Show that a harmonic function $u(x,y)$ is equal at every point $a$ to its average value on any circle centered at $a$

Question

I'm having trouble proving this one. I've let $z = a+re^{i\theta}$ and then found by the Cauchy-Reimann conditions that the average value is $$\oint_C \frac{f(z)}{z-a} = 2\pi i \cdot f(a)$$ However I'm a little confused because a circle on the complex plane is defined to be $|z-a|$ which would mean $f(a) = 0$. I'm unsure if I'm doing something wrong and how to prove it from here. Thanks!

Answer 1

From the formula you wrote, for any holomorphic function $f$, we have \begin{equation} f(a) = \frac{1}{2\pi i}\oint_C\frac{f(z)}{z-a}dz = \frac{1}{2\pi i}\int_0^{2\pi}\frac{f(a + re^{i\theta})}{a+r e^{i\theta} - a} ire^{i\theta}d\theta = \frac{1}{2\pi}\int_0^{2\pi}f(a+re^{i\theta})d\theta \end{equation} Now we know that for any real harmonic function $u(x,y)$, there is another real harmonic function $v(x,y)$ such that $f(x+iy) = u(x,y) + iv(x,y)$ is holomorphic. Letting $a_x = \mathrm{Re}[a]$ and $a_y = \mathrm{Im}[a]$ and applying the above gives \begin{multline} u(a_x,a_y) + i v(a_x,a_y) = \frac{1}{2\pi}\int_0^{2\pi}u(a_x + r\cos\theta,a_y+r\sin\theta)d\theta \\+\frac{i}{2\pi}\int_0^{2\pi}v(a_x +r\cos\theta,a_y+r\sin\theta)d\theta \end{multline} Taking real parts then gives us the desired result $$ u(a_x,a_y)= \frac{1}{2\pi}\int_0^{2\pi}u(a_x + r\cos\theta,a_y+r\sin\theta)d\theta $$