03. Cauchy's Formula and it's Applications

3.1 Cauchy’s Integral Formula

10 - Cauchy's Integral Formula

Cauchy's Integral Formula

Suppose that $f:U\to \mathbb{C}$ is a holomorphic function on an open set $U$
Let $\gamma$ be a simply positively oriented closed curve such that ${\gamma }^{\ast }$ and interior of $\gamma$ are inside $U$

Let $w\in U$

For all $w$ inside of $\gamma$ then

$$f(w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$$

Proof

Fix $w$ inside $\gamma$

Then there exists $r_0$ such that

$$B(w,r_0)\cap {\gamma }^{\ast }=\varnothing$$

So the disc $B(w,r_0)$ is inside $\gamma$

For any $0<r<r_0$ then
Let ${\gamma }_r$ be the positively oriented circle of radius $r$ around $w$

By Homotopy between interior closed paths and interior circle then

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz=\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz$$

As

$$\begin{array}{rl}\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz& =\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)-f(w)}{z-w}dz+\frac{f(w)}{2\pi i}{\int }_{{\gamma }_r}\frac{1}{z-w}dz\\ & =\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)-f(w)}{z-w}+f(w)\end{array}$$

As $f$ is complex differentiable at $z=w$ then

$$\frac{f(z)-f(w)}{z-w}\text{ is bounded as }r\to 0$$

By estimation lemma then the integral over ${\gamma }_r$ tends to $0$

Thus as $r\to 0$ then the path integral around ${\gamma }_r$ tends to $f(w)$

Since $\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$ is independent of $r$ then it must be equal to $f(w)$

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Cauchy Formula for Multiple Curves corollary

Let $U$ be a bounded domain with piecewise $C^1$ boundary with finitely many components
Let $f$ be a function holomorphic in the closure of $U$

Parametrise each boundary component of $U$ by contour ${\gamma }_i$ such that

$$i{\gamma }_i^{\prime }(t)\text{ is an inward normal}$$

Hence the outer boundary is positively oriented (counter clockwise)
So the inner components are negatively oriented (clockwise)

Denoting

$${\int }_{\partial U}=\sum {\int }_{{\gamma }_i}$$

then

$${\int }_{\partial U}f(z)dz=0$$

and

$$\frac{1}{2\pi i}{\int }_{\partial U}\frac{f(z)}{z-w}dz=f(w),w\in U$$

Proof

Idea of proof is to add a few cuts that avoid $w$ and connect inner boundary components to each other and to the outer boundary in such a way that $U$ without those extra curves is simply connected

Let $\gamma$ be the boundary of the domain
Apply Cauchy’s Integral Formula to $\gamma$ so

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}=f(w)$$

As curve $\gamma$ is made up of boundary components ${\gamma }_i$ in the right orientation and cuts ${\eta }_i$
where ${\eta }_i$ appear twice with different orientations so the integral cancels out

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3.2 Homotopy Version of Cauchy’s Theorem (Non-Examinable)

3.2.1 Cycles

Cycle

Cycle $\Gamma$ is a finite formal sum of closed paths so

$$\Gamma =\sum _{i=1}^k{m}_i{\gamma }_i$$

where $m_i\in \mathbb{Z}\setminus \{0\}$ and ${\gamma }_i$ is a closed path for each $i\in \{1,\cdots ,k\}$

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Support of a Cycle

Let $\Gamma$ be a cycle

Suppose of $\Gamma$ is defined as

$${\Gamma }^{\ast }:=\bigcup _{i=1}^k{\gamma }_i^{\ast }$$

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Integral of a Cycle

Let $\Gamma$ be a cycle then

For any piecewise $C^1$ function $f$ defined on ${\Gamma }^{\ast }$ the integral over $\Gamma$ is defined as

$${\int }_{\Gamma }f(z)dz=\sum _{i=1}^k{m}_i{\int }_{{\gamma }_i}f(z)dz$$

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Winding Number of a Cycle

Let $\Gamma$ be a cycle then

$$I(\Gamma ,z_0)=\sum _{i=1}^N{m}_{i}I({\gamma }_i,z_0)$$

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Inside of a Cycle

Let $\Gamma$ be a cycle then

$$\mathrm{ins}(\Gamma ):=\{z\in C:z\notin {\Gamma }^{\ast }\text{ and }I(\Gamma ,z)\ne 0\}$$

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Length of a Cycle

Let $\Gamma$ be a cycle then

$$\ell (\Gamma ):=\sum _{i=1}^k|m_i|\ell ({\gamma }_i)$$

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Removal of a Point in a Cycle lemma

Let $\Gamma$ be a cycle in a domain $U$

Suppose $z_0\in {\Gamma }^{\ast }$ then
There exists cycle $\Sigma$ such that $z_0\notin {\Sigma }^{\ast }$ and

$${\int }_{\Sigma }f={\int }_{\Gamma }f$$

for every holomorphic function $f:U\to C$

Proof

Use induction on $k=k(\Gamma )$ where $k$ is the smallest integer such that $\Gamma$ can be in form

$$\sum _{i=1}^k{m}_i{\gamma }_i$$

Assuming that for all cycles ${\Gamma }^{\prime }$ with $k({\Gamma }^{\prime })<k(\Gamma )$ then

If $\Gamma ={\sum }_{i=1}^k{m}_i{\gamma }_i$ then let ${\Gamma }^{\prime }={\sum }_{i=1}^{k-1}{m}_i{\gamma }_i$
Let ${\Sigma }_1$ be a cycle such that

$$z_0\notin {\Sigma }_1^{\ast }\text{ and }{\int }_{{\Sigma }_1}fdz={\int }_{{\Gamma }_1}fdz$$

for all holomorphic functions $f$ defined on $U$

Let $\gamma ={\gamma }_k$
If $\gamma (t)=z_0$ for all $t$ then let $\Sigma ={\Sigma }^{\prime }$
Otherwise there exists $w\in {\gamma }^{\ast }$ with $w\ne z_0$ so shift parametrisation to get $\gamma (0)=w$

Let $r>0$ such that $B(z_0,r)\subseteq U$ and $|w-z_0|>r$
Since $[0,1]$ is compact then $\gamma$ is uniformly continuous

Hence there exists $N>0$ such that if $|s-t|<\frac{1}{N}$ then

$$\gamma (s)-\gamma (t)<r$$

Hence if image $\gamma \left(\left[\frac{k}{N},\frac{k+1}{N}\right]\right)$ of interval $\left[\frac{k}{N},\frac{k+1}{N}\right]$ contains $z_0$ then

$$\gamma \left(\left[\frac{k}{N},\frac{k+1}{N}\right]\right)\subseteq B(z_0,r)$$

Let $0=x_0<x_1<\cdots <x_m=1$ be the partition of $[0,1]$
By taking $x_{i}s$ to be $\left\{\frac{k}{N}:\gamma \left(\frac{k}{N}\right)\ne z_0,0\le k\le N\right\}$

Suppose $\gamma (t)=z_0$ for some $t\in (x_i,x_{i+1})$ then

$$\gamma {\mid }_{[x_i,x_{x+1}]}\text{ lies in }B(x_0,r)$$

So by Path Deformation Away from a Point
with $a=x_i$ and $b=x_{i+1}$ then replace it with path in $B(x_0,r)\setminus \{x_0\}$

