product \frac{1}{n} (1 + a_n) converge. It converges
to 0 iff: one of the factors is zero.
prop: Suppose {f_n} is a sequence of holo function on open set \Omega if there exist a constant C_n > 0 s.t. \sum C_n < \infty and |f_n| converge uniformly in \Omega to a holo
sequence f(z) with |a_n(z)| \leq C_n since (C_n) are constant
pt: write f_n(z) = |ta_n(z)| with |a_n(z)| \leq C_n since (C_n) are constant
the bounds |f_n - 1| \leq C_n are uniform in \Omega. the product converges
in \Omega to a holo function by the previous lemma.
continue to increase f(z) = 0.
IMG_2622
Thm: we restray infinite products given any sequence {a_n} with |a_n| \rightarrow \infty as n \rightarrow \infty, if an entire function f that vanishes out all the points {a_n} and nowhere else f(z) \in g(z) where g is entire. * any such sufficient function in the form
pt: suppose his a second entire function with the same zeros as f then \frac{1}{f} is entire (every a_n is removable singularities) nowhere vanishing. \Rightarrow \frac{1}{f} = e^{g(z)} for some entire g(z). Remains to find the function vanishing at points {a_n}. Ity \lim_{n \rightarrow \infty} (1 - \frac{z}{a_n}) this makes when convergent want to obtain convergence by inserting expectational terms. def: k \geq 0, ‘canonical factors’, E_0 = 1 - z, E_k(z) = (1 - z) e^{z + ikt} \dots + e^{z + ikt} z^k, k is the degree of canonical factors.
f(z) = z^n \frac{1}{\pi} E_n(z), where n is the order of zero at the origin.
Lemma: If |\frac{z}{a_n}| < \frac{1}{2}, then |1 - E_n(\frac{z}{a_n})| < C |\frac{z}{a_n}|.
pt: when |\frac{z}{a_n}| < \frac{1}{2}, a_n wite 1 - \frac{z}{a_n} = e^{-\frac{z}{a_n} \log(1 - \frac{z}{a_n})}. So E_n(\frac{z}{a_n}) = e^{-\frac{z}{a_n} \log(1 - \frac{z}{a_n})}.
e^{(\log(1 - \frac{z}{a_n}) + \frac{z}{a_n})} = 1 + \frac{z}{n |a_n|^n} = e^{\frac{z}{n |a_n|^n}}, we know \log(1 - \frac{z}{an}) = \frac{z}{a_n} - \frac{z^2}{2a_n^2} \dots \Rightarrow w = -\sum{n=1}^{\infty} \frac{1}{z} (\frac{z}{a_n})^n. Since |\frac{z}{a_n}| < \frac{1}{2}, |w| \leq |\frac{z}{a_n}|^{n+1} \frac{n+1}{2} e^{\frac{z}{n}} \leq 2 |\frac{z}{a_n}|^{n+1}.
|1 - E_n(\frac{z}{a_n})| = \left| 1 - \frac{z}{n |a_n|^n} |1 - e^{\frac{z}{n |a_n|^n}| \right| = \left| 1 - \frac{w}{n |a_n|^n} \right| \leq |w| \leq 2 |\frac{z}{a_n}|^{n+1}. suppose |z| < R and consider terms indicated with |a_n| \leq R and |a_n| \geq R. It remains to check convergence for z \in \mathbb{R} \setminus [0] of the infinite products of these terms when |a_n| \geq R, and |z| < R we have |1 - E_n(\frac{z}{a_n})| \leq C |\frac{z}{a_n}|^{n+1}. then the product converges by the previous prop. so f(z) = z^n is holo in \mathbb{R} \setminus [0]. take R infinitely large then f(z) is entire.
Example: \frac{\sin \pi z}{\pi} = z^{\pi} (1 - \frac{z^2}{\pi^2}). suppose f is entire with order of growth \rho. let k the integer s.t. k = \lceil \rho \rceil. let {a_n} be an entire point of f, then f(z) = e^{k \pi z} is a polyhominal of degree \leq k, and n is the order of zero at z=0.
conformal maps: def: a conformal map, or bij holo is a bij holo function.
If f: U \rightarrow V and V are conformally equivalent or biholo. If f: U \rightarrow V is holo and injective then f(z) \neq 0 for all z \in U. If f: U \rightarrow V is holo and injective then f(z) \neq 0 for all z \in U. If f: U \rightarrow V is holo and injective then f(z) \neq 0 for all z \in U. The particular inverse of f def an entire function.
5. 积分计算、乘积展开、开映射与 Riemann 映射
IMG_2623
the inverse of a conformal map is conformal pt. suppose f is holo. and injective but f(z_0) \neq 0 for z_0 \in U. Then f(z) - f(z_0) = a(z-z_0)^k + b(z) for z near z_0 where a \neq 0, k \neq 0. G has zeros at order k+1 at z_0. choose w \neq 0 sufficiently small s.t. f(z) = a(z-z_0)^k - w has a zero inside C (at least 2 zeros / inside centred on z_0 so f(z) - f(z_0) - f(w)w has at least 2 zeros inside C). we can also choose the radius of C small enough s.t. inside C f(z) \neq 0. \Rightarrow z = z_0 so ho zero of f(z) - f(z_0) - w can be of order 2 \leq 2 inside C because this mould that ensure that f(z) \neq 0 inside C at a point other than z_0 so the f(z) - f(z_0) - w has distinct roots but f is not is not injective
since f(z) \neq 0 for f conformal, implicit differentiation show f^{-1} is holo. Exercise. show that if f: U \rightarrow C is holo and f(z_0) \neq 0 then preserve angles at z_0.
eg. away from 0, z \mapsto \frac{1}{z} preserve angles
pt: calculate Fab Incolinian we cf eg. \Rightarrow J = \lambda I_2 R, R is a rotation.
Ex: ID := unit disk H = \text{upper half plane} f(z) = \frac{z-i}{z+i}, G(w) = \frac{1-w}{1+w}
itm: then f is conformal map with the inverse G pt: each map is holo in the domain. note |f(z)| < 1 for z \in H so f maps H into ID w = u + iv.
\Rightarrow 0 so G(w) \in H check F \circ G(w) = w and G \circ F(z) \geq z.
so the upper half plane and unit disk conformally equivalent. Mapping of type z \mapsto \frac{az+b}{a^2+b^2} and a, b, c, d \in C are called transformational transformation.
Aut(U) = { \text{conformal map of } U \text{ to itself} } is a gp. z \mapsto e^{i i z} dilatation rotations: z \mapsto e^{i i z} invase z \mapsto \lambda z. \lambda \in \mathbb{R}. f(z) = \frac{1+z}{1-z} invase z \mapsto \lambda z. f(z) = \frac{1+iz}{1-iz} in 1st and 2nd drain.
holo on domain |w+1| > |w-1|. in 1st and 2nd drain. G(w) 0. or G(w) 0.
f
vi
p
by
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f(z)
f(z)
inside
def
of U
Ex:
\psi(z)
|z| > 0
ID to
then
f(z) =
pt: z
lensa
in ID
we see
rotation
(ol A
Aut (1)
f is
= f^{-1} \circ \phi
= 1)
for z \in U
then
ly for
pt: first
since Inv
each for
IMG_2624
schwarz lemma - let f: \mathbb{D} \rightarrow \mathbb{D} holo with f(0)=0 then (i) |f(z)| \leq |z| for all z \in \mathbb{D} (ii) If for same z_0 \in \mathbb{D} |f(z_0)| = |z_0| then f is a rotation. (iii) |f’(0)| \leq 1 and if the equality holds, f is a rotation.
pf (ii) since f(0)=0, a_0=0 and \frac{f(z)}{z} has removable singularities at 0, by maximum number then \left| \frac{f(z)}{z} \right| has maximum \leq 1 at on \mathbb{D}. (iii) if it reaches maximum inside \mathbb{D} then \left| \frac{f(z)}{z} \right| is constant, \Rightarrow \frac{f(z)}{z} = C with numbers C. then f(z) = Cz is a rotation. (iiii) If f’(0) \neq 0, then \left| \frac{f(z)}{z} \right| has no numbers inside \mathbb{D} so f is rotation.
def: a conformal map from an open set \Lambda to itself is an automorphism of \Lambda.
