# An entire function whose real part is bounded must be constant.

Greets

This is exercise 15.d chapter 3 of Stein & Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:

Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon>0$ $g(D_{\epsilon>0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.

I would like to know an answer with the the maximum modulus principle.

Thanks

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–  Double AA Nov 19 '14 at 15:52

A simpler way is to use Liouville's Theorem: consider $g(z) = 1/(1+b - f(z))$ where $\text{Re}(f(z)) \le b$.

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As other posters have commented, the standard approach here would be to invoke Liouville's Theorem. One way to do this is to consider the entire function $e^{f(z)}$.

Observe that $|e^{f(z)}| = e^{\Re f(z)}$, which is bounded by our assumption on $\Re f(z)$.

Then $e^{f(z)}$ is an entire bounded function, and hence (by Liouville's Theorem) constant.

From this, we conclude that $f(z)$ is constant as well.

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Another theorem related to this Liouville's theorem.

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@Downvoter: What's the downvote for? –  Mhenni Benghorbal Mar 30 '13 at 12:31
@Moderators: The answer is correct! –  Mhenni Benghorbal Mar 30 '13 at 12:33
@Moderators: This downvoting is misleading, especially without leaving a comment. –  Mhenni Benghorbal Mar 30 '13 at 12:43