A natural proof for Cauchy-Riemann Condition

We know that we can represent a complex number as a matrix

Given by $\begin{array}{c}a+bi\mapsto \left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)\end{array}$$a+bi\mapsto \begin{pmatrix}a&-b\\b&a\end{pmatrix}$

Represent $+$$+$ to matrix add, $\times$$\times$ to matrix multiply, conjugate to transpose, Norm to $\sqrt{detM}$$\sqrt{\det M}$

Consider a function $f(z):\mathbb{C}\to \mathbb{C}$$f(z):\mathbb C\to\mathbb C$$,{\displaystyle \frac{df}{dz}=\underset{z\to z_0}{lim}\frac{f(z)-f(z_0)}{z-z_0}}$$, \displaystyle\frac{df}{dz}=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$

We know that $\mathbb{C}$$\mathbb C$ is a Field and complete metric space, so if the limit exists, it must be a complex number.

And view $f(z)$$f(z)$ as $u(x,y)+iv(x,y)$$u(x,y)+iv(x,y)$, consider the Jacobi Matrix

if $u(x,y)+iv(x,y)$$u(x,y)+iv(x,y)$ is complexly differentiable, then the Jacobi Matrix is a complex number

That means $\left(\begin{array}{cc}{\partial }_{x}u& {\partial }_{y}u\\ {\partial }_{x}v& {\partial }_{y}v\end{array}\right)$$\begin{pmatrix}\part_x u&\part_y u\\\part_x v&\part_y v\end{pmatrix}$ looks like $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ (we denote ${\displaystyle \frac{\partial f}{\partial x}}$$\displaystyle\frac{\part f}{\part x}$ as ${\partial }_{x}f$$\part_xf$)

So we get ${\partial }_{x}u={\partial }_{y}v,{\partial }_{y}u=-{\partial }_{x}v$$\part_x u=\part_yv,\part_y u=-\part_x v$

And for the polar form,

Consider the Jacobi Matrix for $x=r\cos \theta ,y=r\sin \theta$$x= r\cos\theta,y=r\sin\theta$, $\begin{array}{c}J=\left(\begin{array}{cc}\cos x& -r\sin x\\ \sin x& r\cos x\end{array}\right)\end{array}$$J=\begin{pmatrix}\cos x&-r\sin x\\\sin x&r\cos x\end{pmatrix}$

Observe that if we multiply ${\displaystyle \frac{1}{r}}$$\displaystyle\frac{1}{r}$ for the second row, we get a complex number.

Consider the chain rule, $\left(\begin{array}{cc}{\partial }_{x}u& {\partial }_{y}u\\ {\partial }_{x}v& {\partial }_{y}v\end{array}\right)\left(\begin{array}{cc}{\partial }_{r}x& {\partial }_{\theta }x\\ {\partial }_{r}y& {\partial }_{\theta }y\end{array}\right)=\left(\begin{array}{cc}{\partial }_{r}u& {\partial }_{\theta }u\\ {\partial }_{r}v& {\partial }_{\theta }v\end{array}\right)$$\begin{pmatrix}\part_x u&\part_y u\\\part_x v&\part_y v\end{pmatrix}\begin{pmatrix}\part_rx&\part_\theta x\\\part_r y&\part_\theta y\end{pmatrix}=\begin{pmatrix}\part_r u&\part_\theta u\\\part_r v&\part_\theta v\end{pmatrix}$

So $\begin{array}{c}{\displaystyle \left(\begin{array}{cc}{\partial }_{r}u& \frac{1}{r}{\partial }_{\theta }u\\ {\partial }_{r}v& \frac{1}{r}{\partial }_{\theta }v\end{array}\right)}\end{array}$$\displaystyle\begin{pmatrix}\part_r u&\frac{1}{r}\part_\theta u\\\part_r v&\frac{1}{r}\part_\theta v\end{pmatrix}$ is a complex number, So we get the polar version if C-R condition.