01. Complex Differentiability

1.1 Complex Differentiability

Complex Differentiability

Let $a\in \mathbb{C}$

Suppose

$$f:U\to \mathbb{C}\text{ is a function}$$

where $U\subset C$ is an open set containing $a$ then

$f$ is complex differentiable at $a$ if

$$\underset{z\to a}{\lim }\frac{f(z)-f(a)}{z-a}\text{ exists}$$

If the limit exists then it is written as $f^{\prime }(a)$ and is called the derivative of $f$ at $a$

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Properties of Complex Derivatives lemma

Let $a\in \mathbb{C}$
Let $U$ be the neighbourhood of $a$
Let $f:g\to U\to \mathbb{C}$ then

1. Sums, Products
   If $f,g$ are differentiable at $a$ then $f+g$ and $fg$ are differentiable at $a$ with

$$\begin{array}{rl}(f+g{)}^{\prime }(a)& =f^{\prime }(a)+g^{\prime }(a)\\ (fg{)}^{\prime }(a)& =f^{\prime }(a)g(a)+f(a)g^{\prime }(a)\end{array}$$

2. Quotients
   If $f,g$ are differentiable at $a$ and $g(a)\ne 0$ then $f/g$ is differentiable at $a$ with

$${\left(\frac{f}{g}\right)}^{\prime }(a)=\frac{f^{\prime }(a)g(a)-f(a)g^{\prime }(a)}{g(a{)}^2}$$

3. Chain Rule
   If $U$ and $V$ are open subsets of $\mathbb{C}$ with $f:V\to U$ and $g:U\to C$ are functions
   where $f$ is differentiable at $a\in V$ and $g$ is differentiable at $f(a)\in U$ then
   $g\circ f$ is differentiable at $a$ with

$$(g\circ f{)}^{\prime }(a)=g^{\prime }(f(a))f^{\prime }(a)$$

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Expansion of a Function around a Point lemma

Let $a\in \mathbb{C}$
Let $U$ be a neighbourhood of $a$
Let $f:U\to \mathbb{C}$ then

$f$ is differentiable at $a$ with derivative $f^{\prime }(a)$ if and only if

$$f(z)=f(a)+f^{\prime }(a)(z-a)+ϵ(z)(z-a)$$

where $ϵ(z)\to 0$ as $z\to a$

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Holomorphic Function

Let $U\subseteq \mathbb{C}$ be an open set
Let $f:U\to \mathbb{C}$ be a function

If $f$ is complex differentiable at every $a\in U$ then

$$f\text{ is holomorphic on }U$$

Note saying $f$ is holomorphic at a point $a$ means there is an open set $U$ containing $a$ on which $f$ is holomorphic

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1.2 Cauchy-Riemann Equations

01 - Cauchy-Riemann Equations

Cauchy-Riemann Equations

Let $z_0\in \mathbb{C}$
Let $U$ be a neighbourhood of $z_0$

Let $f:U\to C$ be a function which is complex differentiable at $z_0$
Let $u,v:{\mathbb{R}}^2\to \mathbb{R}$ be the components of $f$ then

The four partial derivatives ${\partial }_{x}u,{\partial }_{y}u,{\partial }_{x}v,{\partial }_{y}v$ exist at $z_0$ with Cauchy-Riemann Equations

$${\partial }_{x}u={\partial }_{y}v,{\partial }_{x}v=-{\partial }_{y}u$$

and

$$f^{\prime }(z_0)={\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)$$

Proof

By definition of complex differentiability then

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

By considering a particular way in which $z$ approaches $z_0$
Let $z=z_0+h$ where $h\in \mathbb{R}$ and $h\to 0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)& =\underset{h\to 0}{\lim }\left[\frac{u(x_0+h,y_0)+iv(x_0+h,y_0)-u(x_0,y_0)-iv(x_0,y_0)}{h}\right]\\ & =\underset{h\to 0}{\lim }\left[\frac{u(x_0+h,y_0)-u(x_0,y_0)}{h}\right]+i\underset{h\to 0}{\lim }\left[\frac{v(x_0+h,y_0)-v(x_0,y_0)}{h}\right]\\ & ={\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)\end{array}$$

