# 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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can you please supply more details as to how this proof would go? Thank you. – Jessy Cat Mar 25 at 1:31

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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I was wondering why would $e^{f(z)}$ being constant will imply $f(z)$ is constant... Unlike real exponential, complex exponential is not one one.. It is periodic. So, i was thinking it need not be true.. After some time i realized it is true... Justification is as follows: We have $e^{f(z)}=C$ for all $z\in \mathbb{C}$.. Taking derivative both sides we see that $e^{f(z)}f'(z)=0$ for all $z\in \mathbb{C}$.. As $e^{f(z)}$ is never zero we have $f'(z)=0$ for all $z\in \mathbb{C}$ and so $f(z)$ is constant function.. – cello Apr 7 at 5:56