As alteration to the path on each interval gives a path $\gamma$ which doesn’t pass through $z_0$
Then by Cauchy’s Theorem for Simply Connected Domains for the disk then

$${\int }_{\gamma }fdz={\int }_{\sigma }fdz\text{ for all holomorphic functions }f$$

Hence define

$$\Sigma :={\Sigma }_1+m_k\sigma$$

gives the required cycle

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Path Deformation Away from a Point lemma

If $\gamma :[a,b]\to \mathbb{C}$ be a piece-wise $C^1$ path
where image ${\gamma }^{\ast }$ is contained in open ball $B(z_0,r)$ and $\gamma (a)\ne z_0\ne \gamma (b)$

Then

There exists path ${\gamma }_1:[a,b]\to B(z_0,r)$ with

$${\gamma }_1(a)=\gamma (a)\text{ and }{\gamma }_1(b)=\gamma (b)\text{ and }{\gamma }_1(t)\ne z_0\text{ for all}t\in (a,b)$$

Proof

By translation and scaling assume that $z_0=0$ and $r=1$

If necessary replace $\gamma$ with ${\gamma }^{-}$ (Opposite path) so that $0<|\gamma (a)|\le |\gamma (b)|$

If $|\gamma (t)|\ge |\gamma (a)|$ for all $t\in [0,1]$ then let ${\gamma }_1=\gamma$

Otherwise there exists $s_0\in [a,b]$ with $|\gamma (s_0)|<|\gamma (a)|$ so that for

$$L=\{t\in [a,b]:|\gamma (s)|\ge |\gamma (a)|\forall s\in [a,t]\}\text{ and }U=\{t\in [a,b]:|\gamma (s)|\ge \gamma (a)\forall s\in [t,b]\}$$

Then

$$a\in L\subseteq [a,s_0)\text{ and }b\in U\subseteq (s_0,b]$$

Hence

$$a\le t_0:=\sup (L)<t_1:=\inf (U)\le b$$

As $t\mapsto |\gamma (t)|$ is continuous then

$$|\gamma (t_0)|=|\gamma (a)|=|\gamma (t_1)|$$

If $\alpha ,\beta \in [-\pi ,\pi )$ given by

$$\gamma (t_0)=|\gamma (a)|\cdot e^{i\alpha }\text{ and }\gamma (t_1)=|\gamma (a)|\cdot e^{i\beta }$$

Let $g:[t_0,t_1]\to \mathbb{R}$ and ${\gamma }_1:[0,1]\to \mathbb{C}$ then

$$\gamma (t)=\frac{t_1-t}{t_1-t_0}\alpha +\frac{t-t_0}{t_1-t_0}\beta ,{\gamma }_1(t)=\{\begin{array}{ll}\gamma (t)& t\in [a,t_0]\cup [t_1,b]\\ |\gamma (a)|\exp (ig(t))& t_0\le t\le t_1\end{array}$$

where ${\gamma }_1$ is continuous and $|\gamma (1)|\ge d$ for all $t\in [a,b]$

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3.2.2 The Homology Form of Cauchy’s Theorem

11 - Homology Form of Cauchy's Theorem

Homology Form of Cauchy's Theorem

Let $f:U\to \mathbb{C}$ be a holomorphic function
Let $\Gamma$ be a cycle in $U$ whose inside lies entirely in $U$ so $I(\Gamma ,z)=0$ for all $z\notin U$

Then Cauchy’s Theorem and Integral Formula hold so
1)

$${\int }_{\Gamma }f(\zeta )d\zeta =0$$

$$\frac{1}{2\pi i}{\int }_{f_{\Gamma }}\frac{f(\zeta )}{\zeta -z}d\zeta =I(\Gamma ,z)f(z)\text{ for all }z\in U\setminus {\Gamma }^{\ast }$$

Proof

Let

$$\begin{array}{rl}G(z)& :=\frac{1}{2\pi i}{\int }_{\Gamma }\frac{f(\zeta )}{\zeta -z}d\zeta \text{ for }z\in C\setminus {\Gamma }^{\ast }\\ F(z)& :=G(z)-f(z)I(\Gamma ,z)\end{array}$$

For fixed $z\in U$ define $g_z:U\setminus \{z\}\to \mathbb{C}$ by

$$g_z(\zeta ):=\frac{f(\zeta )-f(z)}{\zeta -z}$$

By Cauchy integral formula for the winding number then

$$\begin{array}{rl}F(z)& =G(z)-f(z)\cdot \frac{1}{2\pi }{\int }_{\Gamma }\frac{1}{\zeta -z}d\zeta \\ & =\frac{1}{2\pi i}{\int }_{\Gamma }\frac{f(\zeta )-f(z)}{\zeta -z}d\zeta \\ & =\frac{1}{2\pi i}{\int }_{\Gamma }{g}_z(\zeta )d\zeta \text{ for }z\in U\setminus {\Gamma }^{\ast }\end{array}$$

By General Cauchy integral formula for the winding number then
Both $g(z)$ and $I(\Gamma ,z)$ are holomorphic for all $z\in \mathbb{C}\setminus {\Gamma }^{\ast }$ so $F(z)$ is holomorphic on $U\setminus {\Gamma }^{\ast }$

For fixed $z$ then $g_z\zeta$ is also clearly a holomorphic function of $\zeta$ on $U\setminus \{z\}$ and as

$$\underset{\zeta \to z}{\lim }{g}_z(\zeta )=f^{\prime }(z)$$

Then it extends to a continuous function on $U$ which also denote by $g_z$ where

$$g_z(z)=f^{\prime }(z)$$

Since it is continuous at $\zeta =z$ then $g_z$ is bounded near $z$

By Riemann’s Removable Singularities Theorem then
Extension is holomorphic as a function of $\zeta$ on all of $U$

But extending the definition of $F$ from $U\setminus {\Gamma }^{\ast }$ to all of $U$ by setting

$$F(z):={\int }_{\Gamma }{g}_z(\zeta )d\zeta$$

If $z_0\in {\Gamma }^{\ast }$ then since $g_z(\zeta )$ is holomorphic then by Removal of a point in a cycle
There is a cycle $\Sigma \subseteq U$ not containing $z_0$ such that

$${\int }_{\Sigma }{g}_z(\zeta )d\zeta ={\int }_{\Gamma }{g}_z(\zeta )d\zeta$$

So replacing $\Gamma$ by $\Sigma$ and applying the same reasoning then $F$ is holomorphic on $U\setminus {\Sigma }^{\ast }$
Most importantly it is holomorphic at $z_0$

Since $z_0$ is arbitrary then $F$ is holomorphic on all of $U$

Let $V=\mathbb{C}\setminus ({\Gamma }^{\ast }\cup \mathrm{ins}(\Gamma )))$
Since $\Gamma (z)$ is holomorphic on all of $\mathbb{C}\setminus {\Gamma }^{\ast }$, it restricts to holomorphic function on $V$
By assumption $\Gamma \cup \mathrm{ins}(\Gamma )\subseteq U$ hence $U\cup V=\mathbb{C}$

Moreover if $z\in U\cap V$ then $z\in U$ but is neither inside $\Gamma$ or on ${\Gamma }^{\ast }$ so by definition

$$F(z)=G(z)$$

So by defining

$$H:\mathbb{C}\to \mathbb{C},H(z):=\{\begin{array}{ll}F(z)& z\in U\\ & \\ G(z)& z\in V\end{array}$$

with $H$ being a well-defined entire function

Clearly $F(z)\equiv 0$ on $U$ if $H(z)\equiv 0$ on $\mathbb{C}$
But $H(z)=G(z)$ for large $z$ so
By General cauchy integral formula for the winding number then

$$G(z)\to 0\text{ as }z\to \infty$$

Hence $H$ defines a bounded entire function which is constant so is $0$ by Liouville’s Theorem