bx: \psi_d(z) = \frac{z-z}{1-jz} is auto of \mathbb{D}. |d| < 1. pf: when |d| < 1, \psi_d(z) holomorphic on \mathbb{D}. \psi_d(e^{i0}) = \frac{z-e^{i0}}{1-je^{i0}} = e^{-i0} - \frac{z-e^{i0}}{z-e^{i0}} so |e^{i0}| = 1, and |-\frac{z-e^{i0}}{z-e^{i0}}| = 1 so it maps \mathbb{D} to \mathbb{D} and it maps \mathbb{D} to itself specially \psi_d(z)=0, \psi_d(0)=z. \psi_d(0)\psi_d(z)=0 by calculation. Thus, if f is an automorphism of \mathbb{D}, z \in \mathbb{C} \setminus \mathbb{D} and an \mathbb{D} \in \mathbb{D} s.t.
f(z) = e^{i0} \frac{z-z}{1-jz} s.t. f(z) = 0. write fg = f \psi_d since g(0) = 0, schwarz pf: z \in \mathbb{D} s.t. f(z) = 0 and |g’(w)| \leq |w|. since g’ is auto lemma \Rightarrow |g(z)| \leq |z| \forall z \in \mathbb{D} \Rightarrow g is rotation. \Rightarrow f = e^{i0} \cdot \psi_d in \mathbb{D} \Rightarrow |g(z)| = |z| \Rightarrow g is rotation. we see that the only conformal map of \mathbb{D} to itself that fixes 0 are rotations.
(ol. Aut (\mathbb{D}) acts transitively on \mathbb{D}.
Aut(\mathbb{H}): denotes are of the form f^{-1}\phi where \phi \in Aut(\mathbb{D}) and f is a conformal between \mathbb{D} and \mathbb{H}. we see f^{-1}\phi of \phi of Aut(\mathbb{H}) \Rightarrow f^{-1}\phi, \phi of Aut(\mathbb{D}) \cong Aut(\mathbb{H}). sl2(\mathbb{R}) = (M \times M){\text{Matrix}}(\mathbb{R}) ad-bc \Rightarrow | \in Aut(\mathbb{H}) denoted = 1 }. M = (a, b) claim z \mapsto \frac{az+b}{z+d}
fm \times (z) = \frac{az+b}{z+d} thus: Every auto of \mathbb{H} takes form fm for some M \in sl_2(\mathbb{R}). conversely, fm is an auto of \mathbb{H}.
pf: first show fm map \mathbb{H} to itself. \bar{fm}(fm(z)) = \bar{m}(z) it implies since fm(z) \neq 0. Next for fm = fmm (by calculation). third: show each fm is an Auto since fm^{-1} = fm^{-1} and fz = id_{\mathbb{H}}. Calculation.
IMG_2625
BLUE GIRL
transitivity of {f_n | M(-SL_2(\mathbb{R}))} obstruct map take z to i
I_n(f_n(z)) = \frac{in(z)}{(z+id)^2} take d=0, CGIR s.t. \frac{in(z)}{(z-i)^2}. we can choose
a=1, b=-\frac{i}{2} \Rightarrow M = (\frac{1}{e-i}, \frac{-i}{e}). then compose with a translation M_2 = (\frac{1}{e+i}, \frac{-i}{e}) that take real part of f_n(z) to 0. where b = -\frac{in(z)}{(z-i)^2} - \frac{in(z)}{(z-i)^2} - \frac{in(z)}{(z-i)^2}. finally, remain to study anti that fix i. let f_C Ant(IH) with f_C(z)
fB = {f{fC} | M(-SL_2(\mathbb{R})), f{BC} = f_B(z) = \beta} g = f_C f_B then f_C f_B f_Ant(IH) is a rotation since it maps z to 0. Missing rep. ration map to rotation
M_0 = (\cos i, -\sin i) maps to id a rotation in id with angle 2\theta \Rightarrow form of i rotation of id f = \underline{\underline{f(z)}} \stackrel{i-z}{=} \underline{\underline{f(z)}} = \underline{\underline{f(z)}} = z^{\frac{i-z}{i+z}}, f^*(z) = z^{\frac{i-z}{i+z}} so f_C f_B f_Ant(IH) is a rotation then g = f_C f_B f_Ant(IH).
What is the kernel of the map SL_2(\mathbb{R}) \rightarrow Ant(IH) kernel is (\pm i) check so we have Ant(IH) \simeq RSL_2(\mathbb{R}).
Cross ratios and fractional linear transformation (M has transformation z \mapsto \frac{az+b}{cz+d}, a, b, c, d \in \mathbb{C} and b \neq 0 is Main transformation)
want to decompose map z \mapsto \frac{az+b}{cz+d} into (1) z \mapsto z+\frac{d}{c} translation
z \mapsto \frac{1}{z} inversion + conjugation z \mapsto \frac{az-bc}{c^2-b^2-c^2} scaling/rotation z \mapsto z+\frac{a}{c} translation
I_k = Inversion through circle K. suppose q is the center of circle and R is radius, then I_k = (C/q) \rightarrow (C/q) \times (q \otimes L) s.t. I_k(q) / I_k(z) is a single line and z and q on the same line.
|z-q| = R.
Lemma: I_k preserves circles (as straight line is a circle with infinite radius)
p: \begin{pmatrix} z_1 \ y_1 \ z_2 \ z_3 \end{pmatrix} \rightarrow \begin{pmatrix} z_1, qz_1 \ -qz_1, qz_1 \ -qz_1, qz_1 \ -qz_1, qz_1 \end{pmatrix} (1). \langle z_1, qz_1 \rangle = \langle I_k(z_1), q I_k(z_2) \rangle and since |I_k(z_1)-q|/|z-q| = \frac{|z_1-q|}{|z-q|} then |I_k(z_1)-q|/|z-q| = \frac{|z_1-q|}{|z-q|} as d \neq 0
so \Delta z_1, qz_1 \sim \Delta I_k(z_1), q I_k(z_2) \Rightarrow |I_k(z_1)-I_k(z_2)| \approx \sqrt{|z_1-z_2|} as d \rightarrow 0
fix z_1, have z_2. I_k(z_1), I_k(z_2) \neq \emptyset so it’s a circle.
b constant. search I_k(z_1) q don’t have so it’s a circle
when L is straight line. we can take state paradigmatical to line
fix the foot of paradigmatical so the single q has a 4\pi^0 angle \Rightarrow
a circle cross q. (3) circle will in the K very similar. all this to go together will give the conclusion
z b I_k = C/k I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
z b I_k \neq 0
IMG_2626
z \mapsto \frac{az+b}{cz+d} fractional linear transformations (Mobius transformation) can be breaken down into translations, rotations, scaling, conjugation and inversion.
I_K := inversion through circle K center on A with radius R. I_K maps z \in C \setminus K to a point on the line through qz and |I_K(z) - q| / |z - q| = R^2.
I_K \circ I_K = \text{id} ihm I_K preserve circle (view straight line as circle) and so is the fractional linear transformation.
P_{\text{to end}}‘s ihm: any a, b, c, d on a circle C then |a-b| / |b-c| + |a-c| / |b-d|.
top: we have lemma the triangle \angle z, qz, z = \angle I_K(z) / |I_K(z)| are similar so if a line L does not pass through K, I_K map L to a circle through q. we have a circle K center on K, with C inside K
then |I_K(b) - I_K(c)| + |I_K(c) - I_K(b)| using similarity of triangles & def of I_K.
|I_K(b) - I_K(c)| = \frac{|I_K(b) - q|}{|a-c|} = \frac{|I_K(b) - q|}{|a-c|} \Rightarrow |I_K(b - I_K(c))| = \frac{R^2 |b-q|}{|a-b| |a-c|}
similarly |I_K(c) - I_K(d)| = \frac{R^2 |c-d|}{|a-c| |a-d|} |I_K(b) - I_K(d)| = \frac{R^2 |b-d|}{|a-b| |a-d|} \Rightarrow
|a-b| / |c-d| + |a-d| / |b-c| = |a-c| / |b-d|
ihm: 3! Mobius transformation reading any 3 distinct point in C to
any other 3 distinct point.
pf: (uniqueness) if not unique with R exception of I_d, Mobius transformation has at least 2 fix point solving z = \frac{az+b}{cz+d}. R different map can’t have 3 fix points (existence) cross ratios! construct a Mobius transformation mapping q rs to q’ if suffices so to do for q’ = 0, r’ = 1, s’ = 0, \bar{b} map q to q’ = 0, r’ = s’ = \infty, we must have z \rightarrow k \frac{zq}{z-s} then let r determine k = \frac{rs}{r-q} so we have z \mapsto \frac{rs}{r-q} \cdot \frac{zq}{z-s}. cross translations: [z, q, r, s] = \frac{(z-q)(r-s)}{(z-s)(r-q)}
comment: we still need normalisation s.t. ad - bc = 1.
