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"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.

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How about considering $g(z)=f(e^{z})$? Then $g$ is constant on vertical lines. –  Hagen von Eitzen Dec 30 '12 at 8:18
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It is Schwarz by the way, not Schwartz. –  WimC Dec 30 '12 at 8:34

2 Answers 2

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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,$$f = C\log r + D = C\log|z| + D.$$

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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 $R$.

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