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3.3 Applications of the Integral Formula

Taylor Series for Holomorphic Functions corollary

Let $f:U\to \mathbb{C}$ be a function

If $f$ is holomorphic on open set $U$ then for any $z_0\in U$ then

$$f(z)\text{ is equal to its Taylor Series at }{z}_0$$

and the Taylor Series converges on any open disk centred at $z_0$ lying in $U$

Derivatives of $f$ at $z_0$ are given by

$$f^{(n)}(z_0)=\frac{n!}{2\pi i}{\int }_{\gamma (z_0,r)}\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

for any $r<R$ where $R$ is such that $B(z_0,R)\subset U$

Note that the integral formulae for the derivatives of $f$ are also referred to as Cauchy’s Integral Formulae

Proof Integral Formula, General cauchy integral formula for the winding number and Derivatives of the general cauchy integral formula

Follows from

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Analytic

Let $U$ be an open subset of $\mathbb{C}$

If $f:U\to \mathbb{C}$ is a function on $U$ then

$f$ is analytic on $U$ if for every $z_0\in U$ there is $r>0$ with $B(z_0,r)\subseteq U$ such that
There exists power series

$$\sum _{k=0}^{\infty }{a}_k(z-z_0{)}^k\text{ with radius of convergence at least }r$$

and

$$f(z)=\sum _{k=0}^{\infty }{a}_k(z-z_0{)}^K$$

Note that an analytic function is holomorphic

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Entire

Function $f:\mathbb{C}\to \mathbb{C}$ is entire if it is complex diffferentiable on the whole complex plane

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![[12 - Liouville's Theorem]]

13 - Fundamental Theorem of Algebra

Fundamental Theorem of Algebra

Suppose that

$$p(z)=\sum _{k=0}^n{a}_k{z}^k$$

is a non-constant polynomial where $a_k\in \mathbb{C}$ and $a_n\ne 0$ then

$$\exists z_0\in \mathbb{C}\text{ such that }p(z_0)=0$$

Proof

By rescaling $p$ then assume that $a_n=1$

If $p(z)\ne 0$ for all $z\in \mathbb{C}$ then

$$f(z)=\frac{1}{p(z)}\text{ is an entire function }$$

since $p$ is entire

Claim that $f$ is bounded
Since $f$ is continuous then it is bounded on any disc $\overline{B}(0,R)$ so need to show

$$|f(z)|\to 0\text{ as }z\to \infty$$

In other words

$$|p(z)|\to \infty \text{ as }z\to \infty$$

However

$$|p(z)|=\left|z^n+\sum _{k=0}^{n-1}{z}_k{z}^k\right|=|z^n|\left(\left|1+\sum _{k=0}^{n-1}\frac{a_k}{z^{n-k}}\right|\right)\ge |z^n|\left(1-\sum _{k=0}^{n-1}\frac{|a_k|}{|z{|}^{n-k}}\right)$$

As $\frac{1}{|z{|}^m}\to 0$ as $|z|\to \infty$ then for sufficiently large $|z|$ so for $|z|>R$

$$1-\sum _{k=0}^{n-1}\frac{|a_k|}{|z{|}^{n-k}}\ge \frac{1}{2}$$

So for $|z|\ge R$ then $|p(z)|\ge \frac{1}{2}|z{|}^n$ so as $|z{|}^n\to \infty$ as $|z|\to \infty$ then $|p(z)|\to \infty$

So $f(z)$ is bounded and hence constant so $p(z)$ is constant which is contradiction

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14 - Morera's Theorem

Morera's Theorem

Suppose that $f:U\to \mathbb{C}$ is a continuous function on an open subset $U\subseteq \mathbb{C}$

If for any closed path $\gamma :[a,b]\to U$ where ${\int }_{\gamma }f(z)dz=0$ then

$$f\text{ is holomorphic}$$

Proof

By Existence of Primitives Theorem then $f$ has primitive $F:U\to \mathbb{C}$
But as $F$ is holomorphic on $U$ then it is infinitely differentiable on $U$ so

$$f=F^{\prime }\text{ is also holomorphic}$$

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Uniform Convergence on Compacts

Let $U$ be an open subset of $\mathbb{C}$

If $(f_n)$ is a sequence of functions defined on $U$ then

$f_n\to f$ uniformly on compacts if
For every compact subset $K$ of $U$ then sequence $(f_{n\mid K})$ converges uniformly to $f_{\mid K}$

Note that $f$ is continuous if $f_n$ are also continuous

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Holomorphic of Uniform Convergence on Compacts

Suppose that $U$ is a domain

Let sequence of holomorphic functions $f_n:U\to \mathbb{C}$
Suppose $f_n$ converges to $f:U\to \mathbb{C}$ uniformly on compacts in $U$ then

$$f\text{ is holomorphic}$$

Proof

Since $U$ is open, for any $w\in U$ there exists $r>0$ such that $B(w,r)\subseteq U$
As $B(w,r)$ is convex then by Cauchy’s Theorem for Simply Connected Domains then
For every closed path $\gamma :[a,b]\to B(w,r)$ whose image lies in $B(w,r)$ then

$${\int }_{\gamma }{f}_n(z)dz=0\text{ for all }n\in N$$

But as ${\gamma }^{\ast }=\gamma ([a,b])$ is a compact subset of $U$ then $f_n\to f$ uniformly on ${\gamma }^{\ast }$ so

$$0={\int }_{\gamma }{f}_n(z)dz\to {\int }_{\gamma }f(z)dz$$

Hence integral of $f$ around any closed path in $B(w,r)$ is zero
So by Morera’s Theorem then $f$ is holomorphic

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3.4 The Identity Theorem

Zeros of Holomorphic Functions

Let $U$ be an open set
Suppose that $g:U\to \mathbb{C}$ is holomorphic on $U$

Let $S=\{z\in U:g(z)=0\}$

If $z_0\in S$ then either $z_0$ is isolated in $S$ or $g=0$ on a neighbourhood of $z_0$
If $z_0$ is isolated in $S$ then there is unique integer $k>0$ and holomorphic function $g_1$ such that

$$g(z)=(z-z_0{)}^k{g}_1(z)$$

where $g_1(z_0)\ne 0$ and $k$ is the multiplicity of zero of $g$ at $z=z_0$
Also known as order of vanishing

Proof

Pick any $z_0\in U$ with $g(z_0)=0$

Since $g$ is analytic at $z_0$ there is $r>0$ such that $\overline{B}(z_0,r)\subseteq U$ then

$$g(z)=\sum _{k=0}^{\infty }{c}_k(z-z_0{)}^k\text{ for all }z\in B(z_0,r)\subseteq U$$

where coefficients $c_k$ are given by Taylor series for holomorphic functions

If $c_k=0$ for all $k$ then $g(z)=0$ for all $z\in \overline{B}(z_0,r)$

Otherwise let $k=\min \{n\in \mathbb{N}:c_n\ne 0\}$
As $g(z_0)=0$ then $c_0=0$ so $k\ge 1$