(or p lies on C iff \text{Im}[p, q, r, s] = 0)
(or p lies on C iff \text{Im}[p, q, r, s] = 0) only
pf: \Sigma [z, q, r, s] maps C to real fine axis since real axis is the ‘circle’ since contain 0, 1, \infty so \bar{a} = i[p, q, r, s] \in \text{real axis}.
IMG_2627
when are two sets conformally equivalent ? i.e. If a conformal map F: U \rightarrow V.
Necessary conditions, If U is connected or simply connected, V must be also (conformal map is a topological homeomorphism, if conserve topological properties). If U = V = \mathbb{C}, V cannot be compact otherwise Liouville theorem imply it’s a constant.
def: U \subseteq \mathbb{C} but U \neq \mathbb{C} or \phi, \lambda is called proper.
Riemann mapping theorem: \lambda is proper and simply connected, if z_0 \in U, \exists ! conformal map f: \lambda \rightarrow \text{ID} s.t. f(z_0) = 0 and f’(z_0) > 0 (real) (or Any two proper simply connected open subset is \mathbb{C} are conformally equivalent.)
def: Let \lambda be an open set of \mathbb{C}. A family F of holo functions on \lambda is called normal if every sequence in F has a subsequence that converges uniformly on every compact subset of \lambda.
def: F is uniformly bounded on compact subset A of \lambda if for each compact subset K \subset \lambda \exists B > 0 s.t. |f(z)| \leq B \forall z \in K and \forall E \subset F.
def: The family F is equicontinuous on compact K if for every \delta > 0 \exists \varepsilon > 0, s.t. for every z, w \in K with |z-w| < \delta, then |f(z) - f(w)| < \varepsilon. What is the formula?
Matel thm: suppose F is a family of holo functions on \lambda that is uniformly bounded on compact subsets of \lambda. Then F is a normal family, and F is equicontinuous on every compact subset of \lambda.
pf: uniformly bounded + holo \Rightarrow equicontinuous: Let K be compact subset of \lambda and choose r > 0 small enough s.t. R D_{3r}(z) \subset \lambda \forall z \in K.
suppose z, w \in K with |z-w| < r that take \gamma = d_{3r}(z)/R cauchy integral formula give us:
uniformly bounded + equicontinuous on compact subset of \lambda \Rightarrow F normal family if F is a sequence {Kn}{n=1}^{\infty} of compact subsets of \lambda is called an exact exhaustion of \lambda if \bigcap_{n=1}^{\infty} K_n is contained in the re interval of K_1.
K_1 is an open set of \mathbb{C} has an exhaustion K_2 and \bigcap_{n=1}^{\infty} K_n is called any open set of \mathbb{C} (exactly any matrix space)
IMG_2628
pt of claim: if \mathcal{N} is bounded, K_C = \text{set of closure of all points in } \mathcal{N} of distance > \frac{1}{C} from d \mathcal{N}. if \mathcal{N} is not bounded, K_C = \text{cat closure of all points in } \mathcal{N} of distance > \frac{1}{C} from d \mathcal{N} (finite) and modulus C. Let {fn}{n=1}^{\infty} be a compact sequence of f_n and k a compact subset of \mathcal{N} {wj}{j=0}^{\infty} a sequence point dense in \mathcal{N}. (eg. the set of all points in \mathcal{N} with rational real and imaginary parts). Uniform boundedness of {f_n} \Rightarrow. \exists a subsequence {f_{in}} \supset \text{set of } f_n s.t. {f_{in}(w_i)} converges similarly \exists a subsequence {f_{in}} \supset \text{set of } f_n s.t. {f_{in}(w_j)} converges in {f_{in}} \Rightarrow a subsequence {f_{jin}} s.t. {f{jin}(w{j+1})} converge. It take Diagonal sequence gn = f{in} \Rightarrow {g_n(w_i)} converge for every i.
claim: equicontinuously \Rightarrow {g_n} converge uniformly on K.
give C > 0, choose \delta from the def of equicontinuous and cover K with finite disk set of disks D\delta(w_i), \dots, D\delta(w_j). For N large enough, \forall n, m > N we have |gn(w_i) - g_m(w_j)| \leq C \cdot V{ij} \cdot III. For z \in K, z \in D_\delta(w_i) for some i \in [II, III], |g_n(z) - g_m(z)| \leq |g_n(z) - g_m(w_i)| + |g_m(w_j) - g_n(w_j)| + |g_n(w_i) - g_n(z)| \leq 3\epsilon.
Remain to prove practice existence of subsequence that converge uniformly for any K_1 in exhaustion {K_1} of \mathcal{N}: (let g_n(g_m) be a subsequence of {f_n} that converge uniformly on K_1, {g_{2n}} be a subsequence of {g_n} that converge uniformly on K_2 similarly {g_{2n}} be a subsequence of {g_{jn}} that converge uniformly on K_j, then {g_{nn}} a subsequence of {f_{in}} converge. uniformly in all every compact set of
lemma If \mathcal{N} is connected open subset of \mathbb{C} and {f_n} a sequence of holo injective functions on \mathcal{N} then that converge uniformly on every compact subset on \mathbb{C} to a holo function, then f is either constant or injective. pf: suppose f is not injective, \exists z_1, z_2 in \mathcal{N} with f(z_1) = f(z_2), z_1 \neq z_2. def g_n := f_n(z_1) - f_n(z_2), g_n has an zeros at z_1, g_n \rightarrow g uniformly since f_n \rightarrow f uniformly g = f(z_1) - f(z_2) if g \neq 0. we have 1 = \frac{1}{z1 \cdot z_2} \int{z_1}^{z_2} g(s) ds since g_n \rightarrow g uniformly since g_n is a circle not containing other zeros around z_2. since \frac{1}{g_n} \rightarrow \frac{1}{g} integral must be integrable so \frac{1}{z1 \cdot z_2} \int{z1}^{z_2} \frac{1}{g_n(s)} ds \rightarrow \frac{1}{g_n(z_1)} \int{z_1}^{z_2} \frac{1}{g_n(s)} ds = 1. since the integral must be integrable so contradicted to g_n has no zeros in \mathcal{N}.
IMG_2629
proof of RMT: ① show the existence of a conformal map from \mathcal{N} into \mathbb{D}
Let d \in \mathbb{R} \setminus \mathbb{C} \setminus \mathcal{N} def f = (\log(d-d)). so f is holo (choose a branch), injective
single value since \mathcal{N} simply connected contains no branch points of f.
If w \in \mathcal{N}, f(z) \neq f(w) + 2\pi i for z in \mathcal{N} since \log and exp are all 1-
if it has otherwise, we have e^{f(z)} = e^{f(w) + 2\pi i} \Rightarrow z > w. contradiction, \exists disk
centered on f(w) + 2\pi i disjoint form f(\mathcal{N}), otherwise there would \exists f(z_i)
\subseteq \mathcal{N}, s.t. f(z_i) \rightarrow f(w) + 2\pi i, take exp, then z_i \rightarrow w. \rightarrow f(z_i) \rightarrow f(w)
by continuity. let F(z) = \underbrace{f(z) - f(w) + 2\pi i}_{\text{transformation}}. then by scaling and
transformation, we get F(\mathcal{N}) \subseteq \mathbb{D} (i) without loss of generality that \mathcal{N}
def is an open subset of \mathbb{D} contain 0 take F = f(\mathcal{N}) \rightarrow \mathbb{D} holo inj and f(z_i)
F is not empty by (i), because it contains \mathbb{D}. (ii) image of F is uniformly branched
since it’s in \mathbb{D} (f(0)) f in F is uniformly branched (cauchy inequality). choose
let S = \sup_{f \in F} |f(x)| (hoose {f_n} \subseteq F s.t. |f_n(x)| \rightarrow S as n \rightarrow \infty). Mortel then
tell us \exists a subsequence of {f_n} converging uniformly on compact sets to a
holo function f on \mathcal{N} ed F so s.t. \Rightarrow f is constant by lemma
f is injective |f(z)| \leq 1 \forall z \in \mathcal{N} since it’s true \forall f \in {f_n} \Rightarrow f is
in F (i) show f is onto \mathbb{D}: suppose f is not onto \mathbb{D} if it let
d be a point in \mathbb{D} \setminus f(\mathcal{N}), U := \psi_d(f(\mathcal{N})) U is simple connected and
o \in U we can def single value function g(w) = |w| = e^{i \ln w} on U
take h = \psi_g(w) o \rightarrow \psi_d of f, h \in F since h is holo, h(0) = 0, h(\mathcal{N}) \subseteq \mathbb{D}.
