# Complex Analysis/Exercises/Sheet 2/Exercise 4
#### Problem (Chain Rule, 5 Points)

Let $f,g\colon \mathbf {C} \to \mathbf {C}$ be continuously differentiable functions. Prove that
${\frac {\partial }{\partial z}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial z}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial z}}$
and
${\frac {\partial }{\partial {\bar {z}}}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial {\bar {z}}}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}$
apply.

#### Solution

We recall (Fischer/Lieb, Page 21 at the bottom): For a differentiable function$f\colon U\to \mathbf {C}$ , the partial derivatives with respect to $z$ and ${\bar {z}}$ are characterized as follows: Let $A_{1},A_{2}\colon U\to \mathbf {C}$ be continuous functions such that
$f(z)=f(z_{0})+A_{1}(z)(z-z_{0})+A_{2}(z)({\bar {z}}-{\bar {z}}_{0})$
so is $\partial _{z}f(z_{0})=A_{1}(z_{0})$ and $\partial _{\bar {z}}f(z_{0})=A_{2}(z_{0})$.
We will use this description of the Wirtinger derivatives. Let $z_{0}\in \mathbf {C}$. There $g$ in $z_{0}$ is differentiable , we have continuous functions$B_{1},B_{2}$ so that
$g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})$
applies.This means
${\begin{array}{rl}(f\circ g)(z)&=f{\biglg(z){\bigr )}\\&=f{\biglg(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\end{array}}$
Now set
${\begin{array}{rl}w&:=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})=g(z)\\w_{0}&:=g(z_{0})\end{array}}$
Da $f$ is $w_{0}$ differentiable ,there exist continuous functions $A_{1}$, $A_{2}$ so,that
$f(w)=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})$
if we insert, this results in
${\begin{array}{rl}(f\circ g)(z)&=f{\biglg(z){\bigr )}\\&=f{\biglg(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\\&=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})\\&=f{\biglg(z_{0}){\bigr )}+A_{1}{\biglg(z){\bigr )}{\biglB_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}+A_{2}{\biglg(z){\bigr )}{\bigl{\overline {B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}}{\bigr )}\\&=(f\circ g)(z_{0})+\underbrace {{\BiglA_{1}{\biglg(z){\bigr )}B_{1}(z)+A_{2}{\biglg(z){\bigr )}{\overline {B_{2}(z)}}{\Bigl )}} _{C_{1}(z):=}(z-z_{0})+\underbrace {{\BiglA_{1}{\biglg(z){\bigr )}B_{2}(z)+A_{2}{\biglg(z){\bigr )}{\overline {B_{1}(z)}}{\Bigl )}} _{C_{2}(z):=}({\bar {z}}-{\bar {z}}_{0})\end{array}}$
Da $C_{1}$ and $C_{2}$ of cotinous functions are continuous,is $f\circ g$ partially differentiable and
${\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\biglg(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\biglg(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\end{array}}$
Continuing above,this lead to
${\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\biglg(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\biglg(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\\end{array}}$
and claimed. Analogously follows
${\begin{array}{rl}{\frac {\partial }{\partial {\bar {z}}}}(f\circ g)(z_{0})&=C_{2}(z_{0})\\&=A_{1}{\biglg(z_{0}){\bigr )}B_{2}(z_{0})+A_{1}{\biglg(z_{0}){\bigr )}{\overline {B_{1}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial z}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\\end{array}}$

### Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
* Source: Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_4 - URL:
https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_4
* Date: 01/14/2024