# 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.

• Note that it doesn't say that $f$ is necessarily holomorphic at $z_0$, only that it is complex differentiable at $z_0$. The former would imply complex differentiable in some neighbourhood around $z_0$, which is stronger. This rules out the cauchy formula, I think, so you'll need to basically show that this limit only holds if the partial derivatives fulfill the Cauchy-Riemann differential equations at $z_0$. – fgp Oct 24 '12 at 2:32
• I think i've figured it out, the main idea is to use Green's theorem to change the loop integral into Cauchy Riemann related expression. Then take the limit LOL – Geralt of Rivia Oct 24 '12 at 14:43

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 $z_0$

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}