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Let $u$ be a nonconstant harmonic function on $\mathbb C$. Show that for any $c\in\mathbb R, u^{-1}(c)$ is unbounded. Hint: $\{|z|>R\}$ is connected for any $R>0$.


It seems like this proof might require some sort of a "trick", because I'm not sure how to attack this one directly. I know what I want to show is that for any potential upper bound $M \in \mathbb{R}$, there exists $z \in \mathbb{C}$ such that $|f(z)|>M$.

However, I don't know much about $u^{-1}$ except that $u$ is harmonic. (Aside: is there a mistake in the question? That is, isn't $u^{-1}(c)$ just a complex number? It seems like it should ask to show the function $u^{-1}$ is unbounded, not $u^{-1}(c)$, but I could be wrong.)

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    $\begingroup$ I think that $u^{-1}(c)$ here means the preimage: $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$ $\endgroup$ – Daniele A Mar 18 '15 at 11:38
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Suppose that $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$ is bounded (that is, $u^{-1}(c)$ is contained in $\{|z|<R\}$). Then, by the continuity of $u$ and the fact that $\{|z|>R\}$ is connected, (i) $u(z)>c$ for all $z\in \{|z|>R\} $ or (ii) $u(z)<c$ for all $z\in \{|z|>R\}$. In case of (i) consider $f(z)$ holomorphic in $\mathbb{C}$ with $\operatorname{Re}f(z)=-u(z)$, and in case of (ii) consider $f(z)$ with $\operatorname{Re}f(z)=u(z)$. We conclude that $f(z)$ is constant in either case since $g(z)=e^{f(z)}$ is holomorphic in $\mathbb{C}$ and bounded.

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  • $\begingroup$ Thank you very much, ts375_zk26! $\endgroup$ – Mathemanic Mar 18 '15 at 23:02

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