h injective since it’s composition of two injective \psi_g(w) \rightarrow \psi_d map \mathbb{D} into
into \mathbb{D} \rightarrow 0 show \psi_d map \mathbb{D} into \mathbb{D} \rightarrow 0 show \psi_d map \mathbb{D} into \mathbb{D}
contradict to \text{thin}(0) \neq \emptyset. |f(x)| is the maximum. At last - totale f to get
f(x) \geq 0
Riemann surfaces
def: function element is a pair (D, f) where D is a region and f: D \rightarrow \mathbb{R}
(\mathbb{R} is a Riemann surface sphere) is a single value nero function on D
Recall: let (b, f), (b, g) be function elements def on the same region
If f = g on a nonempty open set U of D, then f = g on D. (region need to be
simply connected and connected)
call if (D_1, f_1) is a function element and D_2 is a region with intersects D_1 \cap D_2 then there is at most one nero function f_2 on D_2 s.t. f_1 = f_2 on D_1 \cap D_2. (sometimes)
IMG_2630
not exist)
def: when f such as f_2 exist (D_2, f_2) is called a direct nero continuation of (D_1, f_1) written (D_1, f_1) \sim (D_2, f_2).
(D_1, f_1) \sim (D_2, f_2) \Rightarrow (D_1) \cup (D_2), hot: \sim is not an equivalence relation since transitivity doesn’t always hold. (D_1, f_1) \sim (D_2, f_2), (D_2, f_2) \sim (D_3, f_3) \Rightarrow
contrary: (D_1, f_1) \sim (D_2, f_2), (D_2, f_2) \sim (D_3, f_3), then f angles on U, but may disagree on V, hot on whole D \cap D_3, so (D_1, f_1) \sim (D_2, f_2)
example of a direct nero continuation: f1(z) = \sum{n=0}^{\infty} z^n on the interior of D, f_2(z) = \frac{1}{1-z} defined on S is a direct nero morphic continuation def: nero continuation - continuing (D_1, f_1) \sim (D_2, f_2) \sim \sim is such way that single valued nero function on V \cup D_2 is obtained:
the nero continuation on D_2 might depend on the path of continuation you get there eg: (D_1, f_1) \sim (D_2, f_2) \sim (D_3, f_3) and (D_1, f_1) \sim (D_2, f_2) \sim (D_3, f_3) and (D_1, f_1) \sim (D_2, f_2) \sim (D_3, f_3) hence
The intersection of two correx region (eg. disks), and connected def: Analytic continuation via power series. ex: Let D_1 be the large open disk centered on a on which f is analytic. on D_1, f is analytic.
= \sum_{n=0}^{\infty} C_n (z-a)^n, let b \in D_1, let U be the largest disk centered on b with Cn = \sum{n=0}^{\infty} C_n (z-b)^n, f1(z) = \sum{n=0}^{\infty} f_1(z-b)^n, f2(z) = \sum{n=0}^{\infty} f_2(z-b)^n, f3(z) = \sum{n=0}^{\infty} f_3(z-b)^n, f4(z) = \sum{n=0}^{\infty} f_4(z-b)^n, f5(z) = \sum{n=0}^{\infty} f_5(z-b)^n, f6(z) = \sum{n=0}^{\infty} f_6(z-b)^n, f7(z) = \sum{n=0}^{\infty} f_7(z-b)^n, f8(z) = \sum{n=0}^{\infty} f_8(z-b)^n, f9(z) = \sum{n=0}^{\infty} f_9(z-b)^n, f{10}(z) = \sum{n=0}^{\infty} f_{10}(z-b)^n, f{11}(z) = \sum{n=0}^{\infty} f_{11}(z-b)^n, f{12}(z) = \sum{n=0}^{\infty} f_{12}(z-b)^n, f{13}(z) = \sum{n=0}^{\infty} f_{13}(z-b)^n, f{14}(z) = \sum{n=0}^{\infty} f_{14}(z-b)^n, f{15}(z) = \sum{n=0}^{\infty} f_{15}(z-b)^n, f{16}(z) = \sum{n=0}^{\infty} f_{16}(z-b)^n, f{17}(z) = \sum{n=0}^{\infty} f_{17}(z-b)^n, f{18}(z) = \sum{n=0}^{\infty} f_{18}(z-b)^n, f{19}(z) = \sum{n=0}^{\infty} f_{19}(z-b)^n, f{20}(z) = \sum{n=0}^{\infty} f_{20}(z-b)^n, f{21}(z) = \sum{n=0}^{\infty} f_{21}(z-b)^n, f{22}(z) = \sum{n=0}^{\infty} f_{22}(z-b)^n, f{23}(z) = \sum{n=0}^{\infty} f_{23}(z-b)^n, f{24}(z) = \sum{n=0}^{\infty} f_{24}(z-b)^n, f{25}(z) = \sum{n=0}^{\infty} f_{25}(z-b)^n, f{26}(z) = \sum{n=0}^{\infty} f_{26}(z-b)^n, f{27}(z) = \sum{n=0}^{\infty} f_{27}(z-b)^n, f{28}(z) = \sum{n=0}^{\infty} f_{28}(z-b)^n, f{29}(z) = \sum{n=0}^{\infty} f_{29}(z-b)^n, f{30}(z) = \sum{n=0}^{\infty} f_{30}(z-b)^n, f{31}(z) = \sum{n=0}^{\infty} f_{31}(z-b)^n, f{32}(z) = \sum{n=0}^{\infty} f_{32}(z-b)^n, f{33}(z) = \sum{n=0}^{\infty} f_{33}(z-b)^n, f{34}(z) = \sum{n=0}^{\infty} f_{34}(z-b)^n, f{35}(z) = \sum{n=0}^{\infty} f_{35}(z-b)^n, f{36}(z) = \sum{n=0}^{\infty} f_{36}(z-b)^n, f{37}(z) = \sum{n=0}^{\infty} f_{37}(z-b)^n, f{38}(z) = \sum{n=0}^{\infty} f_{38}(z-b)^n, f{39}(z) = \sum{n=0}^{\infty} f_{39}(z-b)^n, f{40}(z) = \sum{n=0}^{\infty} f_{40}(z-b)^n, f{41}(z) = \sum{n=0}^{\infty} f_{41}(z-b)^n, f{42}(z) = \sum{n=0}^{\infty} f_{42}(z-b)^n, f{43}(z) = \sum{n=0}^{\infty} f_{43}(z-b)^n, f{44}(z) = \sum{n=0}^{\infty} f_{44}(z-b)^n, f{45}(z) = \sum{n=0}^{\infty} f_{45}(z-b)^n, f{46}(z) = \sum{n=0}^{\infty} f_{46}(z-b)^n, f{47}(z) = \sum{n=0}^{\infty} f_{47}(z-b)^n, f{48}(z) = \sum{n=0}^{\infty} f_{48}(z-b)^n, f{49}(z) = \sum{n=0}^{\infty} f_{49}(z-b)^n, f{50}(z) = \sum{n=0}^{\infty} f_{50}(z-b)^n, f{51}(z) = \sum{n=0}^{\infty} f_{51}(z-b)^n, f{52}(z) = \sum{n=0}^{\infty} f_{52}(z-b)^n, f{53}(z) = \sum{n=0}^{\infty} f_{53}(z-b)^n, f{54}(z) = \sum{n=0}^{\infty} f_{54}(z-b)^n, f{55}(z) = \sum{n=0}^{\infty} f_{55}(z-b)^n, f{56}(z) = \sum{n=0}^{\infty} f_{56}(z-b)^n, f{57}(z) = \sum{n=0}^{\infty} f_{57}(z-b)^n, f{58}(z) = \sum{n=0}^{\infty} f_{58}(z-b)^n, f{59}(z) = \sum{n=0}^{\infty} f_{59}(z-b)^n, f{60}(z) = \sum{n=0}^{\infty} f_{60}(z-b)^n, f{61}(z) = \sum{n=0}^{\infty} f_{61}(z-b)^n, f{62}(z) = \sum{n=0}^{\infty} f_{62}(z-b)^n, f{63}(z) = \sum{n=0}^{\infty} f_{63}(z-b)^n, f{64}(z) = \sum{n=0}^{\infty} f_{64}(z-b)^n, f{65}(z) = \sum{n=