Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 –  Lorenz Chaos Oct 24 '12 at 14:43

2 Answers 2

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 $z_0$

share|improve this answer

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 $z_0$: $$ \begin{align} u_x(z_0)&=v_y(z_0)\\ u_y(z_0)&=-v_x(z_0) \end{align}\tag{6} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.