Complex Analysis/Complex Functions/Complex Derivatives

Complex differentiability

Let us now define what complex differentiability is.

Example 2.3.2

The function

$f:ℂ\to ℂ,f(z)=\overline{z}$

is nowhere complex differentiable.

Proof

Let $z_0\in ℂ$ be arbitrary. Assume that $f$ is complex differentiable at $z_0$ , i.e. that

$\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in ℂ}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}$

exists.

We choose

$\begin{array}{c}A:=\{z\in ℂ|\mathrm{\Re }z=\mathrm{\Re }{z}_0\}\\ B:=\{z\in ℂ|\mathrm{\Im }z=\mathrm{\Im }{z}_0\}\end{array}$

Due to lemma 2.2.3, which is applicable since of course $ℂ$ is open, we have:

$\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in A}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}=\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in ℂ}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}=\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in B}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}$

But

$\begin{array}{cc}& \underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in A}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}=\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in A}}{\lim }\frac{\mathrm{\Re }(z-z_0)-i\mathrm{\Im }(z-z_0)}{\mathrm{\Re }(z-z_0)+i\mathrm{\Im }(z-z_0)}=-1\\ \\ & \underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in B}}{\lim }\frac{\overline{z}-{\overline{z}}_0}{z-z_0}=\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in B}}{\lim }\frac{\mathrm{\Re }(z-z_0)-i\mathrm{\Im }(z-z_0)}{\mathrm{\Re }(z-z_0)+i\mathrm{\Im }(z-z_0)}=1\end{array}$

a contradiction.$◻$

The Cauchy–Riemann equations

We can define a natural bijective function from $ℂ$ to ${ℝ}^2$ as follows:

$\mathrm{\Phi }(x+yi):=(x,y)$

In fact, $\mathrm{\Phi }$ is a vector space isomorphism between ${ℂ}^1$ and ${ℝ}^2$ .

The inverse of $\mathrm{\Phi }$ is given by

${\mathrm{\Phi }}^{-1}:{ℝ}^2\to ℂ,{\mathrm{\Phi }}^{-1}(x,y)=x+yi$

Proof

1. We prove well-definedness of $u,v$ .

Let $(x,y)\in \mathrm{\Phi }(O)$ . We apply the inverse function on both sides to obtain:

$x+yi\in {\mathrm{\Phi }}^{-1}(\mathrm{\Phi }(O))=O$

where the last equality holds since $\mathrm{\Phi }$ is bijective (for any bijective $f:S_1\to S_2$ we have $f^{-1}(f(S_3))=f(f^{-1}(S_3))=S_3$ if $S_3\subseteq S_1$ ; see exercise 1).

3. We prove differentiability of $u$ and $v$ and the Cauchy-Riemann equations.

We define

$\begin{array}{c}{S}_1:=\{z\in ℂ:\mathrm{\Re }(z)=\mathrm{\Re }(z_0)\}\cap O\\ S_2:=\{z\in ℂ:\mathrm{\Im }(z)=\mathrm{\Im }(z_0)\}\cap O\end{array}$

Then we have:

$\begin{array}{ccc}{\partial }_{x}u(x_0,y_0)& =\underset{x\to x_0}{\lim }\frac{u(x,y_0)-u(x_0,y_0)}{x-x_0}& \\ & =\underset{x\to x_0}{\lim }\frac{\mathrm{\Re }(f(x+y_{0}i))-\mathrm{\Re }(f(x_0+y_{0}i))}{x-x_0}& \\ & =\mathrm{\Re }\left(\underset{x\to x_0}{\lim }\frac{f(x+y_{0}i)-f(x_0+y_{0}i)}{x-x_0}\right)& \text{continuity of }\mathrm{\Re }\\ & =\mathrm{\Re }\left(\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in S_2}}{\lim }\frac{f(z)-f(z_0)}{z-z_0}\right)& \\ & =\mathrm{\Re }\left(\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in S_1}}{\lim }\frac{f(z)-f(z_0)}{z-z_0}\right)& \text{lemma 2.2.3}\\ & =\mathrm{\Re }\left(\underset{y\to y_0}{\lim }\frac{f(x_0+yi)-f(x_0+y_{0}i)}{yi-y_{0}i}\right)& \\ & =\mathrm{\Re }((-i)\underset{y\to y_0}{\lim }\frac{f(x_0+yi)-f(x_0+y_{0}i)}{y-y_0})& i^{-1}=-i\\ & =\mathrm{\Im }\left(\underset{y\to y_0}{\lim }\frac{f(x_0+yi)-f(x_0+y_{0}i)}{y-y_0}\right)& \\ & ={\partial }_{y}v(x_0,y_0)\end{array}$

From these equations follows the existence of ${\partial }_{x}u(x_0,y_0),{\partial }_{y}v(x_0,y_0)$ , since for example

$\underset{\genfrac{}{}{0ex}{}{z\to z_0}{z\in S_2}}{\lim }\frac{f(z)-f(z_0)}{z-z_0}$

exists due to lemma 2.2.3.

The proof for

${\partial }_{y}u(x_0,y_0)=-{\partial }_{x}v(x_0,y_0)$

and the existence of ${\partial }_{y}u(x_0,y_0),{\partial }_{x}v(x_0,y_0)$ we leave for exercise 2.$◻$

Exercises

1. Let $S_1,S_2,S_3$ be sets such that $S_3\subseteq S_1$ , and let $f:S_1\to S_2$ be a bijective function. Prove that $f^{-1}(f(S_3))=f(f^{-1}(S_3))=S_3$ .
2. Let $O\subseteq ℂ$ be open, let $f:O\to ℂ$ be a function and let $z_0=x_0+y_{0}i\in O$ . Prove that if $f$ is complex differentiable at $z_0$ , then ${\partial }_{y}u(x_0,y_0)$ and ${\partial }_{x}v(x_0,y_0)$ exist and satisfy the equation ${\partial }_{y}u(x_0,y_0)=-{\partial }_{x}v(x_0,y_0)$ .

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