# proof to Cauchy-Riemann equations (polar coordinates)

If $f(z)$ is differentialble at $z_{0}$ then the following limit

 $\displaystyle f^{\prime}(z_{0})$ $\displaystyle=$ $\displaystyle\lim_{\xi\to 0}\frac{f(z_{0}+\xi)-f(z_{0})}{\xi}$

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

 $\displaystyle f^{\prime}(z_{0})$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(r_{0}e^{i\theta_{0}}+he^{i\theta_{0}})-f(r_{% 0}e^{i\theta_{0}})}{he^{i\theta_{0}}}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f((r_{0}+h)e^{i\theta_{0}})-f(r_{0}e^{i\theta_% {0}})}{he^{i\theta_{0}}}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{u(r_{0}+h,\theta_{0})+iv(r_{0}+h,\theta_{0})-u% (r_{0},\theta_{0})-iv(r_{0},\theta_{0})}{he^{i\theta_{0}}}$ $\displaystyle=$ $\displaystyle\frac{1}{e^{i\theta_{0}}}\Bigg{[}\lim_{h\to 0}\frac{u(r_{0}+h,% \theta_{0})-u(r_{0},\theta_{0})}{h}+i\lim_{h\to 0}\frac{v(r_{0}+h,\theta_{0})-% v(r_{0},\theta_{0})}{h}\Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{e^{i\theta_{0}}}\Bigg{[}\frac{\partial u}{\partial r}(r_% {0},\theta_{0})+i\frac{\partial v}{\partial r}(r_{0},\theta_{0})\Bigg{]}$

Similarly, if we take the limit along the circle with fixed $r$ equals $r_{0}$. Then

 $\displaystyle f^{\prime}(z_{0})$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(r_{0}e^{i\theta_{0}}+r_{0}e^{i(\theta_{0}+h)% })-f(r_{0}e^{i\theta_{0}})}{r_{0}e^{i\theta_{0}}(e^{ih}-1)}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(r_{0}e^{i(\theta_{0}+h)})-f(r_{0}e^{i\theta_% {0}})}{he^{i\theta_{0}}}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{u(r_{0},\theta_{0}+h)+iv(r_{0},\theta_{0}+h)-u% (r_{0},\theta_{0})-iv(r_{0},\theta_{0})}{he^{i\theta_{0}}}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}e^{i\theta_{0}}}\Bigg{[}\lim_{h\to 0}\frac{u(r_{0}+% h,\theta_{0})-u(r_{0},\theta_{0})}{h}\cdot\frac{h}{e^{ih}-1}+i\lim_{h\to 0}% \frac{v(r_{0}+h,\theta_{0})-v(r_{0},\theta_{0})}{h}\cdot\frac{h}{e^{ih}-1}% \Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}e^{i\theta_{0}}}\Bigg{[}\lim_{h\to 0}\frac{u(r_{0}+% h,\theta_{0})-u(r_{0},\theta_{0})}{h}\cdot\lim_{h\to 0}\frac{h}{e^{ih}-1}+i% \lim_{h\to 0}\frac{v(r_{0}+h,\theta_{0})-v(r_{0},\theta_{0})}{h}\cdot\lim_{h% \to 0}\frac{h}{e^{ih}-1}\Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}e^{i\theta_{0}}}\Bigg{[}\frac{\partial u}{\partial% \theta}(r_{0},\theta_{0})\frac{1}{i}+\frac{\partial v}{\partial\theta}(r_{0},% \theta_{0})\Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}e^{i\theta_{0}}}\Bigg{[}\frac{\partial v}{\partial% \theta}(r_{0},\theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})% \Bigg{]}$

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:

 $\displaystyle\frac{1}{e^{i\theta_{0}}}\Bigg{[}\frac{\partial u}{\partial r}(r_% {0},\theta_{0})+i\frac{\partial v}{\partial r}(r_{0},\theta_{0})\Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}e^{i\theta_{0}}}\Bigg{[}\frac{\partial v}{\partial% \theta}(r_{0},\theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})% \Bigg{]}$ $\displaystyle\Bigg{[}\frac{\partial u}{\partial r}(r_{0},\theta_{0})+i\frac{% \partial v}{\partial r}(r_{0},\theta_{0})\Bigg{]}$ $\displaystyle=$ $\displaystyle\frac{1}{r_{0}}\Bigg{[}\frac{\partial v}{\partial\theta}(r_{0},% \theta_{0})-i\frac{\partial u}{\partial\theta}(r_{0},\theta_{0})\Bigg{]}$

which implies that

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

QED

Title proof to Cauchy-Riemann equations (polar coordinates) ProofToCauchyRiemannEquationspolarCoordinates 2013-03-22 14:06:13 2013-03-22 14:06:13 Daume (40) Daume (40) 5 Daume (40) Proof msc 30E99