Complex Analysis/Exercises/Sheet 2/Exercise 4

Let ${\displaystyle f,g\colon \mathbf {C} \to \mathbf {C} }$ be continuously differentiable functions. Prove that

${\displaystyle {\frac {\partial }{\partial z}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial z}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial z}}}$

and

${\displaystyle {\frac {\partial }{\partial {\bar {z}}}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial {\bar {z}}}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}}$

apply.

We recall (Fischer/Lieb, Page 21 at the bottom): For a differentiable function${\displaystyle f\colon U\to \mathbf {C} }$ , the partial derivatives with respect to ${\displaystyle z}$ and ${\displaystyle {\bar {z}}}$ are characterized as follows: Let ${\displaystyle A_{1},A_{2}\colon U\to \mathbf {C} }$ be continuous functions such that

${\displaystyle f(z)=f(z_{0})+A_{1}(z)(z-z_{0})+A_{2}(z)({\bar {z}}-{\bar {z}}_{0})}$

so is ${\displaystyle \partial _{z}f(z_{0})=A_{1}(z_{0})}$ and ${\displaystyle \partial _{\bar {z}}f(z_{0})=A_{2}(z_{0})}$.

We will use this description of the Wirtinger derivatives. Let ${\displaystyle z_{0}\in \mathbf {C} }$. There ${\displaystyle g}$ in ${\displaystyle z_{0}}$ is differentiable , we have continuous functions${\displaystyle B_{1},B_{2}}$ so that

${\displaystyle g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}$

applies.This means

${\displaystyle {\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\end{array}}}$

Now set

${\displaystyle {\begin{array}{rl}w&:=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})=g(z)\\w_{0}&:=g(z_{0})\end{array}}}$

Da ${\displaystyle f}$ is ${\displaystyle w_{0}}$ differentiable ,there exist continuous functions ${\displaystyle A_{1}}$, ${\displaystyle A_{2}}$ so,that

${\displaystyle f(w)=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})}$

if we insert, this results in

${\displaystyle {\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\\&=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})\\&=f{\bigl (}g(z_{0}){\bigr )}+A_{1}{\bigl (}g(z){\bigr )}{\bigl (}B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}+A_{2}{\bigl (}g(z){\bigr )}{\bigl (}{\overline {B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}}{\bigr )}\\&=(f\circ g)(z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{1}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{2}(z)}}{\Bigl )}} _{C_{1}(z):=}(z-z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{2}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{1}(z)}}{\Bigl )}} _{C_{2}(z):=}({\bar {z}}-{\bar {z}}_{0})\end{array}}}$

Da ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ of cotinous functions are continuous,is ${\displaystyle f\circ g}$ partially differentiable and

${\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\end{array}}}$

Continuing above,this lead to

${\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\\end{array}}}$

and claimed. Analogously follows

${\displaystyle {\begin{array}{rl}{\frac {\partial }{\partial {\bar {z}}}}(f\circ g)(z_{0})&=C_{2}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{2}(z_{0})+A_{1}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{1}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial z}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\\end{array}}}$

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• 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