Similarly by considering $z_0=z_0+ih$ where $h\in \mathbb{R}$ and $h\to 0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)& =\underset{h\to 0}{\lim }\left[\frac{u(x_0,y_0+h)+iv(x_0,y_0+h)-u(x_0,y_0)-iv(x_0,y_0)}{ih}\right]\\ & =-i\underset{h\to 0}{\lim }\left[\frac{u(x_0,y_0+h)-u(x_0,y_0)}{h}\right]+\underset{h\to 0}{\lim }\left[\frac{v(x_0,y_0+h)-v(x_0,y_0)}{h}\right]\\ & =-i{\partial }_{y}u(z_0)+{\partial }_{y}v(z_0)\end{array}$$

As $f^{\prime }(z_0)$ is well-defined then

$${\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)=-i{\partial }_{y}u(z_0)+{\partial }_{y}v(z_0)$$

So by comparing the real and imaginary parts of both we get Cauchy-Riemann Equations

Proof - Alternative

Complex Differentiability implies that

$$f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$$

where $ϵ(z)\to 0$ as $z\to z_0$

By rewriting in the expression in real terms (identifying $\mathbb{C}$ with ${\mathbb{R}}^2$) so

$$f^{\prime }(z_0)=r{e}^{i\theta }=r\cos \theta +ir\sin \theta$$

Hence the derivative term $f^{\prime }(z_0)(z-z_0)$ can be rewritten as

$$\left(\begin{array}{c}r\cos \theta (x-x_0)-r\sin \theta (y-y_0)\\ r\cos \theta (y-y_0)+r\sin \theta (x-x_0)\end{array}\right)=\left(\begin{array}{cc}r\cos \theta & -r\sin \theta \\ r\sin \theta & r\cos \theta \end{array}\right)\left(\begin{array}{c}x-x_0\\ y-y_0\end{array}\right)$$

So then the term is just a linear transformation (rotation by angle $\theta$ and scalar factor $r$)

From Total derivative, if $f:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is real differentiable at $z_0=(x_0,y_0)$ then
There exists linear function $L=d{f}_{z_0}:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that

$$f(z)=f(z_0)+L(z-z_0)+R(z)$$

where $\frac{|R(z)|}{|z-z_0|}\to 0$ as $z\to z_0$ and $L$ is in the standard basis given by matrix

$$\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& {\partial }_{y}u(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{y}v(x_0,y_0)\end{array}\right)$$

where ${\partial }_{x}u,{\partial }_{y}u,{\partial }_{x}v,{\partial }_{y}v$ are partial derivatives of $u$ and $v$

Comparing the two matrices then $f$ is real differentiable and partial derivatives exist and

$$\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& {\partial }_{y}u(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{y}v(x_0,y_0)\end{array}\right)=\left(\begin{array}{cc}r\cos \theta & -r\sin \theta \\ r\sin \theta & r\cos \theta \end{array}\right)$$

Hence we immediately get the Cauchy-Riemann equations and that

$$f^{\prime }(z_0)=r\cos \theta +ir\sin \theta ={\partial }_{x}u(x_0,y_0)+i{\partial }_{x}v(x_0,y_0)$$

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02 - Holomorphic Criterion

Holomorphic Criterion

Suppose that $U\subseteq \mathbb{C}$ is open with function $f:U\to \mathbb{C}$
Let the components of $f$ be $(u,v)$ where $u,v:{\mathbb{R}}^2\to \mathbb{R}$

Suppose that all four partial derivatives ${\partial }_{x}u,{\partial }_{y}u,{\partial }_{x}v,{\partial }_{y}v$ exist and continuous in $U$
and that they satisfy the Cauchy-Riemann Equations then

$$f\text{ is holomorphic on }U\text{ with derivative }{\partial }_{x}u+i{\partial }_{x}v$$

