Proving Complex Function Limit with Green's Theorem

[8daysAweek]

Homework Statement

$$f(z)$$ is a complex function that belongs to $$C^1$$. Prove that:

$$\lim_{r\to{0}}\frac{1}{r^2}\oint_{\tiny{|z-z_0|=r}}{f(z)dz}=2\pi{i}\frac{\partial f}{\partial \overline z}(z_0)$$

The Attempt at a Solution

Using Green's Theorem:
$$\oint_{{C}}{f(z,\overline z)dz}=2{i}\iint\limits_D {\frac{\partial f}{\partial \overline z}} \, dA$$

I got:
$$\lim_{r\to{0}}\frac{1}{r^2}\oint_{\tiny{|z-z_0|=r}}{f(z)dz}= \lim_{r\to{0}}\frac{1}{r^2}2{i}\iint\limits_{|z-z_0| \le r} {\frac{\partial f}{\partial \overline z}} \, dA$$

It would be perfect if $$\iint\limits_{|z-z_0| \le r} {\frac{\partial f}{\partial \overline z}} \, dA = {\pi}{r^2}\frac{\partial f}{\partial \overline z}(z_0)$$ but I don't know how to prove it (acctualy I can't even believe that it is true).

Any directions will be appreciated.

 

[Hurkyl]

It would be perfect if...

This is analysis. You don't need perfect, you just need close enough.