Find all harmonic functions $f:\mathbb{C}\backslash \{0\}\to \mathbb{R}$ that are constant on every circle centered at 0. "Find all harmonic functions $f:\mathbb{C}\backslash\{0\} \to \mathbb{R}$ that are constant on every circle centered at 0." 
This is one of the past qualifying exam problems that I was working on. 
I was thinking to deal with $\frac{1}{f}$ so that $\frac{1}{f}$ is defined at 0 and use Schwarz lemma or something like that.
Any help or guidance would be really appreciated. 
Thank you in advance. 
 A: Here's a hint, which I think will allow you to solve it:
You can represent $f(z)$ in polar form
$$f(z)=f(R,\phi)$$
with $R\in\mathbb{R}$ and $z=Re^{i\phi}$ ($\phi$ is only defined up to $2\pi$, of course). Then write the Cauchy-Riemann equations for $R$ and $\phi$ (they're available here). In this form, you know that $\partial f/\partial \phi=0$, for every $R$.
A: First of all, you should check (this is a good exercise in using the chain rule) that the Laplace operator in polar coordinates is given by
$$\Delta f = \frac{1}{r}\frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}.$$
If $f$ is constant on circles centered at $0$, then $f$ depends only on $r$ (i.e. derivatives with respect to $\theta$ are $0$), so $f$ must satisfy
$$\frac{1}{r}\frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2} = 0.$$
This equation can be viewed as a first order, ordinary differential equation in $u = \frac{\partial f}{\partial r}$:
$$u' + \frac{1}{r} u = 0,$$
whose solutions are
$$u = \frac{C}{r}.$$
Hence,$$f = C\log r + D = C\log|z| + D.$$