Proof Differentiability Criterion then As the partial derivatives are continuous in $U$ then $f$ is real differentiable

By

Hence for $z_0\in U$

$$f(z)=f(z_0)+L(z-z_0)+R(z)$$

where $L=d{f}_{z_0}$ is a real linear transformation with matrix (in standard basis)

$$L=d{f}_{z_0}=\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& {\partial }_{y}u(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{y}v(x_0,y_0)\end{array}\right)$$

and $\frac{|R|}{|z-z_0|}\to 0$

Using the Cauchy-Riemann Equations then it can be rewritten as

$$L=d{f}_{z_0}=\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& -{\partial }_{x}v(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{x}u(x_0,y_0)\end{array}\right)$$

From the Cauchy-Riemann Equations - Alternative Proof then $L$ corresponds to
Complex Multiplication by ${\partial }_{x}u(x_0,y_0)+i{\partial }_{x}v(x_0,y_0)$

Hence

$$f(z)=f(z_0)+({\partial }_{x}u(x_0,y_0)+i{\partial }_{x}v(x_0,y_0))(z-z_0)+R(z)$$

Exactly the same formula as Expansion of a function around a point

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Holomorphic Functions with Zero Derivative Are Constant lemma

Suppose $f:\mathbb{C}\to \mathbb{C}$ and that $f^{\prime }\equiv 0$ then

$$f\text{ is constant}$$

Proof

Let the components of $f$ be $(u,v)$

By Cauchy-Riemann Equations, the partial derivative ${\partial }_{x}u$ exists and is zero

So for fixed $y$ then the function $x\mapsto u(x,y)$ is differentiable with derivative $0$
Similarly ${\partial }_{y}u$ exists and is zero so $u(x,y)$ is constant as a function of $y$ for fixed $x$
Hence for arbitrary $(x,y)$ and $(x^{\prime },y^{\prime })$

$$u(x,y)=u(x^{\prime },y)=u(x^{\prime },y^{\prime })$$

Hence $u$ is constant

By an identical argument then $v$ is constant

Thus $f$ is constant

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1.2.1 Wirtinger Derivatives

Wirtinger Partial Derivatives

Let $f:U\to \mathbb{C}$ be a function with components $(u,v)$ and suppose the partial derivatives exist

Define Wirtinger (partial) derivatives by

$$\begin{array}{rl}{\partial }_{z}f& :=\frac{1}{2}({\partial }_{x}f-i{\partial }_{y}f)=\frac{1}{2}({\partial }_x-i{\partial }_y)u+i\frac{1}{2}({\partial }_x-i{\partial }_y)v\\ {\partial }_{\overline{z}}f& :=\frac{1}{2}({\partial }_{x}f+i{\partial }_{y}f)=\frac{1}{2}({\partial }_x+i{\partial }_y)u+i\frac{1}{2}({\partial }_x+i{\partial }_y)v\end{array}$$

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Relation between Wirtinger Partial Derivatives and Cauchy-Riemann Equations

Let $U$ be an open subset of $\mathbb{C}$
Let $f:U\to \mathbb{C}$ then

$$f\text{ satisfies the Cauchy-Riemann Equations }⟺{\partial }_{\overline{z}}f=0$$

Moreover we also get

$$f^{\prime }={\partial }_{z}f$$

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1.3 Harmonic Functions

Laplacian

Suppose that $u:{\mathbb{R}}^2\to \mathbb{R}$ be a function on some open set $U\subseteq {\mathbb{R}}^2$ which is twice differentiable

Define the Laplacian

$$\Delta u={\partial }_{xx}u+{\partial }_{yy}u$$

where ${\partial }_{xx}u={\partial }_x({\partial }_{x}u)={\partial }_x^{2}u$ and ${\partial }_{yy}u={\partial }_y({\partial }_{y}u)={\partial }_y^{2}u$