0}^{\infty} f_{65}(z-b)^n, f{66}(z) = \sum{n=0}^{\infty} f_{66}(z-b)^n, f{67}(z) = \sum{n=0}^{\infty} f_{67}(z-b)^n, f{68}(z) = \sum{n=0}^{\infty} f_{68}(z-b)^n, f{69}(z) = \sum{n=0}^{\infty} f_{69}(z-b)^n, f{70}(z) = \sum{n=0}^{\infty} f_{70}(z-b)^n, f{71}(z) = \sum{n=0}^{\infty} f_{71}(z-b)^n, f{72}(z) = \sum{n=0}^{\infty} f_{72}(z-b)^n, f{73}(z) = \sum{n=0}^{\infty} f_{73}(z-b)^n, f{74}(z) = \sum{n=0}^{\infty} f_{74}(z-b)^n, f{75}(z) = \sum{n=0}^{\infty} f_{75}(z-b)^n, f{76}(z) = \sum{n=0}^{\infty} f_{76}(z-b)^n, f{78}(z) = \sum{n=0}^{\infty} f_{78}(z-b)^n, f{79}(z) = \sum{n=0}^{\infty} f_{79}(z-b)^n, f{80}(z) = \sum{n=0}^{\infty} f_{80}(z-b)^n, f{81}(z) = \sum{n=0}^{\infty} f_{81}(z-b)^n, f{82}(z) = \sum{n=0}^{\infty} f_{82}(z-b)^n, f{83}(z) = \sum{n=0}^{\infty} f_{83}(z-b)^n, f{84}(z) = \sum{n=0}^{\infty} f_{84}(z-b)^n, f{85}(z) = \sum{n=0}^{\infty} f_{85}(z-b)^n, f{86}(z) = \sum{n=0}^{\infty} f_{86}(z-b)^n, f{87}(z) = \sum{n=0}^{\infty} f_{87}(z-b)^n, f{88}(z) = \sum{n=0}^{\infty} f_{88}(z-b)^n, f{89}(z) = \sum{n=0}^{\infty} f_{89}(z-b)^n, f{90}(z) = \sum{n=0}^{\infty} f_{90}(z-b)^n, f{91}(z) = \sum{n=0}^{\infty} f_{91}(z-b)^n, f{92}(z) = \sum{n=0}^{\infty} f_{92}(z-b)^n, f{93}(z) = \sum{n=0}^{\infty} f_{93}(z-b)^n, f{94}(z) = \sum{n=0}^{\infty} f_{94}(z-b)^n, f{95}(z) = \sum{n=0}^{\infty} f_{95}(z-b)^n, f{96}(z) = \sum{n=0}^{\infty} f_{96}(z-b)^n, f{97}(z) = \sum{n=0}^{\infty} f_{97}(z-b)^n, f{98}(z) = \sum{n=0}^{\infty} f_{98}(z-b)^n, f{99}(z) = \sum{n=0}^{\infty} f_{99}(z-b)^n, f{100}(z) = \sum{n=0}^{\infty} f_{100}(z-b)^n, f{101}(z) = \sum{n=0}^{\infty} f_{101}(z-b)^n, f{102}(z) = \sum{n=0}^{\infty} f_{102}(z-b)^n, f{103}(z) = \sum{n=0}^{\infty} f_{103}(z-b)^n, f{104}(z) = \sum{n=0}^{\infty} f_{104}(z-b)^n, f{105}(z) = \sum{n=0}^{\infty} f_{105}(z-b)^n, f{106}(z) = \sum{n=0}^{\infty} f_{106}(z-b)^n, f{107}(z) = \sum{n=0}^{\infty} f_{107}(z-b)^n, f{108}(z) = \sum{n=0}^{\infty} f_{108}(z-b)^n, f{109}(z) = \sum{n=0}^{\infty} f_{109}(z-b)^n, f{109}(z) = \sum{n=0}^{\infty} f_{109}(z-b)^n, f{110}(z) = \sum{n=0}^{\infty} f_{110}(z-b)^n, f{111}(z) = \sum{n=0}^{\infty} f_{111}(z-b)^n, f{112}(z) = \sum{n=0}^{\infty} f_{112}(z-b)^n, f{113}(z) = \sum{n=0}^{\infty} f_{113}(z-b)^n, f{114}(z) = \sum{n=0}^{\infty} f_{114}(z-b)^n, f{115}(z) = \sum{n=0}^{\infty} f_{115}(z-b)^n, f{116}(z) = \sum{n=0}^{\infty} f_{116}(z-b)^n, f{117}(z) = \sum{n=0}^{\infty} f_{117}(z-b)^n, f{118}(z) = \sum{n=0}^{\infty} f_{118}(z-b)^n, f{119}(z) = \sum{n=0}^{\infty} f_{119}(z-b)^n, f{120}(z) = \sum{n=0}^{\infty} f_{120}(z-b)^n, f{121}(z) = \sum{n=0}^{\infty} f_{121}(z-b)^n, f{122}(z) = \sum{n=0}^{\infty} f_{122}(z-b)^n, f{123}(z) = \sum{n=0}^{\infty} f_{123}(z-b)^n, f{124}(z) = \sum{n=0}^{\infty} f_{124}(z-b)^n, f{125}(z) = \sum{n=0}^{\infty} f_{125}(z-b)^n, f{126}(z) = \sum{n=0}^{\infty} f_{126}(z-b)^n, f{127}(z) = \sum{n=0}^{\infty} f_{127}(z-b)^n, f{128}(z) = \sum{n=0}^{\infty} f_{128}(z-b)^n, f{129}(z) = \sum{n=0}^{\infty} f_{129}(z-b)^n, f{130}(z) = \sum{n=0}^{\infty} f_{130}(z-b)^n, f{131}(z) = \sum{n=0}^{\infty} f_{131}(z-b)^n, f{132}(z) = \sum{n=0}^{\infty} f_{132}(z-b)^n, f{133}(z) = \sum{n=0}^{\infty} f_{133}(z-b)^n, f{134}(z) = \sum{n=0}^{\infty} f_{134}(z-b)^n, f{135}(z) = \sum{n=0}^{\infty} f_{135}(z-b)^n, f{136}(z) = \sum{n=0}^{\infty} f_{136}(z-b)^n, f{137}(z) = \sum{n=0}^{\infty} f_{137}(z-b)^n, f{138}(z) = \sum{n=0}^{\infty} f_{138}(z-b)^n, f{139}(z) = \sum{n=0}^{\infty} f_{139}(z-b)^n, f{140}(z) = \sum{n=0}^{\infty} f_{140}(z-b)^n, f{141}(z) = \sum{n=0}^{\infty} f_{141}(z-b)^n, f{143}(z) = \sum{n=0}^{\infty} f_{143}(z-b)^n, f{144}(z) = \sum{n=0}^{\infty} f_{144}(z-b)^n, f{145}(z) = \sum{n=0}^{\infty} f_{145}(z-b)^n, f{146}(z) = \sum{n=0}^{\infty} f_{146}(z-b)^n, f{147}(z) = \sum{n=0}^{\infty} f_{147}(z-b)^n, f{148}(z) = \sum{n=0}^{\infty} f_{148}(z-b)^n, f{149}(z) = \sum{n=0}^{\infty} f_{149}(z-b)^n, f{150}(z) = \sum{n=0}^{\infty} f_{150}(z-b)^n, f{151}(z) = \sum{n=0}^{\infty} f_{151}(z-b)^n, f{152}(z) = \sum{n=0}^{\infty} f_{152}(z-b)^n, f{153}(z) = \sum{n=0}^{\infty} f_{153}(z-b)^n, f{154}(z) = \sum{n=0}^{\infty} f_{154}(z-b)^n, f{155}(z) = \sum{n=0}^{\infty} f_{155}(z-b)^n, f{156}(z) = \sum{n=0}^{\infty} f_{156}(z-b)^n, f{157}(z) = \sum{n=0}^{\infty} f_{157}(z-b)^n, f{158}(z) = \sum{n=0}^{\infty} f_{158}(z-b)^n, f{159}(z) = \sum{n=0}^{\infty} f_{159}(z-b)^n, f{160}(z) = \sum{n=0}^{\infty} f_{160}(z-b)^n, f{161}(z) = \sum{n=0}^{\infty} f_{161}(z-b)^n, f{162}(z) = \sum{n=0}^{\infty} f_{162}(z-b)^n, f{163}(z) = \sum{n=0}^{\infty} f_{163}(z-b)^n, f{164}(z) = \sum{n=0}^{\infty} f_{164}(z-b)^n, f{165}(z) = \sum{n=0}^{\infty} f_{165}(z-b)^n, f{166}(z) = \sum{n=0}^{\infty} f_{166}(z-b)^n, f{167}(z) = \sum{n=0}^{\infty} f_{167}(z-b)^n, f{168}(z) = \sum{n=0}^{\infty} f_{168}(z-b)^n, f{169}(z) = \sum{n=0}^{\infty} f_{169}(z-b)^n, f{170}(z) = \sum{n=0}^{\infty} f_{170}(z-b)^n, f{171}(z) = \sum{n=0}^{\infty} f_{171}(z-b)^n, f{172}(z) = \sum{n=0}^{\infty} f_{172}(z-b)^n, f{173}(z) = \sum{
IMG_2631
Def. by def. \exists a hero function continuation (E, g) \sim (D, f) with C \in E so for z \in D \cap E we have f(z) = g(z) \rightarrow g(c) \in E as z \in C.