Let $g_1(z)=(z-z_0{)}^{-k}g(z)$
So $g_1(z)$ holomorphic on $U\setminus \{z_0\}$ and

$$g_1(z)=\sum _{n=0}^{\infty }{c}_{k+n}(z-z_0{)}^n\text{ for all }z\in B(z_0,r)$$

Hence

$$g_1(z_0)=c_k\ne 0$$

Hence $g_1(z)$ is holomorphic on all of $U$

Since $g_1$ is continuous at $z_0$ and $g_1(z_1)\ne 0$ there exists $ϵ>0$ such that

$$g_1(z)\ne 0\text{ for all }z\in B(z_0,ϵ)$$

However $(z-z_0{)}^k$ vanishes only at $z_0$ hence

$$g(z)=(z-z_0{)}^k{g}_1(z)\text{ is non zero on }B(z_0,ϵ)\setminus \{z_0\}$$

so $z_0$ is isolated

In order to show $k$ is unique
Suppose

$$g(z)=(z-z_0{)}^k{g}_1(z)=(z-z_0{)}^l{g}_2(z)$$

with $g_1(z_0)\ne 0$ and $g_2(z_0)\ne 0$

If $k<l$ then

$$\frac{g(z)}{(z-z_0{)}^k}=(z-z_0{)}^{l-k}{g}_2(z)\text{ for all }z\ne z_0$$

Hence as $z\to z_0$ then $\frac{g(z)}{(z-z_0{)}^k}\to 0$ which contradicts assumption that $g_1(z)\ne 0$
So by symmetry then $k\ngeqq l$ so $k=l$

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15 - Identity Theorem

Identity Theorem

Let $U$ be a domain
Suppose $f_1,f_2$ are holomorphic functions defined on $U$

If $S=\{z\in U:f_1(z)=f_2(z)\}$ has a limit point in $U$ then

$$S=U$$

In other words

$$f_1(z)=f_2(z)\text{ for all }z\in U$$

Proof

Let $g=f_1-f_2$ so $S=g^{-1}(\{0\})$

Since $g$ is clearly holomorphic in $U$ then by Zeros of holomorphic functions then
If $z_0\in S$ then either $z_0$ is an isolated point of $S$ or it lies in an open ball contained in $S$

So

$$S=V\cup T$$

where $T=\{z\in S:z\text{ is isolated}\}$ and $V=\text{int}(S)$

As $g$ is continuous then $S=g^{-1}(\{0\})$ is closed in $U$ thus $V\cup T$ is closed
So ${\mathrm{Cl}}_U(V)$ which is the closure of $V$ in $U$ lies in $V\cup T$

However by definition, no limit point of $V$ can lie in $T$ so ${\mathrm{Cl}}_U(V)=V$
Hence $V$ is open and closed in $U$ so as $U$ is connected then $V=\varnothing$ or $V=U$

If $V=\varnothing$ then all zeros of $g$ are isolated so $S$ has no limit points
If $V=U$ then $V=S=U$

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3.5 Isolated Singularities

Singular Point

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

If $f$ is not holomorphic at $z_0$ then

$$z_0\text{ is a singular}$$

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Isolated Singularity

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

If $f$ is holomorphic on $B(z_0,r)\setminus \{z_0\}$ for some $r>0$ then

$$z_0\text{ is an isolated singular}$$

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Meromorphic

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

$f$ is a meromorphic function on $U$ if

$$f\text{ has only isolated singularities all of which are poles}$$

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Types of Isolated Singularities

Let $f:U\to \mathbb{C}$ be a function where $U$ is open

Let $z_0$ be an isolated singularity of function $f$ then

1. Removable Singularity - Exists function $g$ holomorphic in $B(z_0,r)$ with $r>0$ such that

$$f(z)=g(z)\text{ in }B(z_0,r)\setminus \{z_0\}$$

2. Pole of Order $n$ - Exists function $g$ holomorphic in $B(z_0,r)$ with $r>0$ such that

$$g(z_0)\ne 0\text{ and }f(z)=(z-z_0{)}^{-n}g(z)\text{ in }B(z_0,r)zzz\{z_0\}$$

3. Essential Singularity - Otherwise

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16 - Laurent's Theorem

Laurent's Theorem

Suppose $0<r<R$ and

$$A=A(z_0,r,R)=\{z:r<|z-z_0<R|\}$$

where $A$ is an annulus centred at $z_0$

If $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ which contains $\overline{A}$ then exists $c_n\in \mathbb{C}$ such that

$$f(z)=\sum _{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n\forall z\in A$$

Series is known as the Laurent Series of $f$ and

1. Converges for all $z\in A$

2. Converges uniformly for all $z\in A(x_0,r^{\prime },R^{\prime })$ where $r<r^{\prime }<R^{\prime }<R$

Moreover $c_n$ is unique and given by

$$c_n=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

where $s\in [r,R]$ and for any $s>0$ then ${\gamma }_s(t):=z_0+s{e}^{2\pi it}$

Proof

By Cauchy formula for multiple curves for any $w\in A$ then

$$f(w)=\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{f(z)}{z-w}dz-\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz$$

Note that both boundary components are counter-clockwise oriented
However in Cauchy formula for multiple curves, inner component is clockwise oriented
So it is compensated by the minus sign in front of second integral

Fixing $w$ then for $|w|<R=|z|$ then

$$\frac{1}{z-w}=\sum _{n=0}^{\infty }\frac{w^n}{z^{n+1}}$$

and for $|w|<R-ϵ=|z|-ϵ$ then series converges uniformly $z$ for any $ϵ>0$ thus

$${\int }_{{\gamma }_R}\frac{f(z)}{z-w}={\int }_{{\gamma }_R}\sum _{n=0}^{\infty }\frac{f(z)w^n}{z^{n+1}}=\sum _{n=0}^{\infty }\left({\int }_{{\gamma }_R}\frac{f(z)}{z^{n+1}}dz\right)w^n$$

for all $w\in A$

Similarly for $|z|=r<|w|$ then

$$\frac{1}{w-z}=\sum _{n=0}^{\infty }\frac{z^n}{w^{n+1}}=\sum _{n=-1}^{-\infty }\frac{w^n}{z^{n+1}}$$

which again converges uniformly when $|z|<r<|w|-ϵ$ for $ϵ>0$ then

$${\int }_{{\gamma }_r}\frac{f(z)}{w-z}dz={\int }_{{\gamma }_r}\sum _{n=-1}^{-\infty }\frac{f(z)w^n}{z^{n+1}}dz=\sum _{n=-1}^{-\infty }\left({\int }_{{\gamma }_r}\frac{f(z)}{z^{n+1}}dz\right)w^n$$

Thus by taking $(c_n{)}_{n\in \mathbb{Z}}$ to be the statement in the theorem then

$$f(w)=\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{f(z)}{z-w}dz-\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz=\sum _{n=-\infty }^{\infty }{c}_n{w}^n$$

as required

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Laurent Series in a Punctured Disc corollary

If $f:U\to \mathbb{C}$ is a holomorphic function and $z_0$ is an isolated singularity then

$$f\text{ has a Laurent Expansion on punctured disc }B(z_0,R)\setminus \{x_0\}$$

for any $R$ such that $B(z_0,R)\setminus \{z_0\}\subset U$

Proof

Let $r$ be such that $0<r<R$ and apply Laurent’s Theorem to $A=A(z_0,r,R)$
As the coefficients $c_n$ can be written in terms of integrals along ${\gamma }_R$ then

$${\gamma }_R\text{ is independent of }r$$

By sending $r$ to zero then Laurent Series converges in

$$B(z_0,R)=\bigcup _{o<r<R}A(z_0,r,R)$$

and converges uniformly in $A(z_0,r,R^{\prime })$ for any $0<r<R^{\prime }<R$

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Principal Part of the Laurent's Series