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Harmonic

Supposez $u:{\mathbb{R}}^2\to \mathbb{R}$ be a function on some open set $U\subseteq {\mathbb{R}}^2$ which is twice differentiable

$u$ is harmonic if

$$\Delta u=0$$

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03 - Holomorphic Implies Harmonic

Holomorphic Implies Harmonic

Let $U\subseteq \mathbb{C}$ be open, and suppose that $f:U\to \mathbb{C}$ is holomorphic
Let components of $f$ be $(u,v)$, and suppose both are twice continuously differentiable then

$$u\text{ and }v\text{ are harmonic}$$

Proof

From the Holomorphic Implies Harmonic then

$${\partial }_{xx}u={\partial }_{xy}v(={\partial }_x{\partial }_{y}v),{\partial }_{yy}u=-{\partial }_{yx}v$$

However under the stated conditions, the symmetry property of partial derivatives apply

$${\partial }_{xy}v={\partial }_{yx}v$$

Hence the result follows

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Harmonic Conjugates

Let $u,v:{\mathbb{R}}^2\to \mathbb{R}$ be harmonic functions

If $f(z)=u(z)+iv(z)$ is holomorphic then

$$u\text{ and }v\text{ are harmonic conjugates}$$

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1.4 Power Series

Radius of Convergence

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series
Let $S$ be the set of $z\in \mathbb{C}$ at which it converges

Radius of Convergence of the Power Series is

$$\sup \{|z|:z\in \mathcal{S}\}$$

or $\infty$ is the set $S$ is unbounded

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Property of the Radius of Convergence

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series
Let $S$ be the set of $z\in \mathbb{C}$ at which it converges and $R$ be it’s radius of convergence then

$$B(0,R)\subseteq S\subseteq \overline{B}(0,R)$$

With series converging absolutely on $B(0,R)$ and if $0\le r\le R$ then it converges uniformly on $\overline{B}(0,r)$ with

$$\frac{1}{R}=\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}$$

Proof

Containment $S\subseteq \overline{B}(0,R)$ is immediate from definition of the radius of convergence

Other containment $B(0,R)\subseteq S$ with series converging absolutely on $B(0,R)$
is a consequence of the series converging uniformly on $\overline{B}(0,r)$ when $0\le r<R$ as

$$B(0,R)=\bigcup _{r<R}\overline{B}(0,r)$$

By definition of $R$ there is some $w$ with $|w|>r$ such that

$$\sum _{n=0}^{\infty }{a}_n{w}^n\text{ converges}$$

Hence the terms of the sums are bounded

$$|a_n{w}^n|\le M\text{ for some }M$$

But then if $|z|\le r$ then

$$|a_n{z}^n|=|a_n{w}^n|{\left|\frac{z}{w}\right|}^n\le M{\left|\frac{r}{w}\right|}^n$$

As the geometric series ${\sum }_{n=0}^{\infty }{\left|\frac{r}{w}\right|}^n$ converges since $|w|>r$
Therefore by Weierstrass Test

$$\sum _{n=0}^{\infty }{a}_n{z}^n\text{ converges uniformly for }|z|<r$$

Suppose the radius of convergence is $R$
Let $0\le r<R$ then by the above there is $w$ with $|w|>r$ such that

$$|a_n{w}^n|\le M\text{ for all }n$$

Assume that $M\ne 0$ then by taking $n$th roots

$$|a_n{|}^{1/n}|w|\le M^{1/n}$$

Since $M^{1/n}\to 1$ as $n\to \infty$ then

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\le \frac{1}{|w|}<\frac{1}{r}$$

