proof to Cauchy-Riemann equations (polar coordinates)

If $f(z)$ is differentialble at $z_0$ then the following limit

$f^{\prime }(z_0)$$=$$\underset{\xi \to 0}{lim}{\displaystyle \frac{f(z_0+\xi )-f(z_0)}{\xi }}$

will remain the same approaching from any direction. First we fix $\theta$ as ${\theta }_0$ then we take the limit along the ray where the argument is equal to ${\theta }_0$. Then

$f^{\prime }(z_0)$$=$$\underset{h\to 0}{lim}{\displaystyle \frac{f(r_0{e}^{i{\theta }_0}+h{e}^{i{\theta }_0})-f(r_0{e}^{i{\theta }_0})}{h{e}^{i{\theta }_0}}}$

$=$$\underset{h\to 0}{lim}{\displaystyle \frac{f((r_0+h)e^{i{\theta }_0})-f(r_0{e}^{i{\theta }_0})}{h{e}^{i{\theta }_0}}}$

$=$$\underset{h\to 0}{lim}{\displaystyle \frac{u(r_0+h,{\theta }_0)+iv(r_0+h,{\theta }_0)-u(r_0,{\theta }_0)-iv(r_0,{\theta }_0)}{h{e}^{i{\theta }_0}}}$

$=$${\displaystyle \frac{1}{e^{i{\theta }_0}}}\left[\underset{h\to 0}{lim}{\displaystyle \frac{u(r_0+h,{\theta }_0)-u(r_0,{\theta }_0)}{h}}+i\underset{h\to 0}{lim}{\displaystyle \frac{v(r_0+h,{\theta }_0)-v(r_0,{\theta }_0)}{h}}\right]$

$=$${\displaystyle \frac{1}{e^{i{\theta }_0}}}\left[{\displaystyle \frac{\partial u}{\partial r}}(r_0,{\theta }_0)+i{\displaystyle \frac{\partial v}{\partial r}}(r_0,{\theta }_0)\right]$

Similarly, if we take the limit along the circle with fixed $r$ equals $r_0$. Then

$f^{\prime }(z_0)$$=$$\underset{h\to 0}{lim}{\displaystyle \frac{f(r_0{e}^{i{\theta }_0}+r_0{e}^{i({\theta }_0+h)})-f(r_0{e}^{i{\theta }_0})}{r_0{e}^{i{\theta }_0}(e^{ih}-1)}}$

$=$$\underset{h\to 0}{lim}{\displaystyle \frac{f(r_0{e}^{i({\theta }_0+h)})-f(r_0{e}^{i{\theta }_0})}{h{e}^{i{\theta }_0}}}$

$=$$\underset{h\to 0}{lim}{\displaystyle \frac{u(r_0,{\theta }_0+h)+iv(r_0,{\theta }_0+h)-u(r_0,{\theta }_0)-iv(r_0,{\theta }_0)}{h{e}^{i{\theta }_0}}}$

$=$${\displaystyle \frac{1}{r_0{e}^{i{\theta }_0}}}\left[\underset{h\to 0}{lim}{\displaystyle \frac{u(r_0+h,{\theta }_0)-u(r_0,{\theta }_0)}{h}}\cdot {\displaystyle \frac{h}{e^{ih}-1}}+i\underset{h\to 0}{lim}{\displaystyle \frac{v(r_0+h,{\theta }_0)-v(r_0,{\theta }_0)}{h}}\cdot {\displaystyle \frac{h}{e^{ih}-1}}\right]$

$=$${\displaystyle \frac{1}{r_0{e}^{i{\theta }_0}}}\left[\underset{h\to 0}{lim}{\displaystyle \frac{u(r_0+h,{\theta }_0)-u(r_0,{\theta }_0)}{h}}\cdot \underset{h\to 0}{lim}{\displaystyle \frac{h}{e^{ih}-1}}+i\underset{h\to 0}{lim}{\displaystyle \frac{v(r_0+h,{\theta }_0)-v(r_0,{\theta }_0)}{h}}\cdot \underset{h\to 0}{lim}{\displaystyle \frac{h}{e^{ih}-1}}\right]$

$=$${\displaystyle \frac{1}{r_0{e}^{i{\theta }_0}}}\left[{\displaystyle \frac{\partial u}{\partial \theta }}(r_0,{\theta }_0){\displaystyle \frac{1}{i}}+{\displaystyle \frac{\partial v}{\partial \theta }}(r_0,{\theta }_0)\right]$

$=$${\displaystyle \frac{1}{r_0{e}^{i{\theta }_0}}}\left[{\displaystyle \frac{\partial v}{\partial \theta }}(r_0,{\theta }_0)-i{\displaystyle \frac{\partial u}{\partial \theta }}(r_0,{\theta }_0)\right]$

Note: We use l’Hôpital’s rule to obtain the following result used above ${lim}_{h\to 0}\frac{h}{e^{ih}-1}=\frac{1}{i}$.

Now, since the limit is the same along the circle and the ray then they are equal:

${\displaystyle \frac{1}{e^{i{\theta }_0}}}\left[{\displaystyle \frac{\partial u}{\partial r}}(r_0,{\theta }_0)+i{\displaystyle \frac{\partial v}{\partial r}}(r_0,{\theta }_0)\right]$$=$${\displaystyle \frac{1}{r_0{e}^{i{\theta }_0}}}\left[{\displaystyle \frac{\partial v}{\partial \theta }}(r_0,{\theta }_0)-i{\displaystyle \frac{\partial u}{\partial \theta }}(r_0,{\theta }_0)\right]$

$\left[{\displaystyle \frac{\partial u}{\partial r}}(r_0,{\theta }_0)+i{\displaystyle \frac{\partial v}{\partial r}}(r_0,{\theta }_0)\right]$$=$${\displaystyle \frac{1}{r_0}}\left[{\displaystyle \frac{\partial v}{\partial \theta }}(r_0,{\theta }_0)-i{\displaystyle \frac{\partial u}{\partial \theta }}(r_0,{\theta }_0)\right]$

which implies that

${\displaystyle \frac{\partial u}{\partial r}}$$=$${\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial v}{\partial \theta }}$

${\displaystyle \frac{\partial v}{\partial r}}$$=$$-{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u}{\partial \theta }}$

QED