Prove that $f=u+iv$ is differentiable if and only if $\lim_{r→0} \frac{1}{πr^2 } \int_{C(z_0,r)}f(z)dz=0$ Suppose that $u,v$ are real-valued function that having continuous partial derivative of first order in the neighborhood of  $z_0=x_0+iy_0$ . Prove that $f=u+iv$ is differentiable if and only if 
$$\lim_{r→0} \frac{1}{πr^2 } \int_{C(z_0,r)}f(z)dz=0$$

Theorem: Let $G$ be open in $\mathbb C$ and let $\gamma$ be rectifiable path in $G$ initial and end at $\alpha$ and $\beta$ respectively. If $f:G \to \mathbb C$ is continuous with a primative $F:G \to \mathbb C$  then $\int_\gamma f= F(\beta)-F(\alpha)$

Here is what I got so far
$\Rightarrow$
Assume that $u,v$ are real-valued function that having continuous partial derivative of first order in the neighborhood of  $z_0=x_0+iy_0$. Let $G$ be open in $\mathbb C$ such that $C(z_0,r)$ in $G$. Suppose $f$ is differentiable in $G$ then there exist a primitive $F:G \to \mathbb C$ of $f$ in the neighborhood of $z_0$. Note that $C(z_0,r)$ is rectifiable, moreover it's a closed curve, so by the above theorem
$$ \int_{C(z_0,r)}f(z)dz=0$$
thus
$$\lim_{r→0} \frac{1}{πr^2 } \int_{C(z_0,r)}f(z)dz=0$$
$\Leftarrow$
Assume that
$$\lim_{r→0} \frac{1}{πr^2 } \int_{C(z_0,r)}f(z)dz=0$$
Let $\epsilon >0$, there exist $\delta >0$ such that $|r|<\delta$ implies $|\frac{1}{πr^2 } \int_{C(z_0,r)}f(z)dz|<\epsilon$. 
Now I want to show that $\lim_{h \to 0} \frac{|f(z+h)-f(z)|}{|h|}$ exist but I can't find any way to approach this direction. I wonder if anyone would please give me a hand.
 A: Let's expand the real and imaginary parts of $f$ around $z_0$. That is,
\begin{align}
u(z_0+z) &= u(z_0) + u_x(z_0)\Re\{z\} + u_y(z_0)\Im\{z\} + \mathcal{O}(z)\\
v(z_0+z) &= v(z_0) + v_x(z_0)\Re\{z\} + v_y(z_0)\Im\{z\} + \mathcal{O}(z)\\
f(z_0+z) &= u + iv\\
&= u(z_0) + iv(z_0) + [u_x(z_0) + iv_x(z_0)]\Re\{z\} + [u_y(z_0) + iv_y(z_0)]\Im\{z\} + \mathcal{O}(z)\\
&=f(z_0) + f_x(z_0)\Re\{z\} + f_y(z_0)\Im\{z\} + \mathcal{O}(z)
\end{align}
Let $z=re^{it}$ and $\gamma(t) = z_0 + z$ so $\gamma'(t) = ire^{it}$.
\begin{align}
\int_{\gamma}f(z)dz & = \int_0^{2\pi}f(\gamma(t))\gamma'(t)dt\\
&=irf(z_0)\int_0^{2\pi}e^{it}dt + irf_x(z_0)\int_0^{2\pi}\Re\{z\}e^{it}dt + irf_y(z_0)\int_0^{2\pi}\Im\{z\}e^{it}dt + \mathcal{O}(r^2)\\
&= irf(z_0)\int_0^{2\pi}e^{it}dt + ir^2f_x(z_0)\int_0^{2\pi}\cos(t)e^{it}dt + ir^2f_y(z_0)\int_0^{2\pi}\sin(t)e^{it}dt + \mathcal{O}(r^2)\\
&= i[f_x(z_0)+if_y(z_0)]r^2\pi + \mathcal{O}(r^2)\\
\lim_{r\to 0}\frac{1}{r^2\pi}\int_{\gamma}f(z)dz &= i[f_x(z_0)+if_y(z_0)] + \lim_{r\to 0}\frac{\mathcal{O}(r^2)}{r^2\pi}
\end{align}
Now
$$
\lim_{r\to 0}\frac{\mathcal{O}(r^2)}{r^2\pi} = 0
$$
since $\mathcal{O}(r^2)$ all have terms with a power of $r$ greater than $2$. 
$$
i[f_x(z_0)+if_y(z_0)]=0\iff f_x(z_0)+if_y(z_0) =0\tag{1}
$$
What is equation $(1)$?