Since $r<R$ was arbitrary then it follows that

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\le \frac{1}{R}$$

Note that it when $R=\infty$ with $\frac{1}{R}=0$ in this case

For the other direction, suppose that ${\mathrm{limsup}}_n|a_n{|}^{1/n}=L$ and that $L\in (0,\infty )$
If $L^{\prime }>L$ then

$$|a_n{|}^{1/n}\le L^{\prime }\text{ for all sufficiently large }n$$

Therefore

$$|a_n{z}^n|\le |L^{\prime }z{|}^n\text{ for sufficiently large }n$$

By the geometric series formula then

$$\sum _{n=0}^{\infty }{a}_n{z}^n\text{ converges provided }|z|<\frac{1}{L^{\prime }}$$

Therefore

$$R\ge \frac{1}{L^{\prime }}$$

Since $L^{\prime }>L$ was arbitrary then

$$R\ge \frac{1}{L}$$

Hence

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\ge \frac{1}{R}$$

Minimal changes to argument for when $L=0$ or $L=\infty$

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Combining Power Series lemma

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ and ${\sum }_{n=0}^{\infty }{b}_n{z}^n$ be power series of radii of convergence $R_1$ and $R_2$ respectively

For $|z|<\min \{R_1,R_2\}$ let $s(z),t(z)$ be the function to which these series converge to then

1. ${\sum }_{n=0}^{\infty }(a_n+b_n)z^n$ converges for $|z|<\min \{R_1,R_2\}$ to

$$s(z)+t(z)$$

2. ${\sum }_{n=0}^{\infty }\left({\sum }_{k+l=n}{a}_k{b}_l\right)z^n$ converges for $|z|<\min \{R_1,R_2\}$ to

$$s(z)t(z)$$

Note that $\min \{R_1,R_2\}$ is only a lower bound for the radii of convergence in each case0

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Differentiation of Power Series

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series, with radius of convergence $R$
Let $s(t)$ be the function to which this series converges on $B(0,R)$ then

Power Series

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

also has radius of convergence $R$

Additionally on $B(0,R)$ then power series $s$ is complex differentiable with

$$s^{\prime }(z)=t(z)$$

In particular, a power series is infinitely differentiable within its radius of convergence

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1.4.1 Power Series about Other Points

Power Series about other Points

Power Series about points $z_0$ are functions given by an expression of form

$$f(z)=\sum _{n=0}^{\infty }{a}_n(z-z_0{)}^n$$

with

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

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Taylor Series of Power Series

Let $f(z)={\sum }_{n=0}^{\infty }{a}_n(z-z_0{)}^n$ be a function given by power series with radius of convergence $R$

As radius of convergence for the derivative $f^{\prime }(z)={\sum }_{n=1}^{\infty }n{a}_n(z-z_0{)}^{n-1}$ is at least $R>0$ then

$$f^{\prime }\text{ is also holomorphic in }B(z_0,R)$$

By induction then all derivatives of $f$ as well

Moreover as $f(z_0)=a_0,f^{\prime }(z_0)=a_1$ etc then

$$f(z)=\sum _{n=0}^{\infty }\frac{f^{(n)}(z_0)}{n!}(z-z_0{)}^n$$

so $f$ is given by it’s Taylor Series and is called analytic

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1.4.2 The Exponential and Trigonometric Functions

Exponential Function

$$e^z=\exp (z)=\sum _{n=0}^{\infty }\frac{z^n}{n!}$$

with

$${\exp }^{\prime }=\exp \text{ and }{e}^{iz}=\exp (iz)=\cos (z)+i\sin (z)$$

Holomorphic on all of $\mathbb{C}$ and derivatives given by term-by-term differentiation of series

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Cosine Function

$$\cos (z)=\sum _{n=0}^{\infty }(-1{)}^n\frac{z^{2n}}{(2n)!}$$

with

$${\cos }^{\prime }=-\sin$$

Holomorphic on all of $\mathbb{C}$ and derivatives given by term-by-term differentiation of series

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Sine Function

$$\sin (z)=\sum _{n=0}^{\infty }(-1{)}^n\frac{z^{2n+1}}{(2n+1)!}$$

with

$${\sin }^{\prime }=\cos$$

Holomorphic on all of $\mathbb{C}$ and derivatives given by term-by-term differentiation of series