Plunk: this means essential singularities are a/way singlar
Richman
Richman surfaces: Last time we def hero continuation region D : D \rightarrow \mathbb{C} region D_1 : f_1: D_1 \rightarrow \mathbb{C}, if f = f_1 on D \cap D_1, we can its a hero continuation def: hero continuation constructing (D_1, f_1) \sim (D_2, f_2) —- go in such a way that a single value hero function V_i is obtained in the new textbook, a singular point is the obstacle of hero continuation and regular is point not singular (which can construct hero continuation eg. a pole is regular).
With the def lemma: let (D, f) be a function of \mathbb{C} with C \in D of \mathbb{C} regular then f(z) approach a limit in Remain sphere a z \rightarrow z \in C pt: by def \exists a hero continuation (E, g) \sim (D, f) with C \in E so for z \in D \cap E f = g f(z) = g(z) \rightarrow g(c) \in E as z \rightarrow c
Least: the argument we use to have help \Rightarrow analytic (candy then), we know the radius of convergence of the power series is given by the distance from center to nearest won removable singularities.
Thus: If a power series f(z) = \sum_{n=0}^{\infty} a_n (z-a)^n has a radius of convergence \rho = 0, then there is at least one point c on cycle where analytic continuation is propossible
pr. Assume there is a analytic continuation pass every point on boundary of disk radius \rho contain on a. use this to construct an analytic continuation, then contradiction
A path \Gamma: I \rightarrow \Sigma is close interval, \Gamma continuous. Image of f is compact and connected f is a path from a to b. if f(a) = i: [0, 1] \rightarrow \mathbb{C}, f(b) = a f(1) = b \Gamma is cloth path (loop) if \Gamma(0) = \Gamma(1). \Gamma is simple if f(1) = f(0)
f = f(s_1) s_1 \in S_2 \Rightarrow s_1 = 0, s_2 = 0
def: Hero continuation along path of (D, f) along path if is fixed sequence of direct hero continuation (D, f_1) \sim (D_2, f_2) \sim \dots \sim (D_n, f_n) s.t. (i) each region is open disk in \Sigma, and D_1 \subset D_2 \subset \dots \subset D_n a sub division of \Sigma. (sn){n=1}^{\infty} \subseteq D_n.
6. Riemann 曲面、分支点与覆盖
IMG_2632
If all the function elements are analytic, it’s call one continuation.
eg. let f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 want to continue log z and along f. let U_1 = D_1(1), U_2 = D_2(-i), U_3 = D_3(-i), U_4 = D_4(-i), so we have s_1 = 0, s_2 = \frac{1}{4}, s_3 = \frac{1}{2}, s_4 = \frac{1}{2} subdivision
recall the def of homotopy path to and f from a to b in a topological space X (a retspace have ratio of open set and close set) are homotopic if there exists a continues map f: I^2 \rightarrow X call homotopy) s.t. f(s, 0) = f(0, 1) f(s, t) = f(s, 0) f(0, t) = a, f(1, t) = b so homotopy is a equivalence relation on path from a to b. Equivalence class call homotopy classes. this let X \subseteq \Sigma, a, b \in X and let T be a homotopy in X between f and f’ from a to b if (D_1 f) is a function direct that can be nero continued as along each path f: S \rightarrow f(Sf) (Sf) \cap (I^2 \cap S, f(-I)). the the nero function at b resulting from the continuation along to and to are identically equal in neighbourhood of b.
pf: for t, t’ \in I write t \sim t’ if nero continuation of (D_1 f) along f and f’ give function are equal on or neighbourhood of b. Deck \sim is an equivalence relation on I. Enough to show equivalence class is open, because from the complement of an equivalence class is a union of copies equivalence class is open. Since E is concated it can’t be covaried by disjoint open sets to prove openness; suffices to show that for any t^, t’^ \in (0, 1) I \subseteq \mathbb{R}^2, s.t. every t \in (t^ - \varepsilon), t^ + \varepsilon is in the same equivalence class. see the picture they pass same region and intersection for sufficiently small \varepsilon i.e. \forall s \in (t^ - \varepsilon), t^ + \varepsilon (D_1 f, f’, s) is the same function el.
see the picture they pass same region and intersection for sufficiently small \varepsilon i.e. \forall s \in (t^ - \varepsilon), t^ + \varepsilon if it’s path connected and all realized paths are homotopic to the existent path to constant path f (f(t) > a take t \in I). cal (Monodramy) let E be simply connected region in E and D_1 f be a function of t with DC \cap DS \cap E if (D_1 f) can be continued continued nero along all paths in E straighting across the there is a direct nero continuation (E, g) of (D_1 f).
Ex: Suppose (D_1 f) is given by a branch of log z on the disk of radius r connected on l. let E be (C \setminus {-Ve}) real disk anisotropic. then monodramy
IMG_2633
gives a branch of \log z
def: fundamental group: if t_1 and t_2 are the paths connected together topological space X with f_n(0) = t_1(0) \dots t_n(1) then the product t_1 t_2 \dots t_n f_n(0) is a dense space. check the product of paths can be extended to equivalent dense spaces: f(1-s) equivalent containing f(t_1 \dots t_n) = f(t_1 \dots t_n)
bl = [f_n] is a contact path. Fundamental group \pi_1(X, \alpha) = {\text{honology}}
classes of double paths f(0) = f(1) = 0 with multiplication \pi_1(X, \alpha) call fundamental group of X. when X is path connected \pi_1(X, \alpha) \cong \pi_1(X, b) \forall b \in D
def: concorring map X, Y Euplerical space A continuous map f: X \rightarrow Y for which every y \in Y has a neighborhood V s.t. f^{-1}(V) consist of disjoint open ret napped horizontal contours by f Hoge X is the concorring space of Y.
eg. [P^2] is the concorring space of an infinite cylinder f(x, y) = in \text{modl}(y).
def: branch concorring this is a way of deducing discrete ret of points b_i in Y called branch point. The point b_i is a branch point of order k. if b_i has a neighborhood V s.t. f|_{f^{-1}(V)} is given by z^{k+1} in appropriate direction of coordinate.
Finding Riemann surface (expl for multivalue function (x, 0)). eg. the Riemann surface of \log z. \pi_1(C \setminus {0}) \cong \mathbb{Z} using winding humber (i.e. winding numbers can decide branch).
The Riemann surface is the space concorring space of C \setminus {0} by glue tappers. Infinitely many copies of C \setminus {-Ve \text{ real axis}} is a cover of C \setminus {0}.
Riemann surface S, \psi concorring map. def s_0 = \phi^{-1}(0), sn = \phi^{-1}(s{n-1}) for s \in C \setminus {0} contains \psi^{-1}(z) is containable set of maps for z \in C \setminus {0}.
