01 - 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 $a$

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

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

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 $z\to a$

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

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

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

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

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

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

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 ${\partial }_{yy}u={\partial }_y({\partial }_{y}u)={\partial }_y^{2}u$

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

$$\Delta u=0$$

Harmonic Conjugates

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

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

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

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

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

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

Entire

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