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1.4.3 Properties of Exponential and Trigonometric Functions

Properties of Exponential and Trigonometric Functions

$$\sin (-z)=\sin (z)\cos (-z)=\cos (z)$$

with

$$\cos (z)=\frac{e^{iz}+e^{-iz}}{2}\sin (z)=\frac{e^{iz}-e^{-iz}}{2}$$

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Property of Exponential Multiplication

For $z,w\in \mathbb{C}$

$$\exp (z+w)=\exp (z)\exp (w)$$

Proof

For fixed $a\in \mathbb{C}$ then
Consider

$$f(z)=\exp (z)\exp (a-z)$$

By differentiation and using product rule then

$$f^{\prime }(z)=\exp (z)\exp (a-z)-\exp (z)\exp (a-z)=0$$

So by Holomorphic functions with zero derivative are constant then

$$f(z)=f(a)=\text{constant}$$

Hence

$$\exp (z)\exp (z-a)=\exp (a)$$

Substituting $a=z+w$ gets the final result

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Property of Complex Exponentials

For $x,y\in \mathbb{R}$ then

$$e^{x+iy}=e^x(\cos y+i\sin y)$$

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1.4.4 Logarithms and Powers

Simplified Inverse Function Theorem

Let $f:U\to \mathbb{C}$ be a holomorphic function
Let $z_0$ be a point in $U$ and assume that there is $V\subset U$ such that

$$f\text{ is one-to-one on }V\to f(V)=W$$

and that the inverse function

$$g=f^{-1}:W\to V\text{ is continuous}$$

Then

$g$ is differentiable at $w_0=f(z_0)$ with

$$g^{\prime }(w_0)=\frac{1}{f^{\prime }(z_{0}0)}$$

Proof

Consider $f(z)=w\to w_0=f(z_0)$ so

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

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Log Function

Let

$$D=\mathbb{C}\setminus (-\infty ,0]$$

Define function $\mathrm{Log}:D\to \mathbb{C}$
If $z=|z|e^{i\theta }$ for $\theta \in (-\pi ,\pi ]$ then define

$$\mathrm{Log}(z):=\log |z|+i\theta$$

So $\mathrm{Log}$ is holomorphic on $D$ and

$${\mathrm{Log}}^{\prime }(z)=\frac{1}{z}$$

Proof - Derivative

As $|z|$ and $\log$ is continuous then the real part of $\mathrm{Log}$ is continuous

Need to show that $\theta =\theta (z)$ is a continuous function of $z$
By the law of cosines then

$$\cos (\theta (z+h)-\theta (z))=\frac{|z{|}^2+|z+h{|}^2-|h{|}^2}{2|z||z+h|}$$

For all $|z|>0$ we have that

$$\cos (\theta (z+h)-\theta (z))\to 1\text{ as }h\to 0$$

Since $z\in D$ and $\theta \in (-\pi ,\pi )$ then this implies that

$$\theta (z+h)\to \theta (z)$$

Hence $\mathrm{Log}$ is continuous

Exclude negative real line from $D$ as argument would be $\pi$ so $\theta$ would not be continuous

By Simplified inverse function theorem then $\mathrm{Log}$ is differentiable and it’s inverse if

$$\frac{1}{{\exp }^{\prime }(\mathrm{Log}(z))}=\frac{1}{\exp (Log(z))}=\frac{1}{z}$$

Since this works for any $z\in D$ then $\mathrm{Log}$ is holomorphic in $D$ and it’s derivative is

$$\frac{1}{z}$$

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Principal Value

Values $\theta \in (-\pi ,\pi ]$ such that

$$z=|z|e^{i\theta }$$

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Principal Value of Logarithm

$\mathrm{Log}$ used in Log function

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Principal Value of $z^{\alpha }$

Let $\alpha ,z\in \mathbb{C}$ and $z\ne 0$ then

Principal Value of $z^{\alpha }$ is

$$\exp (\alpha \mathrm{Log}(z))$$

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1.5 Branch Cuts and Multifunctions

