Complex integration with a $\mathcal{C}^1$ function $f$ This is the problem from the book Notes on Complex Function Theory, Sarason. 

Let the complex valued function $f$ defined and $\mathcal{C}^1$ in the disk $|z-z_0|<R.$ For $0<r<R$ let $C_r$ denote the circle $|z-z_0|=r$, with the counterclockwise orientation. Prove that $$ \lim_{r\to 0}\frac{1}{r^2}\int_{C_r}{f(z)}\ dz=2\pi i\ \frac{\partial f}{\partial \bar{z}}(z_0) .$$

I found the answer to my question here. But I wanted to do this by basic theory of complex integration. Any help will be appreciated. 
Thanks.
 A: Presumably you're happy that $f(x+iy) = f(z_0) + (x-x_0)\partial_x f(z_0) + (y-y_0) \partial_y f(z_0) + o(\sqrt{x^2+y^2}) $ (this is what $C^1$ means, essentially). Then putting $z-z_0=re^{i\theta}$, the line integral becomes
$$ \int_{C_r} f(z) \, dz = \int_0^{2\pi} f(r\cos{\theta}+ir\sin{\theta})ire^{i\theta} \, d\theta, $$
and we can expand this using the power series, replacing $x$ and $y$:
\begin{align}
 &\int_0^{2\pi} f(z_0 + r\cos{\theta}+ir\sin{\theta})ire^{i\theta} \, d\theta \\
&= \int_0^{2\pi} \left( f(z_0) + r\cos{\theta} \, \partial_x f(z_0) + r\sin{\theta}  \, \partial_y f(z_0) + o(r) \right) ie^{i\theta} \, d\theta \\
&= rf(z_0) \int_0^{2\pi} ie^{i\theta} \, d\theta + r^2\left( \partial_x f(z_0) \int_0^{2\pi} ie^{i\theta}\cos{\theta} \, d\theta + \partial_y f(z_0) \int_0^{2\pi} ie^{i\theta}\sin{\theta} \, d\theta \right) + o(r^2).
\end{align}
The first term vanishes, and the second two integrals are $\pi$ and $\pi i$, so we find
$$ \int_{C_r} f(z) \, dz = r^2 (2\pi i) \frac{1}{2}( \partial_x f(z_0) + i \partial_y f(z_0) ) + o(r^2), $$
and we recognise the last part of the first term as the Wirtinger derivative $\partial_{\bar{z}} f(z_0)$. The result follows by dividing by $r^2$ and taking the limit.

Alternatively, one could use $f(z) = f(z_0) + (z-z_0) \partial_zf(z_0) + \overline{z-z_0} \partial_{\bar{z}}f(z_0) +o(\lvert z \rvert + \lvert \bar{z} \rvert)$ and use that $\int_{\partial A} \bar{z} dz$ calculates the signed area of the interior of $A$.
A: Hint: WLOG, $z_0=0.$ Let $A=\partial f/\partial x(0), B = \partial f/\partial y(0).$ Then
$$f(z)= f(0) + Ax +By +o(|z|) = f(0) + A\frac{z+\bar z}{2} +B\frac{z-\bar z}{2i} +o(|z|).$$