Let $z_0$ be an isolated singularity of $f$ with Laurent’s Expansion

$$\sum _{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n$$

Principal Part of $f$ at $z_0$ is the sum of terms with negative powers denoted by

$$P_{z_0}f(z)=\sum _{n=-\infty }^{-1}{c}_n(z-z_0{)}^n=\sum _{n=1}^{\infty }{c}_{-n}(z-z_0{)}^{-n}$$

Link to original

Residue

Let $z_0$ be an isolated singularity of $f$ then

Residue of $f$ at $z_0$ is defined as the coefficient of $c_{-1}$ of the Lareunt Expansion denoted by

$${\mathrm{Res}}_{z_0}f\text{ or }\mathrm{Res}(f,z_0)$$

Reason for such Definition

Consider function $f$ given by series $\sum c_n(z-z_0{)}^n$
which converges uniformly in an annulus containing ${\gamma }_s$ then integrate term by term

For $n\ne 1$ then $(z-z_0{)}^n$ has a well defined primitive $\frac{(z-z_0{)}^{n+1}}{n+1}$ so integrate to $0$ on ${\gamma }_s$

However for $n=-1$ then the integral of $1/(z-z_0)$ along ${\gamma }_s$ is $2\pi i$ so

$$\frac{1}{2\pi i}{\int }_{{\gamma }_s}f(z)dz=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\sum _{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\frac{c_{-1}}{z-z_0}dz=c_{-1}$$

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17 - Characterisation of Isolated Singularities

Characterisation of Isolated Singularities

Let $z_0$ be an isolated singularity of $f$
Let ${\sum }_{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n$ be the Laurent Expansion of $f$ then

$z_0$ is classified as

1. Removable Singularity: If $c_n=0$ for all $n<0$

2. Pole of Order $n$: If $c_{-n}\ne 0$ and $c_k=0$ for all $k<n$

3. Essential Singularity: There is arbitrary large $n$ such that $c_n\ne 0$

Equivalent Statements for Principal

1. Principal Part vanishes

2. Principal Part is non-trivial but contains a finite number of non-zero terms

3. Principal Part contains infinitely many non-zero terms

Proof

If there is no principal part then $f$ has a ordinary power series for it’s Laurent Expansion
$f$ would converge to function $g$ which is analytic in some disc $B(z_0,R)$ around $z_0$
where $f(z)=g(z)$ for all $0<|z|<R$ so $z_0$ is removable

If $z_0$ is removable then there is function $g$ which is holomorphic at $z_0$
with $g$ coinciding with $f$ in $B(z_0,R)\setminus \{z_0\}$
Hence Laurent Expansion coefficients are equal, but for holomorphic function $g$ then
All coefficients with $n<0$ are given by integrals of holomorphic function $(z-z_0{)}^{-n}g(z)$
However by Cauchy’s Theorem for Simply Connected Domains then integrals vanish

If Laurent Expansion is in form

$$f(z)=\frac{c_{-n}}{(z-z_0{)}^n}+\cdots +c_0+c_1(z-z_0)+\cdots$$

with $c_{-n}\ne 0$ then $f(z)=(z-z_0{)}^{-n}g(z)$ where

$$g(z)=c_{-n}+c_{-n+1}(z-z_0)+\cdots$$

which is analytic in some neighbourhood of $z_0$

If $z_0$ is a pole of order $n$ then there is holomorphic $g$ such that

$$g(z_0)\ne 0\text{ and }f(z)=(z-z_0{)}^{-n}g(z)$$

Writing the Taylor Series of $g$ then the Laurent Expansion of $f$ is in form

$$f(z)=\frac{g(z_0)}{(z-z_0{)}^n}+\frac{g^{\prime }(z_0)}{(z-z_0{)}^{n-1}}\cdots$$

where the last part follows trivially from the last two

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18 - Riemann's Removable Singularity Theorem

Riemann's Removable Singularity Theorem

Suppose $U$ is an open subset of $\mathbb{C}$ and $z_0\in U$
Suppose $f:U\setminus \{z_0\}\to \mathbb{C}$ is holomorphic then

$z_0$ is a removable singularity if and only if $f$ is bounded near $z_0$

Proof

Proof $⟹$
If $z_0$ is removable then there is holomorphic $g$ such that $g(z)=f(z)$ on $B(z_0,r)$ for $r>0$
So $f(z)=g(z)\to g(z_0)$ as $z\to z_0$ so $f$ is bounded

Proof $⟸$
Assume that $f$ is bounded and define $h(z)$ by

$$h(z)=\{\begin{array}{ll}(z-z_0{)}^{2}f(z)& \text{ if }z\ne z_0\\ & \\ 0& \text{ if }z=z_0\end{array}$$

so $h(z)$ is holomorphic on $U\setminus \{z_0\}$ using $f$ is complex differentiable
If $z=z_0$ then

$$\frac{h(z)-h(z_0)}{z-z_0}=(z-z_0)f(z)\to 0$$

as $z\to z_0$ since $f$ is bounded near $z_0$ by assumption

So $f$ is holomorphic everywhere in $U$
However for $r>0$ such that $\overline{B}(z_0,r)\subset U$ so by Taylor series for holomorphic functions
Then it is equal to its Taylor Series centred at $z_0$ thus

$$h(z)=\sum _{k=0}^{\infty }{a}_k(z-z_0{)}^k$$

Since $h(z_0)=h^{\prime }(z_0)=0$ then $a_0=a_1=0$ hence

$$\sum _{k=0}^{\infty }{a}_{k+2}(z-z_0{)}^k$$

defines a holomorphic function in $B(z_0,r)$

Since it agrees with $f(z)$ on $B(z_0,r)\setminus \{0\}$ then extend $f$ to holomorphic function on $U$

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Poles and Zeroes of Reciprocal lemma

Let $f$ be a holomorphic function in a neighbourhood of $z_0$ then

$z_0$ is a pole if and only if $|f(z)|\to \infty$ as $z\to z_0$

In this case function

$$h(z)=\{\begin{array}{ll}\frac{1}{f(z)}& \text{ if }z\ne z_0\\ 0& \text{ if }z=z_0\end{array}$$

is holomorphic on neighbourhood of $z_0$, multiplicity of zero at $z_0$ is equal to order of pole of $f$

Proof

Assume $f$ has pole of order $n$ so by Characterisation of Isolated Singularities then
$f(z)$ has a Laurent Expansion of

$$f(z)=\frac{c_{-n}}{(z-z_0{)}^n}+\cdots +c_0+c_1(z-z_0)+\cdots$$

which can be rewritten as

$$f(z)=\frac{1}{(z-z_0{)}^n}(c_{-n}+c_{-n+1}(z-z_0)+\cdots )$$

where the series converges to a function analytic at $z_0$ denoted by $g$ then

$$\frac{1}{f}=(z-z_0{)}^n\frac{1}{g}$$

Since $g(z_0)=c_{-n}\ne 0$ then function $h=\frac{1}{g}$ is holomorphic in neighbourhood of $z_0$ and

$$h(z_0)\ne 0$$

Thus $\frac{1}{f}$ has a removable singularity and after removing has a zero multiplicity of $n$

