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I have to solve following problem

If $\Re (f)$ is bounded above or below for a function $f$ holomorphic on $\mathbb{C}$ then $f$ is constant.

My attempt: If there is $M$ such that $\Re(f) \le M$, then $\|e^{f}\|=e^{\Re(f)}\le e^{M}$. From Liouville's theorem the entire function $e^f$ is constant, that is $0=(e^f)'=f' e^f$. This means $f' =0$, so $f$ is constant. If $\Re(f)$ is bounded below, we consider $e^{-f}$ and proceed the same way.

Am I correct? Is there a solution using maximum modulus principle?

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    $\begingroup$ Yes, your proof is correct. The maximum modulus principle applies to functions on a bounded domain, so I think a solution using that would be quite artificial. $\endgroup$ – Daniel Fischer Feb 2 '14 at 0:35
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    $\begingroup$ Can we exactly repeat everything for $Im(f)$ bounded ? $\endgroup$ – SMath Mar 31 '15 at 9:19
  • $\begingroup$ Awesome...Proof...But i have a little doubt that how ||e^f|| = e^ R(f)...please explain it to me.. $\endgroup$ – Sam Christopher May 10 '15 at 5:32
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    $\begingroup$ @SMath for that you have to consider $e^{if}$ $\endgroup$ – tattwamasi amrutam May 10 '15 at 5:44
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    $\begingroup$ Very elegant proof, by the way! I like it. $\endgroup$ – Fimpellizieri Jun 2 '17 at 22:27
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Suppose $\operatorname{re} f(z) \ge \alpha$ and let $g(z) = {1 \over f(z)-\alpha+1 }$. Since $\operatorname{re} (f(z)-\alpha+1) \ge 1$, we have $|g(z)| \le 1$ for all $z$ hence $g$ is a (non zero) constant. Since $f(z) = \alpha-1+{1 \over g(z)}$, we see that $f$ is constant.

If $\operatorname{re} f(z) \le \alpha$, then repeat the above with $-f$ (and $-\alpha$, of course).

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Suppose $f$ is nonconstant. If $f$ does not have an essential singularity at $\infty$, then $f$ is a polynomial, which is a surjective function from $\Bbb{C}$ to itself by the fundamental theorem of algebra. Therefore $\Re (f)$ is unbounded.

If $f$ has an essential singularity at $\infty$, its range is dense in $\Bbb{C}$ by the Casorati-Weierstrass theorem, and $\Re (f)$ is unbounded.

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