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.