Assume $|f|\to \infty$ then

$$\frac{1}{f}\to 0$$

so by Characterisation of Isolated Singularities then
It can be extended to holomorphic function $h$ with $0$ at $z_0$

As the $0$ must be of finite order so $n\ge 1$ then

$$h(z)=(z-z_0{)}^{n}g(z)$$

where $g$ is holomorphic with $g(z_0)\ne 0$ then

$$f(z)=(z-z_0{)}^{-n}\cdot \frac{1}{g(z)}$$

so $f$ has a pole of order $n$

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19 - Casorati-Weierstrass or Weierstrass Theorem

Casorati-Weierstrass or Weierstrass Theorem

Let $U$ be an open subset of $\mathbb{C}$
Let $a\in \mathbb{C}$

Suppose $f:U\setminus \{a\}\to \mathbb{C}$ is a holomorphic function with isolated essential singularity at $a$ then

For all $\rho >0$ with $B(a,\rho )\subseteq U$ then

$$f(B(a,\rho )\setminus \{a\})\text{ is dense in }\mathbb{C}$$

In other words the closure of $f(B(a,\rho )\setminus \{a\})$ is all of $\mathbb{C}$

Proof

Suppose for contradiction that there is some $\rho >0$ such that

$$z_0\in \mathbb{C}\text{ is not a limit point of }f(B(a,\rho )\setminus \{a\})$$

Then function

$$g(z)=\frac{1}{f(z)-z_0}$$

is bounded and non-vanishing on $B(a,\rho )\setminus \{a\}$

Hence by Riemann’s Removable Singularity Theorem then
$g$ can be extended to a holomorphic function on all of $B(a,\rho )$

But then

$$f(z)=z_0+\frac{1}{g(z)}$$

has at most a pole at $a$ which is a contradiction

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20 - Residue Theorem

Residue Theorem

Suppose that $U$ is an open set in $\mathbb{C}$
Suppose $\gamma$ is a closed curve that is contained in $U$ together with its inside

Suppose $f$ is holomorphic on $U\setminus S$ where $S$ is a finite set of isolated singularities of $f$

Assume that $f$ has no singularity on ${\gamma }^{\ast }$ so that $S\cap {\gamma }^{\ast }=\varnothing$ then

$$\frac{1}{2\pi i}{\int }_{\gamma }f(z)dz=\sum _{a\in S}I(\gamma ,a){\mathrm{Res}}_a(f)$$

Proof

For each $a\in S$ let

$$P_a(f)(z)=\sum _{n=-1}^{-\infty }{c}_n(a)(z-a{)}^n$$

where $P_a(f)(z)$ is the principal part of $f$ at $a$ so is a holomorphic function on $\mathbb{C}\setminus \{a\}$

By definition of $P_a(f)$ then difference $f-P_a(f)$ is holomorphic at $a\in S$ thus

$$g(z)=f(z)-\sum _{a\in S}{P}_a(f)$$

is holomorphic on all of $U$

By Homotopy Cauchy’s Theorem as $U$ contains $\gamma$ and it’s inside then $\gamma$ is null homotopic

$${\int }_{\gamma }g(z)dz=0$$

Hence

$${\int }_{\gamma }f(z)dz=\sum _{a\in S}{\int }_{\gamma }{P}_a(f)(z)dz$$

By Convergence of the principal part then series $P_a(f)$ converges uniformly on ${\gamma }^{\ast }$ so

$$\begin{array}{rl}{\int }_{\gamma }{P}_a(f)dz& ={\int }_{\gamma }\sum _{n=-1}^{-\infty }{c}_n(a)(z-a{)}^n\\ & =\sum _{n=1}^{\infty }\frac{c_{-n}(a)}{(z-a{)}^n}dz\\ & ={\int }_{\gamma }\frac{c_{-1}(a)}{z-a}dz\\ & =2\pi iI(\gamma ,a){\mathrm{Res}}_a(f)\end{array}$$

as for $n>1$ then function $(z-a{)}^{-n}$ has a primitive on $\mathbb{C}\setminus \{a\}$

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3.6 The Argument Principle

Residue of the Logarithmic Derivative lemma

Suppose $f:U\to \mathbb{C}$ is meromorphic and has zero of order $k$ or pole of order $k$ at $z_0\in U$ then

$$\frac{f^{\prime }(z)}{f(z)}\text{ has a simple pole at }{z}_0\text{ with residue }k\text{ or }-k\text{ respectively}$$

Proof

If $f(z)$ has a zero of order $k$ then

$$f(z)=(z-z_0{)}^{k}g(z)$$

where $g(z)$ is holomorphic near $z_0$ and $g(z_0)\ne 0$ then

$$\frac{f^{\prime }(z)}{f(z)}=\frac{k}{z-z_0}+\frac{g^{\prime }(z)}{g(z)}$$

since $g(z)\ne 0$ near $z_0$ then $\frac{g^{\prime }(z)}{g(z)}$ is holomorphic near $z_0$ so result follows

Case is similar where $f$ has pole at $z_0$

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21 - Argument Principle

Argument Principle

Suppose $f$ is meromorphic on $U$
Let $\gamma$ be simply positively oriented contour such that $\gamma$ and it’s inside is contained in $U$

Assume $f$ has no zeroes or poles on ${\gamma }^{\ast }$

Let $N$ is the number of zeros (counted with multiplicity) of $f$ inside $\gamma$
Let $P$ be the number of poles (counted with multiplicity) of $f$ inside $\gamma$ then

$$N-P=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\prime }(z)}{f(z)}dz$$

Moreover it is also the winding number of path $\Gamma =f\circ \gamma$ about the origin

Proof

Curve is simple so winding number around any point inside is $1$
By Residue of the logarithmic derivative then
$\frac{f^{\prime }(z)}{f(z)}$ has simple poles at zeros and poles of $f$ with residues given by their order
So result follows from Residue Theorem

For the last part then

$${\int }_{f\circ \gamma }\frac{1}{w}dw={\int }_0^1\frac{1}{f(\gamma (t))}{f}^{\prime }(\gamma (t)){\gamma }^{\prime }(t)dt={\int }_{\gamma }\frac{f^{\prime }(z)}{f(z)}dz$$

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22 - Rouché's Theorem

Rouché's Theorem

Suppose $f$ and $g$ are holomorphic functions on an open set $U$ in $\mathbb{C}$
Let $\gamma$ be a simple contour that is contained inside of $U$ together with it’s interior

If $|f(z)|>|g(z)|$ for all $z\in {\gamma }^{\ast }$then

$$f\text{ and }f+g$$

have the same change in argument around $\gamma$

Hence they have the same number of $0$s (counted with multiplicity) inside of $\gamma$

Proof

Consider function

$$h=\frac{f+g}{f}=1+\frac{g}{f}$$

then the zeros of $h$ are zeros of $f+g$ and poles are zeros of $f$

Note that the assumption $|f(z)|>|g(z)|$ implies that $h$ has no zeroes or poles on ${\gamma }^{\ast }$

By the argument principle then
Difference between number of zeroes and poles is equal to winding number of

$$\Gamma (t)=h(\gamma (t))\text{ about zero}$$

By assumption for $z\in {\gamma }^{\ast }$ then $|g(z)|<|f(z)|$ so $\left|\frac{g(z)}{f(z)}\right|<1$

