01 - Cauchy-Riemann Equations

Cauchy-Riemann Equations

Let $z_0\in \mathbb{C}$
Let $U$ be a neighbourhood of $z_0$

Let $f:U\to C$ be a function which is complex differentiable at $z_0$
Let $u,v:{\mathbb{R}}^2\to \mathbb{R}$ be the components of $f$ then

The four partial derivatives ${\partial }_{x}u,{\partial }_{y}u,{\partial }_{x}v,{\partial }_{y}v$ exist at $z_0$ with Cauchy-Riemann Equations

$${\partial }_{x}u={\partial }_{y}v,{\partial }_{x}v=-{\partial }_{y}u$$

and

$$f^{\prime }(z_0)={\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)$$

Proof

By definition of complex differentiability then

$$f^{\prime }(z_0)=\underset{z\to z_0}{\lim }\frac{f(z)-f(z_0)}{z-z_0}$$

By considering a particular way in which $z$ approaches $z_0$
Let $z=z_0+h$ where $h\in \mathbb{R}$ and $h\to 0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)& =\underset{h\to 0}{\lim }\left[\frac{u(x_0+h,y_0)+iv(x_0+h,y_0)-u(x_0,y_0)-iv(x_0,y_0)}{h}\right]\\ & =\underset{h\to 0}{\lim }\left[\frac{u(x_0+h,y_0)-u(x_0,y_0)}{h}\right]+i\underset{h\to 0}{\lim }\left[\frac{v(x_0+h,y_0)-v(x_0,y_0)}{h}\right]\\ & ={\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)\end{array}$$

Similarly by considering $z_0=z_0+ih$ where $h\in \mathbb{R}$ and $h\to 0$ then

$$\begin{array}{rl}{f}^{\prime }(z_0)& =\underset{h\to 0}{\lim }\left[\frac{u(x_0,y_0+h)+iv(x_0,y_0+h)-u(x_0,y_0)-iv(x_0,y_0)}{ih}\right]\\ & =-i\underset{h\to 0}{\lim }\left[\frac{u(x_0,y_0+h)-u(x_0,y_0)}{h}\right]+\underset{h\to 0}{\lim }\left[\frac{v(x_0,y_0+h)-v(x_0,y_0)}{h}\right]\\ & =-i{\partial }_{y}u(z_0)+{\partial }_{y}v(z_0)\end{array}$$

As $f^{\prime }(z_0)$ is well-defined then

$${\partial }_{x}u(z_0)+i{\partial }_{x}v(z_0)=-i{\partial }_{y}u(z_0)+{\partial }_{y}v(z_0)$$

So by comparing the real and imaginary parts of both we get Cauchy-Riemann Equations

Proof - Alternative

Complex Differentiability implies that

$$f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$$

where $ϵ(z)\to 0$ as $z\to z_0$

By rewriting in the expression in real terms (identifying $\mathbb{C}$ with ${\mathbb{R}}^2$) so

$$f^{\prime }(z_0)=r{e}^{i\theta }=r\cos \theta +ir\sin \theta$$

Hence the derivative term $f^{\prime }(z_0)(z-z_0)$ can be rewritten as

$$\left(\begin{array}{c}r\cos \theta (x-x_0)-r\sin \theta (y-y_0)\\ r\cos \theta (y-y_0)+r\sin \theta (x-x_0)\end{array}\right)=\left(\begin{array}{cc}r\cos \theta & -r\sin \theta \\ r\sin \theta & r\cos \theta \end{array}\right)\left(\begin{array}{c}x-x_0\\ y-y_0\end{array}\right)$$

So then the term is just a linear transformation (rotation by angle $\theta$ and scalar factor $r$)

From Total derivative, if $f:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is real differentiable at $z_0=(x_0,y_0)$ then
There exists linear function $L=d{f}_{z_0}:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that

$$f(z)=f(z_0)+L(z-z_0)+R(z)$$

where $\frac{|R(z)|}{|z-z_0|}\to 0$ as $z\to z_0$ and $L$ is in the standard basis given by matrix

$$\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& {\partial }_{y}u(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{y}v(x_0,y_0)\end{array}\right)$$

where ${\partial }_{x}u,{\partial }_{y}u,{\partial }_{x}v,{\partial }_{y}v$ are partial derivatives of $u$ and $v$

Comparing the two matrices then $f$ is real differentiable and partial derivatives exist and

$$\left(\begin{array}{cc}{\partial }_{x}u(x_0,y_0)& {\partial }_{y}u(x_0,y_0)\\ {\partial }_{x}v(x_0,y_0)& {\partial }_{y}v(x_0,y_0)\end{array}\right)=\left(\begin{array}{cc}r\cos \theta & -r\sin \theta \\ r\sin \theta & r\cos \theta \end{array}\right)$$

Hence we immediately get the Cauchy-Riemann equations and that

$$f^{\prime }(z_0)=r\cos \theta +ir\sin \theta ={\partial }_{x}u(x_0,y_0)+i{\partial }_{x}v(x_0,y_0)$$