I'm trying to prove VI.8.4 from Sarson's Complex Functions Theory:
Let $f \in C^1$ be a complex-valued function defined and continuous on the disk $|z-z_0| < R$. For $0 < r < R $ let $C_r$ denote the circle $|z-z_0| = r$ with positive orientation.
I want to prove that $f$ satisfies:
$$\lim_{r\rightarrow 0} \frac{1}{r^2} \int_{C_r} f(z)dz = 2\pi i \frac{\partial f}{\partial \bar z}(z_0) $$
I tried to use the result from VI.8.3 that $$\int_{C_r} \frac{f'(z)}{z-z_0}dz = 2\pi i f'(z_0)$$
But I couldnt understand how to proceed.. A direction will be helpful. Thanks.
I think it is better to write $f$ as a function of both $z$ and $\bar{z}$. Then the limit becomes
$$\begin{align}\lim_{r \to 0} \frac1{r^2} \oint_{|z-z_0|=r} dz \, f(z,\bar{z}) &= \lim_{r \to 0} \frac{i}{r} \int_0^{2 \pi} dt \, e^{i t} \, f(z_0+r e^{i t},\bar{z_0}+r e^{-i t}) \\ &= \lim_{r \to 0} \frac{i}{r} f(z_0) \int_0^{2 \pi} dt \, e^{i t} + \lim_{r \to 0} i \left [ \frac{\partial f}{\partial z}\right ]_{z=z_0} \int_0^{2 \pi} dt \, e^{2 i t} \\ &+ \lim_{r \to 0} i \left [ \frac{\partial f}{\partial \bar{z}}\right ]_{z=\bar{z_0}} \int_0^{2 \pi} dt \, + O(r)\end{align}$$
The only term to not vanish is the piece with the derivative with respect to $\bar{z}$. The result is
$$\lim_{r \to 0} \frac1{r^2} \oint_{|z-z_0|=r} dz \, f(z,\bar{z}) = i 2 \pi \left [ \frac{\partial f}{\partial \bar{z}}\right ]_{z=\bar{z_0}}$$
Note that this differs from the OP's result.