Hence image of $\Gamma$ lies entirely in $B(1,1)$ and thus lies in half plane $\{z:\mathrm{Re}(z)>0\}$
So picking the principal branch of $[\log ]$ defined on the half plane then integral

$${\int }_{\Gamma }\frac{1}{z}dz=\mathrm{Log}(h(\gamma (1)))-\mathrm{Log}(h(\gamma (0)))=0$$

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23 - Open Mapping Theorem

Open Mapping Theorem

Suppose $f:U\to \mathbb{C}$ is holomorphic and non-constant on a domain $U$ then

For any open set $V\subset U$ then set $f(V)$ is also open

Proof

Suppose $w_0\in f(V)$

Consider $z_0$ such that $f(z_0)=w_0$ so
Let $g(z):=f(z)-w_0$ such that it has a zero at $z_0$

Since $g$ is non-constant then the zero is isolated

Thus there exists some $r>0$ such that $g(z)$ has no other zeros inside $\overline{B}(z_0,r)\subset U$

Since $\partial B(z_0,r)$ is compact then there is $\delta >0$ such that

$$|g(z)|\ge \delta >0\text{ on }\partial B(z_0,r)$$

Consider any $w$ such that $|w-w_0|<\delta$ then

$$|w-w_0|<\delta \le |g|\text{ on }\partial B(z_0,r)$$

Hence by Rouché’s Theorem then

$$g(z)\text{ and }g(z)+w_0-w=f(w)-w$$

have the same number of zeros (counting multiplicities) inside $B(z_0,r)$

Since $g(z)$ has a zero at $z_0$ of multiplicity of at least one then
Equation $f=w$ has at least one solution inside $B(z_0,r)$

Thus

$$B(w_0,\delta )\subset f(B(z_0,r))$$

and $f(U)$ is open

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24 - Inverse Function Theorem

Inverse Function Theorem

Suppose $f:U\to \mathbb{C}$ is injective and holomorphic and $f^{\prime }(z)\ne 0$ for all $z\in U$

If $g:f(U)\to U$ is the inverse of $f$ then $g$ is holomorphic with

$$g^{\prime }(w)=\frac{1}{f^{\prime }(g(w))}$$

Proof

By the Open Mapping Theorem| then $g$ is continuous and
If $V$ is open in $f(U)$ then $g^{-1}(V)=f(V)$ is open by the theorem

Fix $w_0\in f(U)$
Let $z_0=g(w_0)$ then since $f$ and $g$ are continuous then

$$w-w_0⟹z=g(w)\to z_0$$

Thus

$$\underset{w\to w_0}{\lim }\frac{g(w)-g(w_0)}{w-w_0}=\underset{z\to z_0}{\lim }\frac{z-z_0}{f(z)-f(z_0)}=\frac{1}{f^{\prime }(z_0)}$$

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3.7 Applications of the Residue Theorem

Residue of a Simple Positively Oriented Contour

Let $\gamma$ be a simple positively oriented contour

Let $f$ be holomorphic on and inside of $\gamma$ except at a finite set $S$ of isolated singularities
where none of the singularities lie on ${\gamma }^{\ast }$ itself then

$${\int }_{\gamma }f(z)dz=2\pi i\sum _{z_0\in S}{\mathrm{Res}}_{z_0}(f)$$

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3.7.1 On the Computation of Residue

Residue of Poles

Suppose $f$ has a pole of order $m$ at $z_0$ then

$${\mathrm{Res}}_{z_0}(f)=\underset{z\to z_0}{\lim }\frac{1}{(m-1)!}\frac{d^{m-1}}{d{z}^{m-1}}((z-z_0{)}^{m}f(z))$$

Residue of Poles for Ratios $f=\frac{g}{h}$ is a ratio of two holomorphic functions defined on domain $U\subseteq \mathbb{C}$ where $h$ is non-constant then

Suppose

$f$ is meromorphic with poles at zeros of $h$

If $h$ has a simple zero at $z_0$ and $g$ is non-vanishing then $f$ has a simple pole at $z_0$
Since the zero of $h$ is simple at $z_0$ then $h^{\prime }(z_0)\ne 0$ hence

$${\mathrm{Res}}_{z_0}(f)=\underset{z\to z_0}{\lim }\frac{g(z)(z-z_0)}{h(z)}=\underset{z\to z_0}{\lim }g(z)\underset{z\to z_0}{\lim }\frac{z-z_0}{h(z)-h(z_0)}=\frac{g(z_0)}{h^{\prime }(z_0)}$$

Proof

Since $f$ has a pole of order $m$ at $z_0$ then

$$f(z)=\sum _{n\ge -m}{c}_n(z-z_0{)}^n$$

where $z$ is sufficiently close to $z_0$ hence

$$(z-z_0{)}^{m}f(z)=c_{-m}+c_{-m+1}(z-z_0)+\cdots +c_{-1}(z-z_0{)}^{m-1}+\cdots$$

where the result follows from the formula derivatives of a power series

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3.7.2 Residue Calculus

Boundary of Half-Disk Contour Integral

Suppose we want to find integral of

$${\int }_0^{\infty }f(x)dx$$

Define contour ${\Gamma }_R$ as the concatenation of paths ${\gamma }_1:[-R,R]\to \mathbb{C}$ and ${\gamma }_2:[0,1]\to \mathbb{C}$ where

$${\gamma }_1(t)=-R+t\text{ and }{\gamma }_2(t)={\mathrm{Re}}^{i\pi t}$$

such that ${\Gamma }_R={\gamma }_2\star {\gamma }_1$ traces boundary of half-disk

In many cases it can be shown that

$${\int }_{{\gamma }_2}f(z)dz\to 0\text{ as }R\to \infty$$

So by calculating the residues inside contour ${\Gamma }_R$ then the
Integral of $f$ on $(-\infty ,\infty )$ can be deduced

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3.7.3 Jordan’s Lemma and Applications

Concavity of a Function lemma

Let $g:\mathbb{R}\to \mathbb{R}$ be a differentiable function then

If $[a,b]$ is an interval where $g^{\prime \prime }(x)<0$ then function $g$ is concave on $[a,b]$ if

For all $x<y\in [a,b]$ then

$$g(tx+(1-t)y)\ge tg(x)+(1-t)g(t)\forall t\in [0m1]$$

In other words the chords between any two points on the graph lies below the graph itself

Proof

For $x,y\in [a,b]$ and $t\in [0,1]$

Let $\xi =tx+(1-t)y$ be a point in interval between $x,y$

Slope of chord between $(x,g(x))$ and $(\xi ,g(\xi ))$ by MVT is equal to

$$g^{\prime }(s_1)\text{ where }{s}_1\text{ lies between }x\text{ and }\xi$$

whilst slope of chord between $(\xi ,g(\xi ))$ and $(y,g(y)$ is equal to

$$g^{\prime }(s_2)\text{ where }{s}_2\text{ lies between }\xi \text{ and }y$$

If $g(\xi )<tg(x)+(1-t)g(y)$ then

$$g^{\prime }(s_1)<0\text{ and }{g}^{\prime }(s_2)>0$$

So by MVT for $g^{\prime }(x)$ applied to $s_1$ and $s_2$ then exists $s\in (S_1,s_2)$ with

$$g^{\prime \prime }(s)=\frac{g^{\prime }(s_2)-g^{\prime }(s_1)}{s_2-s_1}>0$$

which contradicts assumption that $g^{\prime \prime }(x)$ is negative on $(a,b)$

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Jordon's Lemma lemma

let $f$ be a continuous function on ${\gamma }_R^{\ast }$ where

$${\gamma }_R(t)=R{e}^{it}\text{ where }t\in [0,\pi ]$$

Then for all positive $\alpha$

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\right|\le \frac{\pi }{\alpha }{M}_R\text{ where }{M}_R=\underset{t\in [0,\pi ]}{\max }|f({\mathrm{Re}}^{it}])|$$

