[Math] Showing that if $u$ is a real-valued harmonic function then for any real $c$ we have that $u^{-1}(c)$ is unbounded

complex-analysisharmonic functions

I have the following homework question:

Let $u$ be a non-constant real-valued harmonic function in
$\mathbb{C}$. Prove that $u^{-1}(c)$ is unbounded for every real
number $c$

There is a hint that say we may use the result of the following exercise:

Prove that a positive harmonic function in $\mathbb{C}$ is
non-constant

This exercise is taken from

Donald Sarason Notes on Complex Function Theory (page 89)

What I tried:

I don't really know how to start, but a reasonable beginning can be:

Assume by negation that there is $c\in\mathbb{R}$ s.t $$u^{-1}(c)\subseteq D(0,M)$$
where $$D(0,M)=\{z\in\mathbb{C}:\,|z|<M\}$$

We know that there is a harmonic conjugate $v$ s.t $$f=u+iv\in H(D(0,M))$$

I don't know how to move forward, or how to use the hint (since in
this question $u$ need not be a positive function).

Can someone please direct me in the right direction ?

Best Answer

I think you mean a positive harmonic function is constant. See Does there exist a harmonic function in the whold plane that is postive everywhere?

Also, you should be able to use the fact that $u^{-1}(c)$ is bounded to conclude that, by continuity, everything falls on one side of $c$. This can be manipulated into giving you a positive harmonic function.

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