Complex Analysis/Exercises/Sheet 2/Exercise 4

Problem (Chain Rule, 5 Points)

Let $f,g:𝐂\to 𝐂$ be continuously differentiable functions. Prove that

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

and

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

apply.

Solution

We recall (Fischer/Lieb, Page 21 at the bottom): For a differentiable function$f:U\to 𝐂$ , the partial derivatives with respect to $z$ and $\overline{z}$ are characterized as follows: Let $A_1,A_2:U\to 𝐂$ be continuous functions such that

$f(z)=f(z_0)+A_1(z)(z-z_0)+A_2(z)(\overline{z}-{\overline{z}}_0)$

so is ${\partial }_{z}f(z_0)=A_1(z_0)$ and ${\partial }_{\overline{z}}f(z_0)=A_2(z_0)$.

We will use this description of the Wirtinger derivatives. Let $z_0\in 𝐂$. There $g$ in $z_0$ is differentiable , we have continuous functions$B_1,B_2$ so that

$g(z)=g(z_0)+B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0)$

applies.This means

$\begin{array}{cc}(f\circ g)(z)& =f(g(z))\\ & =f(g(z_0)+B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0))\end{array}$

Now set

$\begin{array}{cc}w& :=g(z_0)+B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0)=g(z)\\ w_0& :=g(z_0)\end{array}$

Da $f$ is $w_0$ differentiable ,there exist continuous functions $A_1$, $A_2$ so,that

$f(w)=f(w_0)+A_1(w)(w-w_0)+A_2(w)(\overline{w}-{\overline{w}}_0)$

if we insert, this results in

$\begin{array}{cc}(f\circ g)(z)& =f(g(z))\\ & =f(g(z_0)+B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0))\\ & =f(w_0)+A_1(w)(w-w_0)+A_2(w)(\overline{w}-{\overline{w}}_0)\\ & =f(g(z_0))+A_1(g(z))(B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0))+A_2(g(z))\left(\overline{B_1(z)(z-z_0)+B_2(z)(\overline{z}-{\overline{z}}_0)}\right)\\ & =(f\circ g)(z_0)+\underset{C_1(z):=}{\underbrace{\left(A_1\right(g(z))B_1(z)+A_2(g(z)\left)\overline{B_2(z)}\right)}}(z-z_0)+\underset{C_2(z):=}{\underbrace{\left(A_1\right(g(z))B_2(z)+A_2(g(z)\left)\overline{B_1(z)}\right)}}(\overline{z}-{\overline{z}}_0)\end{array}$

Da $C_1$ and $C_2$ of cotinous functions are continuous,is $f\circ g$ partially differentiable and

$\begin{array}{cc}\frac{\partial }{\partial z}(f\circ g)(z_0)& =C_1(z_0)\\ & =A_1(g(z_0))B_1(z_0)+A_2(g(z_0))\overline{B_2(z_0)}\\ & =\frac{\partial f}{\partial z}(w_0)\frac{\partial g}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(w_0)\overline{\frac{\partial g}{\partial \overline{z}}(z_0)}\end{array}$

Continuing above,this lead to

$\begin{array}{cc}\frac{\partial }{\partial z}(f\circ g)(z_0)& =C_1(z_0)\\ & =A_1(g(z_0))B_1(z_0)+A_2(g(z_0))\overline{B_2(z_0)}\\ & =\frac{\partial f}{\partial z}(w_0)\frac{\partial g}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(w_0)\overline{\frac{\partial g}{\partial \overline{z}}(z_0)}\\ & =\frac{\partial f}{\partial z}(w_0)\frac{\partial g}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}(w_0)\frac{\partial \overline{g}}{\partial z}(z_0)\\ & =\frac{\partial f}{\partial z}\circ g(z_0)\frac{\partial g}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}\circ g(z_0)\frac{\partial \overline{g}}{\partial z}(z_0)\end{array}$

and claimed. Analogously follows

$\begin{array}{cc}\frac{\partial }{\partial \overline{z}}(f\circ g)(z_0)& =C_2(z_0)\\ & =A_1(g(z_0))B_2(z_0)+A_1(g(z_0))\overline{B_1(z_0)}\\ & =\frac{\partial f}{\partial z}(w_0)\frac{\partial g}{\partial \overline{z}}(z_0)+\frac{\partial f}{\partial \overline{z}}(w_0)\overline{\frac{\partial g}{\partial z}(z_0)}\\ & =\frac{\partial f}{\partial z}(w_0)\frac{\partial g}{\partial \overline{z}}(z_0)+\frac{\partial f}{\partial \overline{z}}(w_0)\frac{\partial \overline{g}}{\partial \overline{z}}(z_0)\\ & =\frac{\partial f}{\partial z}\circ g(z_0)\frac{\partial g}{\partial z}(z_0)+\frac{\partial f}{\partial \overline{z}}\circ g(z_0)\frac{\partial \overline{g}}{\partial \overline{z}}(z_0)\end{array}$

Translation and Version Control

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• Source: Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_4 - URL:

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• Date: 01/14/2024

de:Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_4