Multi-Valued Function / Multifunction

Multi-valued function or Multifunction on a subset $U\subseteq \mathbb{C}$ is map

$$f:U\to \mathcal{P}(\mathbb{C})$$

assigning to each point in $U$ a subset of the complex numbers (refer to Power set)

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Branch

Let $f:U\to \mathcal{P}(\mathbb{C})$ be a multifunction

Branch of $f$ on subset $V\subseteq U$ is a function

$$g:V\to \mathbb{C}\text{ such that }g(z)\in f(z)\text{ for all }z\in V$$

If $g$ is continuous on $V$ then it is a continuous branch of $f$
If $g$ is holomorphic on $V$ then it is a holomorphic branch of $f$

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Branch Point

Suppose that $f:U\to \mathcal{P}(\mathbb{C})$ is a multi-valued function defined on an open subset $U$ of $\mathbb{C}$

$z_0\in U$ is not a Branch Point of $f$ if there is an open disk $D\subseteq U$ containing $z_0$ such that
There exists a holomorphic branch of $f$ defined on $D\setminus \{z_0\}$

Otherwise $z_0$ is called a Branch Point

When $\mathbb{C}\setminus U$ is bounded then $f$ does not have a branch point at $\infty$ if
There is a holomorphic branch of $f$ defined on $\mathbb{C}\setminus B(0,R)\subseteq U$ for some $r>0$
Otherwise $\infty$ is a branch point of $f$

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Power Set

$\mathcal{P}(X)$ is the power set of $X$ so the set of all subsets of $X$

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Notation for Multi-functions

Use square brackets to denote multifunction

$z^{1/2}$ Consider $z\mapsto z^{1/2}$ as a multifunction then write

$$[z^{1/2}]=\{w\in \mathbb{C}:w^2=z\}$$

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Properties of Log and Powers

For $\alpha ,z,w\in \mathbb{C}$ with $z,w\ne 0$ then

$$\begin{array}{rl}[\log z]+[\log w]& =[\log zw]\\ [z^{\alpha }][w^{\alpha }]& =[(zw{)}^{\alpha }]\end{array}$$

Note that in general $[z^{\alpha }][z^{\beta }]\ne [z^{\alpha +\beta }]$

Proof

Let $\theta$ and $\varphi$ be some values of $\arg (z)$ and $\arg (w)$ then

$$\begin{array}{rl}[\log z]& =\{\log |z|+i\theta +2\pi in:n\in \mathbb{Z}\}\\ [\log w]& =\{\log |w|+i\varphi +2\pi ik:k\in \mathbb{Z}\}\end{array}$$

By adding these sets term by term then

$$\begin{array}{rl}[\log z]+[\log w]& =\{\log |z|+\log |w|+i(\theta +\varphi )+2\pi i(k+n):k,n\in \mathbb{Z}\}\\ & =\{\log |zw|+i(\theta +\varphi )+2\pi im:m\in \mathbb{Z}\}\\ & =[\log (zw)]\end{array}$$

Similarly

$$\begin{array}{rl}[z^{\alpha }]& =\{|z{|}^{\alpha }\exp (ia\theta +i\alpha 2\pi in)\}\\ [w^{\alpha }]& =\{|w{|}^{\alpha }\exp (ia\varphi +i\alpha 2\pi ik)\}\end{array}$$

Multiplying these sets term by term then

$$\begin{array}{rl}[z^{\alpha }][w^{\alpha }]& =\{|zw{|}^{\alpha }\exp (i\alpha (\theta +\varphi )+i\alpha 2\pi i(n+k)):k,n\in \mathbb{Z}\}\\ & =\{|zw{|}^{\alpha }\exp (i\alpha (\theta +\varphi )+i\alpha 2\pi im):m\in \mathbb{Z}\}\\ & =[(zw{)}^{\alpha }]\end{array}$$

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