Daily Chaos: Proof Cauchy Riemann Equations

Cauchy Riemann equations are necessary conditions but not sufficient conditions to check if a function is analytic or not. According to the definition of derivative we have

$f'(z_0)=\lim\limits_{\Delta z\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta z}$

One important fact to take notice is $f(z)=f(x+iy)=f(x,y)$

$\Rightarrow f'(z_0)=\lim\limits_{\Delta x+i \Delta y \to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta x+ i\Delta y}$

For the derivative to exists at point $z_0$ it should be same no matter what path we take to reach $z_0$

$\Rightarrow f'(z_0)=\lim\limits_{x_0\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta x}$

Lets go around only x axis which means $\Delta y=0$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x \to 0}\dfrac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)+iv(x_0+\Delta x,y_0)-(u(x_0,y_0)+iv(x_0,y_0))}{\Delta x}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)-u(x_0,y_0)+iv(x_0+\Delta x,y_0)-iv(x_0,y_0)}{\Delta x}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)-u(x_0,y_0)}{\Delta x}+\lim\limits_{\Delta x\to 0}\dfrac{iv(x_0+\Delta x,y_0)-iv(x_0,y_0)}{\Delta x}$

$\Rightarrow f'(z_o)=u_x+iv_x$

Now let’s go along only y axis i.e $\Delta x = 0$

$\Rightarrow f'(z_0)=\lim\limits_{\Delta y\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{i\Delta y}$

$\Rightarrow f'(x_0+iy_0)=\lim\limits_{\Delta y\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{i\Delta y}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y \to 0}\dfrac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{i\Delta y}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)+iv(x_0,y_0+\Delta y)-(u(x_0,y_0)+iv(x_0,y_0))}{i\Delta y}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)+iv(x_0,y_0+\Delta y)-iv(x_0,y_0)}{i\Delta y}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)}{i\Delta y}+\lim\limits_{\Delta x\to 0}\dfrac{iv(x_0,y_0+\Delta y)-iv(x_0,y_0)}{i\Delta y}$

$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}-i\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)}{\Delta y}+\lim\limits_{\Delta x\to 0}\dfrac{v(x_0,y_0+\Delta y)-v(x_0,y_0)}{\Delta y}$