In particular, suppose $f$ is holomorphic on $\mathbb{H}\setminus S$ where

$$\mathbb{H}=\{z\in \mathbb{C}:\mathrm{Im}(z)>0\}\text{ is the upper half-plane}$$

and $S$ be the finite set of isolated singularities

Suppose $f(z)\to 0$ as $z\to \infty$ in $\mathbb{H}$ then

$${\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\to 0\text{ as }R\to \infty \text{ for all }\alpha >0$$

Proof

Apply Concavity of a function to function $g(t)=\sin (t)$ with $x=0$ and $y=\frac{\pi }{2}$
So

$$\sin (t)\ge \frac{2}{\pi }t\text{ for }t\in \left[0,\frac{\pi }{2}\right]$$

and similarly

$$\sin (\pi -t)\ge \frac{2(\pi -t)}{\pi }\text{ for }t\in \left[\frac{\pi }{2},\pi \right]$$

Thus

$$|e^{i\alpha z}|\le e^{-\alpha R\sin (t)}\le \{\begin{array}{ll}{e}^{-2\alpha Rt/\pi }& \text{ if }t\in \left[0,\frac{\pi }{2}\right]\\ & \\ e^{-2\alpha R(\pi -t)/\pi }& \text{ if }t\in \left[\frac{\pi }{2},\pi \right]\end{array}$$

But then

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}\right|\le 2R{M}_R{\int }_0^{\frac{\pi }{2}}{e}^{-2\alpha Rt/\pi }dt=M_R\frac{\pi }{\alpha }(1-e^{-\alpha R})<M_R\frac{\pi }{\alpha }$$

Fix $ϵ>0$
Since $f\to 0$ as $z\to \infty$ then as $S$ is finite
There exists $R_0$ such that $M_R<ϵ$ for all $R>R_0$ hence

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\right|\le ϵ\frac{\pi }{\alpha }\text{ for }R>R_0$$

Link to original

Residue from Small Circle Integrals lemma

Let $f:U\to \mathbb{C}$ be a meromorphic function with a simple pole at $a\in U$
Let ${\gamma }_{ϵ}:[\alpha ,\beta ]\to \mathbb{C}$ be path defined by

$${\gamma }_{ϵ}(t)=a+ϵe^{it}$$

then

$$\underset{ϵ\to 0}{\lim }{\int }_{{\gamma }_{ϵ}}f(z)dz={\mathrm{Res}}_a(f)(\beta -\alpha )i$$

Proof

Since $f$ has a simple pole at $a$ then

$$f(z)=\frac{c}{z-a}+g(z)$$

where $g(z)$ is holomorphic near $z$ and $c={\mathrm{Res}}_a(f)$

Note that $\frac{c}{z-a}$ is the principal part of $f$ at $a$

As $g$ is holomorphic at $a$ then it is continuous at $a$ and hence bounded
Let $M,r>0$ such that

$$|g(z)|\le M\text{ for all }z\in B(a,r)$$

Then if $0<ϵ<r$ then

$$\left|{\int }_{{\gamma }_{ϵ}}g(z)dz\right|\le \ell ({\gamma }_e)M=(\beta -\alpha )ϵM$$

which tends to $0$ as $ϵ\to 0$

However also

$${\int }_{{\gamma }_{ϵ}}\frac{c}{z-a}dz={\int }_{\alpha }^{\beta }\frac{c}{ϵe^{it}}iϵe^{it}dx={\int }_{\alpha }^{\beta }(ic)dt=ic(\beta -\alpha )$$

Since

$${\int }_{{\gamma }_{ϵ}}f(z)dz={\int }_{{\gamma }_{ϵ}}\frac{c}{z-a}dz+{\int }_{{\gamma }_{ϵ}}g(z)dz$$

then result follows

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3.7.4 Summation of Infinite Series

Contour Bound for $\cot (\pi z)$

Let $f(z)=\cot (\pi z)$

Let ${\Gamma }_N$ denote the square path with vertices $\left(N+\frac{1}{2}\right)(\pm i\pm i)$

Then there is constant $C$ independent of $N$ such that

$$|f(z)|\le C\text{ for all }z\in {\Gamma }_N^{\ast }$$

Proof

Consider horizontal and vertical sides of square separately

Note

$$\cot (\pi z)=\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi z}}$$

So on horizontal sides of ${\Gamma }_N$ where $z=x\pm \left(N+\frac{i}{2}\right)i$ for $-\left(N+\frac{1}{2}\right)\le x\le \left(N+\frac{1}{2}\right)$

$$\begin{array}{rl}|\cot (\pi z)|& =\left|\frac{e^{i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}+e^{-\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}}{e^{i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}-e^{-i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}}\right|\\ & \le \frac{e^{\pi (N+1/2)}+e^{-\pi (N+1/2)}}{e^{\pi (N+1/2)}-e^{-\pi (N+1/2)}}\\ & =\coth \left(\pi \left(N+\frac{1}{2}\right)\right)\end{array}$$

As $\coth (x)$ is a decreasing function for $x\ge 0$ then on horizontal sides of ${\Gamma }_n$ then

$$|\cot (\pi z)|\le \coth \left(\frac{3}{2}\pi \right)$$

On vertical sides then $x=\pm \left(N+\frac{1}{2}\right)+iy$ where $-N-\frac{1}{2}\le y\le N+\frac{1}{2}$
As $\cot (z+N\pi )=\cot (z)$ for any integer $N$ and $\cot \left(z+\frac{\pi }{2}\right)=-\tan (z)$ then

If $z=\pm \left(N+\frac{1}{2}\right)+iy$ for any $y\in \mathbb{R}$ then

$$|\cot (\pi z)|=|-\tan (iy)|=|-\tanh (y)|{\le }_1$$

So set

$$C=\max \left\{1,\coth \left(\frac{3\pi }{2}\right)\right\}$$

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3.7.5 Keyhole Contours

Keyhole Contours

Constructed from two circular paths of radius $ϵ$ and $R$
where $R$ becomes arbitrarily large and $ϵ$ arbitrarily small

Join the two circles by line segments with narrow neck in between
Explicitly if $0<ϵ<R$ then pick $\delta >0$ with

$$\begin{array}{rl}{\eta }_{+}(t)& =t+i\delta \\ {\eta }_{-}(t)& =(R-t)-i\delta \end{array}$$

where $t$ runs over closed interval with endpoints such that
Endpoints of ${\ell }_{\pm }$ lie on circles of radius $ϵ$ and radius $R$ about origin

Letting ${\gamma }_R$ be positively oriented path on circle of radius $R$ joining endpoints of ${\eta }_{+}$ and ${\eta }_{-}$ on circle of radius $R$

Similarly let ${\gamma }_{ϵ}$ be path on circle of radius $ϵ$ which is negatively oriented and joining endpoints of ${\eta }_{+}$ and ${\eta }_{-}$ on circle of radius $ϵ$

Set

$${\Gamma }_{R,ϵ}={\ell }_{+}\star {\gamma }_R\star {\ell }_{-}\star {\gamma }_{ϵ}$$

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