Wirtinger derivatives and Cauchy integral

Question

Suppose a complex valued function $f$ is of class $C^1$ defined on the disk $|z-z_0|<R,$ and let $C_r$ denote the circle $|z-z_0|=r$ with $0<r<R$. Prove that $$\lim_{r \to 0}\frac{1}{2\pi i r^2} \oint_{C_r} f(z)\mathrm{d}z={\frac{\partial f}{\partial \bar{z}}(z_0)}.$$ My idea is to estimate $\oint_{C_r} \frac{1}{r^2}f(z)\mathrm{d}z-\oint_{C_r}\frac{\partial f}{\partial \bar{z}}/(z-z_0)dz$ by writting in a parametric form. But I got stuck when I tried to estiamte $f(z)(z-z_0)-r^2\frac{\partial f}{\partial \bar{z}}.$ Can you give me some hints to estimate that quantity? Like can I use talor expansion in polar coordiante?

Answer 1

For $f(z,\bar z)$ and $z=z_0+re^{i\theta}$, we have

$$\frac{d}{dr}f(z_0+re^{i\theta},\bar z_0+re^{-i\theta})=\left.\left(e^{i\theta}\frac{\partial f(z,\bar z)}{\partial z}+e^{-i\theta}\frac{\partial f(z,\bar z)}{\partial \bar z}\right)\right|_{(z,\bar z)=(z_0+re^{i\theta},\bar z_0+re^{-i\theta})}$$

and $dz=ire^{i\theta}\,d\theta$ for $\theta\in [0,2\pi]$.

Then, applying L'Hospital's Rule yields

$$\begin{align} \lim_{r\to0}\frac{\oint_{|z-z_0|=r}f(z,\bar z)\,dz}{2\pi ir^2}&=\lim_{r\to0}\frac{\int_0^{2\pi}f(z_0+re^{i\theta},\bar z_0+re^{-i\theta})\,e^{i\theta}\,d\theta}{2\pi r}\\\\ &=\lim_{r\to 0}\frac1{2\pi}\int_0^{2\pi}\left.\left(e^{i2\theta}\frac{\partial f(z,\bar z)}{\partial z}+\frac{\partial f(z,\bar z)}{\partial \bar z}\right)\right|_{(z,\bar z)=(z_0+re^{i\theta},\bar z_0+re^{-i\theta})}\,d\theta\\\\ &=\frac1{2\pi}\left(\frac{\partial f(z_0,\bar z_0)}{\partial z_0}\right)\int_0^{2\pi}e^{i2\theta}\,d\theta+\frac1{2\pi}\left(\frac{\partial f(z_0,\bar z_0)}{\partial \bar z_0}\right)\int_0^{2\pi}(1)\,d\theta\\\\ &=\frac{\partial f(z_0,\bar z_0)}{\partial \bar z_0} \end{align}$$

as was to be shown!