Complex Analysis/Complex Functions/Complex Derivatives

Complex differentiability
Let us now define what complex differentiability is.

The function
 * Example 2.3.2
 * $$f:\Complex\to\Complex,f(z)=\bar z$$

is nowhere complex differentiable.

Let $$z_0\in\Complex$$ be arbitrary. Assume that $$f$$ is complex differentiable at $$z_0$$, i.e. that
 * Proof
 * $$\lim_{z\to z_0\atop z\in\Complex}\frac{\bar z-\bar{z}_0}{z-z_0}$$

exists.

We choose
 * $$\begin{align}A:=\{z\in\Complex|\Re z=\Re z_0\}\\B:=\{z\in\Complex|\Im z=\Im z_0\}\end{align}$$

Due to lemma 2.2.3, which is applicable since of course $$\Complex$$ is open, we have:
 * $$\lim_{z\to z_0\atop z\in A}\frac{\bar z-\bar{z}_0}{z-z_0}=\lim_{z\to z_0\atop z\in\Complex}\frac{\bar z-\bar{z}_0}{z-z_0}=\lim_{z\to z_0\atop z\in B}\frac{\bar z-\bar{z}_0}{z-z_0}$$

But
 * $$\begin{align}&\lim_{z\to z_0\atop z\in A}\frac{\bar z-\bar{z}_0}{z-z_0}=\lim_{z\to z_0\atop z\in A}\frac{\Re(z-z_0)-i\Im(z-z_0)}{\Re(z-z_0)+i\Im(z-z_0)}=-1\\\\

&\lim_{z\to z_0\atop z\in B}\frac{\bar z-\bar{z}_0}{z-z_0}=\lim_{z\to z_0\atop z\in B}\frac{\Re(z-z_0)-i\Im(z-z_0)}{\Re(z-z_0)+i\Im(z-z_0)}=1\end{align}$$ a contradiction.

The Cauchy–Riemann equations
We can define a natural bijective function from $$\Complex$$ to $$\R^2$$ as follows:
 * $$\Phi(x+yi):=(x,y)$$

In fact, $$\Phi$$ is a vector space isomorphism between $$\Complex^1$$ and $$\R^2$$.

The inverse of $$\Phi$$ is given by
 * $$\Phi^{-1}:\R^2\to\Complex,\Phi^{-1}(x,y)=x+yi$$

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

Let $$(x,y)\in\Phi(O)$$. We apply the inverse function on both sides to obtain:
 * $$x+yi\in\Phi^{-1}(\Phi(O))=O$$

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

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

We define
 * $$\begin{align}S_1:=\{z\in\Complex:\Re(z)=\Re(z_0)\}\cap O\\S_2:=\{z\in\Complex:\Im(z)=\Im(z_0)\}\cap O\end{align}$$

Then we have:
 * $$\begin{align}

\partial_xu(x_0,y_0)&=\lim_{x\to x_0}\frac{u(x,y_0)-u(x_0,y_0)}{x-x_0}&\\ &=\lim_{x\to x_0}\frac{\Re\bigl(f(x+y_0i)\bigr)-\Re\bigl(f(x_0+y_0i)\bigr)}{x-x_0}&\\ &=\Re\left(\lim_{x\to x_0}\frac{f(x+y_0i)-f(x_0+y_0i)}{x-x_0}\right)&\text{continuity of }\Re\\ &=\Re\left(\lim_{z\to z_0\atop z\in S_2}\frac{f(z)-f(z_0)}{z-z_0}\right)&\\ &=\Re\left(\lim_{z\to z_0\atop z\in S_1}\frac{f(z)-f(z_0)}{z-z_0}\right)&\text{lemma 2.2.3}\\ &=\Re\left(\lim_{y\to y_0}\frac{f(x_0+yi)-f(x_0+y_0i)}{yi-y_0i}\right)&\\ &=\Re\left((-i)\lim_{y\to y_0}\frac{f(x_0+yi)-f(x_0+y_0i)}{y-y_0}\right)&i^{-1}=-i\\ &=\Im\left(\lim_{y\to y_0}\frac{f(x_0+yi)-f(x_0+y_0i)}{y-y_0}\right)&\\ &=\partial_yv(x_0, y_0) \end{align}$$ From these equations follows the existence of $$\partial_xu(x_0,y_0),\partial_yv(x_0,y_0)$$, since for example
 * $$\lim_{z\to z_0\atop z\in S_2}\frac{f(z)-f(z_0)}{z-z_0}$$

exists due to lemma 2.2.3.

The proof for
 * $$\partial_yu(x_0,y_0)=-\partial_xv(x_0,y_0)$$

and the existence of $$\partial_yu(x_0,y_0),\partial_xv(x_0,y_0)$$ we leave for exercise 2.

Exercises

 * 1) Let $$S_1,S_2,S_3$$ be sets such that $$S_3\sube S_1$$, and let $$f:S_1\to S_2$$ be a bijective function. Prove that $$f^{-1}\bigl(f(S_3)\bigr)=f\bigl(f^{-1}(S_3)\bigr)=S_3$$.
 * 2) Let $$O\sube\Complex$$ be open, let $$f:O\to\Complex$$ be a function and let $$z_0=x_0+y_0i\in O$$ . Prove that if $$f$$ is complex differentiable at $$z_0$$, then $$\partial_yu(x_0,y_0)$$ and $$\partial_xv(x_0,y_0)$$ exist and satisfy the equation $$\partial_yu(x_0,y_0)=- \partial_xv(x_0,y_0)$$.

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