\psi solves \phi for z \in C \setminus {0} contains \psi^{-1}(z) is containable set of maps for z \in C \setminus {0}. This contains set to the different values of \log z in C \setminus {0, 0}. Riemann surface of z is similarly. f_n(s_0) = \phi^{-1}(0) so f_n(z) = \phi^{-1}(z) for z \in C \setminus {0, 0}.
S \setminus {P1, \dots, P_n} = F{n+1} similarly, we have C \setminus {P_1, \dots, P_n} is a branch point of order j if b_j has a neighborhood V on which f_j(f_1(x) \dots f_j(y)) is given by z^{j+1} \mapsto z for an appropriate set of coordinates.
A branch cover is a concorring space among from a discrete set of points {b_1, \dots, b_k} in Y (there are branch points). The point b_j is a branch point of order j if b_j has a neighborhood V on which f_j(f_1(x) \dots f_j(y)) is given by z^{j+1} \mapsto z for an appropriate set of coordinates.
IMG_2634
Ex: Riemann Riemann sphere of \log z T_1(\mathbb{C} \setminus {0}) \cong \mathbb{Z}. \psi is a covariant map for z \in \mathbb{C} \setminus {0}. \psi^*(z) covar infinitely many point (a contable set) to the \phi map this set to the \infty contable value of \log z.
Ex: the Riemann surface of z\bar{z}, q \in \mathbb{N}. def. s_0 = \phi^{-1}(0), s_{iv} = \phi^{-1}(\infty). Riemann surface of z\bar{z} is TV{s0, s{iv}}, is this a branch covar with branch points of order q-1 at 0 and \infty. eg. q > 1. the \phi map z \mapsto z \cdot z^{-1} is a branch covar with branch points of order q-1 at 0 and \infty.
Riemann surfaces of \overline{p(z)} p(z) is a poly.
eg 2. p(z) = z^{-a}. branch point: a and \infty. of order 1.
eg 2. p(z) = (z-a)(z-b). a & b. branch point. a and b. order (not \infty, because a loop contain a and b. cancel out wind number, thus can def an a single value function on it). then monodromy then \Rightarrow 2 nero branch of \overline{p(z)}.
now we obtained a 2-sheeted branch covar, with branch points of order 1 above a and b.
eg 3. p(z) = (z-a)^2. we can have two branch \overline{p(z)} and \overline{p(z-a)^2}. there are no branch point so there is a double sheet covar but not a branch covar, the Riemann surface is 2 disjoint copies of \Sigma. eg 4. p(z) = z \cdot (z-a)(z-b)(z-c). a.b.c all distinct. Branch point a.b.c or branch cut like \begin{pmatrix} a & \infty \ c & \infty \ b & c \end{pmatrix} is not simply connected so we can’t connect the branch cut like \begin{pmatrix} a & \infty \ c & \infty \ b & c \end{pmatrix}. but there are 2 nonordromy thin branch points and we do direct nero integration into the function. If we can still def generators of \pi(\mathbb{C} \setminus {L_1, L_2}) and we do direct nero integration into both generators and get back the to the function. If we can still def a branch of \overline{p(z)} on \Sigma \setminus {L_1, L_2}. the Riemann surface is a loop around branch point b like this.
the Riemann surface is torus, s \rightarrow z is a 2-sheeted branch covar. eg 5. p(z) = z^{-1}(z-a). the Riemann surface is a branch disjoint such have no branch covar, the Riemann surface is a branch disjoint such surface.
IMG_2635
Suppose we have at least l a_j is odd, with p = q \sqrt{k}, r with all multiplicity l in q is single value, the Riemann surface only rely on r. r = \frac{\pi}{j} (z - a_j). if m is even the branch point a_1, \dots, a_m is odd. branch point a_1, \dots, a_m, \infty, and branch point a_1, \dots, a_m, \infty, and branch point a_1, \dots, a_m, \infty are pair wide disjoint. L_1, \dots, L_k are disjoint branch cuts are k = { \frac{\pi}{j} m \text{ in odd} \text{ or } \frac{\pi}{j} m \text{ in even} \text{ or odd} }. so Max continuation possible along any path is \Sigma = { L_1, \dots, L_k }.
\Sigma is a two sheet cover of \Sigma, with branch point of order 1 and {a_1, \dots, a_m} or {a_1, \dots, a_m, \infty}
when h=5 or, we get Remann surface of genus 2. (上有 有 有 有 有) genus of se surface = \lfloor \frac{m-1}{2} \rfloor easy to check.
Riemann surface are man be not compact. eg. to (a, \infty). \rightarrow Non orientable surfaces cannot be realised as Riemann surface eg. Michius bands, Klonik klein bottle.
The monodrony group : suppose f is a multivalue function, with Riemann surface is a branch cover of \Sigma with old sheets, and branch points a_1, \dots, a_m let T_i be fundamental gp of \Sigma \setminus {a_1, \dots, a_m}. T_1, \dots, T_m are elliptic determiners a permutation of the sheets of the cover, in other words, it determines an element of S_k so there exist homo h: T_i \rightarrow S_k the image of this, homo is the monodrony group k is number of sheets so T_i \setminus {0, \infty} = \mathbb{Z} (~~k~~ T_i S_k) widening number.
T_i \setminus {0, \infty} = \mathbb{Z} (~~k~~ T_i S_k) widening number.
to find figure out monodrony gp. want to figure out the image of R around 0 f = h(r) = (1, 2). h(r^2) = (1, 2) h(r^3) = I
eg. f = \sqrt{(z-1)(z-2)(z-3)(z-4)} \cong F_3 = \text{i.e. Riemann surface like } h(r) = (1, 1)
these 3 loop are generators. h(r) = (1, 1), h(r^2) = (1, 2), h(r^3) = I
eg. f(z) = \sqrt{1+z^2} branch point 0, 1, \infty 4 branches choose a point on surface I. f_1(z) = \sqrt{1+z}, f_2(z) = \sqrt{1+z}, f_3(z) = \sqrt{1+z}, f_4(z) = -\sqrt{1+z}
go
(on the left)
def:
if what is AVD equivalent def:
def:
def:
for an extension of the metric Note to chart iff C_4 of def: lolo if \neq \emptyset doesn’t then surface
IMG_2636
go around 0, I then with the orientation r, \delta, so curve go around \phi is (\delta r)^{-1}, we only need to determine image of r and r\delta h(r) = (1/2) |3\varphi| the covering map has 2 branch points above 0, each of order 1, f_1, f_2 go around 0, but f_1, f_2 don’t so h(\delta) = (2/4). h(\delta^{-1} \delta) = h(r)^{-1} h(\delta), h(\delta \delta)^{-1} = h(\delta^{-1}) h(r)^{-1}
def: A surface S is a handdott topological space s.t. at every point SGS there exist an open set U containing S homeomorphic to an open subset of \mathbb{C}. (U \cup U_1 \cup U_2 \subseteq S, \exists V_1, V_2 \subseteq S \text{ open, } s.t. U_1 \in V_1, U_2 \subseteq V_2 \text{ and } V_1 \cap V_2 = \emptyset).
def: an atlas A is a set of pairs {(U_i, \phi_i)} U_i open, \phi_i: U_i \rightarrow \mathbb{C} s.t. ① {U_i} cover S. ② \phi_i: U_i \rightarrow \mathbb{C} is a homeomorphism. (continuous and has continuous inverse) Each pair (U_i, \phi_i) is a chart the coordinate transition function \phi_i \circ \phi_i^{-1}: \phi_i(U_i \cap U_i) \rightarrow \phi_i(U_i \cap U_i)
Atlas A is analytic if all coordinate transition functions are analytic def: Two analytic atlas A = {(U_i, \phi_i)} and B = {(V_i, \psi_i)} are compatible if whenever U_i \cap V_j \neq \emptyset then \phi_i \circ \psi_j: \psi_j(U_i \cap V_j) \rightarrow \phi_i(U_i \cap V_j) \Rightarrow. A \cup B is an analytic atlas easy to check: compatibility is an equivalence reflection
def: A complex structure is an equivalence class of analytic atlas.
def: An abstract Riemann surface is a surface with a complex structure
def: suppose S is a Riemann surface. A function f: S \rightarrow \mathbb{C} is analytic for every chart (U, \phi) on S, the function f \circ \phi^{-1}: \phi(U) \rightarrow \mathbb{C} is analytic.
