Complex differentiability at one point and loop integral around it Let $u(x, y)$ and $v(x, y)$ be differentiable real functions with continuous partial derivatives in a neighbourhood of $z_0$. Prove that $f = u+ iv$ is
complex differentiable at $z_0$ if and only if
                $$\displaystyle\lim_{r\to 0}\frac{1}{\pi r^2}\oint_{C(z_0,r)}f(z)dz=0$$
Where  $C(z_0, r)$ is the circle of radius r centered at $z_0$. 
I think this maybe related to Cauchy formula but the condition seems to point toward Cauchy Riemann equations.  
 A: We have
$f(z)dz=f(x+iy)d(x+iy)=(u+iv)(dx+idy)=udx+iudy+ivdx-vdy$
Thus we can write the contour integral as the line integral of a differential form
$$\displaystyle\oint_{C(z_0,r)}f(z)dz=\displaystyle\oint_{C(z_0,r)}udx-vdy+i\displaystyle\oint_{C(z_0,r)}vdx+udy$$
Since $u,v$ are both differentiable by assumption, we can apply Green's Theorem and get
$$\displaystyle\oint_{C(z_0,r)}f(z)dz=\displaystyle\iint_{D(z_0,r)}(-\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})dxdy+i\displaystyle\iint_{D(z_0,r)}(-\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})dxdy$$
Now apply mean value thm for double integral
$$\displaystyle\oint_{C(z_0,r)}f(z)dz=\pi r^2[-\frac{\partial v}{\partial x}(z_0+\rho e^{i\theta})-\frac{\partial u}{\partial y}(z_0+\rho e^{i\theta})]+i\pi r^2
[-\frac{\partial u}{\partial x}(z_0+\rho' e^{i\theta'})-\frac{\partial v}{\partial y}(z_0+\rho' e^{i\theta'})]$$
for suitable $0<\rho,\rho'<r$ and $0<\theta,\theta'<2\pi$
As $r$ tends to $0^{+}$, $z_0+\rho e^{i\theta}$ and $z_0+\rho' e^{i\theta'}$ tend to $z_0$, uniformly w.r.t $\theta$ and $\theta'$.
Hence the RHS of last equation (divided by $\pi r^2$)goes to zero iff $u,v$ satisfy CR-equation in $z_0$, i.e iff $f$ is differentiable in the complex sense at $z_0$
A: Since $f$ is differentiable, we have
$$
f(z)=f(z_0)+f_x(z_0)(x-x_0)+f_y(z_0)(y-y_0)+o(z-z_0)\tag{1}
$$
and along $C(z_0,r)$,
$$
\mathrm{d}z=r(-\sin(\theta)+i\cos(\theta))\mathrm{d}\theta\tag{2}
$$
Therefore
$$
\begin{align}
&\frac1{\pi r^2}\oint_{C(z_0,r)}f(z)\,\mathrm{d}z\\
&=\frac1{\pi r^2}\int_0^{2\pi}{\Large(}f(z_0)+f_x(z_0)r\cos(\theta)+f_y(z_0)r\sin(\theta)+o(r){\Large)}\,r(-\sin(\theta)+i\cos(\theta))\mathrm{d}\theta\\
&=if_x(z_0)-f_y(z_0)+\frac1\pi\int_0^{2\pi}\frac{o(r)}{r}(-\sin(\theta)+i\cos(\theta))\mathrm{d}\theta\\[6pt]
&\to if_x(z_0)-f_y(z_0)\\[6pt]
&=iu_x(z_0)-v_x(z_0)-u_y(z_0)-iv_y(z_0)\tag{3}
\end{align}
$$
Thus,
$$
\lim_{r\to0}\frac1{\pi r^2}\oint_{C(z_0,r)}f(z)\,\mathrm{d}z
=iu_x(z_0)-v_x(z_0)-u_y(z_0)-iv_y(z_0)\tag{4}
$$
Equation $(4)$ says that the condition given above
$$
\lim_{r\to0}\frac1{\pi r^2}\oint_{C(z_0,r)}f(z)\,\mathrm{d}z=0\tag{5}
$$
is precisely the Cauchy-Riemann conditions at $z_0$:
$$
\begin{align}
u_x(z_0)&=v_y(z_0)\\
u_y(z_0)&=-v_x(z_0)
\end{align}\tag{6}
$$