the meromorphic function is almost the same Note that if a point p is in the intersection of the domains of chart (U, \phi) and (V, \psi) f \circ \phi^{-1} on \phi(U \cap V) is an analytic function iff f \circ \psi^{-1} on \psi(U \cap V) is an analytic function because f \circ \phi^{-1} = f \circ \psi^{-1}.
iff (\psi \circ \phi^{-1}) \circ f: S_1 \rightarrow S_2 is a Riemann surface. A continuous function f: S_1 \rightarrow S_2 is analytic if f is a Riemann surface on S_1 and S_2 with U_1 \cap f(U_2) \neq \emptyset def: let S_1 and S_2 be Riemann surfaces. A continuous function f: S_1 \rightarrow S_2 is analytic if whenever (U_1, \phi_1) and (U_2, \phi_2) and chart on S_1 and S_2 with U_1 \cap f(U_2) \neq \emptyset holo if whenever (U_1, \phi_1) and (U_2, \phi_2) and chart on S_1 and S_2 is analytic if the function \phi_2 \circ \phi_1^{-1}: \phi_1(U_1 \cap f(U_2)) \rightarrow \phi_2(U_1 \cap f(U_2)) is analytic again, the holo if whenever (U_1, \phi_1) and (U_2, \phi_2) and chart on S_1 and S_2 is analytic if the function \phi_2 \circ \phi_1^{-1}: \phi_1(U_1 \cap f(U_2)) \rightarrow \phi_2(U_1 \cap f(U_2)) is analytic again, then if f: S_1 \rightarrow S_2 is analytic if f doesn’t depend on what chart we use since the holo homomorphism between Riemann surfaces is then get: S_1 \rightarrow S_2 is holo/mero easy to show by exchange diagram
IMG_2637
if f and g are nero functions on S, for s \in S, we can def f_s(z) = f(s) \pm g(s) first f_g(z) = f(s) \cdot g(s), f/g(s) = f(s)/g(s) when g(s) \neq 0 all this function are nero, the nero function form a field but do function is just a ring.
The sheaf of germs of nero functions. Let T be (unbranched) fienan surface of function w = f(z) does not exist branch point. covering map \psi: T \rightarrow \Sigma set T with \psi(s) \in C has an open neighborhood wrapped by \psi homog onto an open ret subset of C, take the chart to be (U, \psi|_U) unless it \psi(s) = \infty then take the chart to be (U, J_0 \psi|_U), J(z) = \frac{1}{z}. for a point s_1 wrapped by \psi to a branch point C of order s-1, we map neighborhood of s_1 to neither heightor of C if could be |z-C|/q. since all transition raps all holo.
idea: A branched cover of a fienanan surface is a fienanan surface choose a \in \Sigma, f_a := the set of functions nero on so open set contain a for f_a \in F_a, f_a is an open set contain a, is an equivalent relation (easy to deck).
def: the equivalent relation of f in F_a is note by [f]_a called the germ of f at a so we have [f]_a = [g]_b a > b and f=g on a neighborhood of a
def: sheaf of germs of nero function M, M is the set of germs [f]_a for all a \in \Sigma
Germs close to [f]_a (f is nero on so disk D \subset M) are the germs [f]_a for b close to a
A surface we want to def topology on M:
def: D = {z \in C \mid |z-a| < \epsilon} for a \in C D = {z \in C \mid |z-a| < \epsilon} if a = 0
D_a := set of disk centered at a, ordered by inclusion
if m = [f]_a, f is nero on a disk D \subset D_a write this as m = [D, f]_a
where m determines f uniquely on D
def: D-neighborhood D_m of m in M is D(m) := {[f]_b \mid b \in D} A \subset M is open if for each m \in A, \exists a \in D-neighborhood D_m in SA.
inn: this gives a topology on M. (i.e. we have a notion of open set s.t. D_m is open)
IMG_2638
open union of open sets \cup are open. finite intersections of open sets are open.
if: M, P is open (obvious). Union of open sets are open (obvious), finite intersection. suppose m \in A \cap B. m = [D, f]_a, D \in D_a and D(m) \subseteq A, m = [f, y]_a equal and nero so f \in D or D \subseteq [f, y]_a so \exists f \in D on which f and g are non-nero so f \in D.
we want to show M has the structure of a Riemann surface for m = [f, y]_a \in M, def \psi(m) = a. then \phi(m) = f(a).
(entria: \psi is continuous.)
pf: this we want to show the pre-image of open set is open. Let U \subseteq S be an open set containing a \exists D \in D_a \subset U, s.t. m = [D, f]_a, A element of D(m) is of the form [f, y]_a for b \in P_1 so \psi([f, y]_a) = b \in D_a \subset U. since D(m) \subset \psi^{-1}(U) then and so \psi^{-1}(m) is open.
(entria: M is handoff.)
pf: let m = [f, y]_a and n = [g, h]_b are disjoint el.t. since m \neq n, either a \in U, or b \in V. s.t. f \in g. If a \in U, choose a open subset S of S, a \in U, b \in V, then \psi^{-1}(U) and \psi^{-1}(V) are open and disjoint in M. f(a) = b, \exists D \in D_a if and g are nero and f \in g on V. so D(m) and D(n) are open neighborhoods of m and n on M. if not fig on so D(m) \cap D(n) = \emptyset.
is D(m) \cap D(n) \neq \emptyset so
(entria: M is Riemann surface)
pf: since m \in M is handoff, remains to construct charts. \psi(D(m)) = D(m) \rightarrow D \subseteq S (open) def charts X(D(m)), \psi(D(m)) if \psi(m) \neq \emptyset or (D(m), T) \psi(D(m)) if \psi(m) = \emptyset the transition function are either identity or J (restriction).
they are analytic covariant map \phi D(m) \subset M \xrightarrow{\phi} S \psi(D(m)) \subset M \xrightarrow{\phi \circ \psi^{-1}(\psi(m)) = f} S \xrightarrow{f} T
let \Gamma: I \rightarrow S be a transition function with initial point \psi(m), m = [f, y]_a and do nero continuation of f along \Gamma, at a point \Gamma(t), this continuation is \tilde{\Gamma}(t) \in M which \psi(\tilde{\Gamma}(t)) = \Gamma(t). note: \tilde{\Gamma}(t) only depends on m, \Gamma, T.
(entria: \tilde{\Gamma}: I \rightarrow M is continuous)
pf: need to show pre-image of open sets under \tilde{\Gamma} to be open. Let A \subset M be an open subset where m = \Gamma(t_0) \in A. D(m) \subset A for some D \subset D_a for sufficiently large D_a, where m = \Gamma(t_0).
IMG_2639
small t, f(t) \in D, so \tilde{f}(t) \in D(m), t \in f^{-1}(A) so f is a path conversely any path \tilde{f} in M starting at m \in M project to a path f = \psi \circ \tilde{f}(t) in \Sigma starting at \psi(h). Macever, the years f(H) give a nero continuation of m along f(t). We get equivalence relations on M, much if \exists a path \tilde{f} for m in M to h equivalent object position M into distinct path connected pieces. Each equivalent class is open (by continuation) and is closed since component is other equivalence classes’ union. A component (connected) of topological space X is the set that A connected component M(n) containing m is called the unbrached Riemann surface of m. Add in a brach point to get a larger space f whose connected components are brached Riemann surfaces M \subset X f. This is identical to the way we given back on brach brach point in example. Harder to prove X f is handoff and that the brach point are a discrete set in f.
\Sigma z^n g rational to get \frac{1}{g} GN, if \frac{1}{g} = \frac{m}{n}. brach points 0, \infty brach outs (i_0) the \Sigma-brach outs is simply connected, has n sheets. nonodromy group is the cyclic dgp of order n. so (i_0) \rightarrow (1) the Riemann surface is the n-sheeted cover with brach point of order m-1 over \infty and 0. Need to explain how many sheets the cover has where are the brach points and what is the order and sometimes describe nonodromy gp. g irrational brach point is 0 and \infty order is \infty that take \infty copies of (i_0) nonodromy group is Z. Riemann surface is the infinite sheafed brach over 0, \infty, with logarithmic brach points 0 and \infty. \Sigma g unreal. \Sigma z^n z^n = z^n. z^n a could be rational/irrational follow above let b rational b = \frac{m}{n}. z^{\frac{m}{n}} = e^{\frac{m}{n} \log z} logarithm brach point at 0 and \infty.
