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.



3 Answers 3


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.

  • 27
    $\begingroup$ 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.. $\endgroup$
    – user312648
    Apr 7, 2016 at 5:56
  • $\begingroup$ @user312648: Thanks! $\endgroup$ Aug 3, 2018 at 16:30
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    $\begingroup$ The "standard" way of noting that $\exp \circ f$ being constant implies $f$ is so is to note that the preimage of every singleton is discrete. Indeed, if $\exp \circ f$ is constant, then $f$ must take values in some set of the form $c + 2 \pi i \Bbb Z$. Since $f$ is continuous and $\Bbb C$ connected, $f$ must be constant. $\endgroup$ Oct 31, 2021 at 12:55
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    $\begingroup$ @AryamanMaithani Thanks! $\endgroup$ Nov 6, 2021 at 19:28

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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    $\begingroup$ can you please supply more details as to how this proof would go? Thank you. $\endgroup$
    – user100463
    Mar 25, 2016 at 1:31
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    $\begingroup$ @ALannister Note that $g(z)$ is a bounded entire function, by Liouville's Theorem we have $g(z)$ is constant and so is $f(z)$. $\endgroup$
    – Bach
    Jul 12, 2019 at 4:49

Let $f(z) $ be an entire function with real part bounded,then

$f(z)$ is entire$\implies \phi(z)=e^{f(z)}$ entire

$\implies \vert\phi(z)\vert=\vert e^{f(z)}\vert= e^{Re(f(z))} \le e^M$ ,Where $Re(z) \le M$ for some fixed $M \in \mathbb R$

$\implies \phi(z)$ is constant,by Liouville's theorem

$\implies\phi'(z)=e^{f(z)}f'(z)=0\forall z\in \mathbb C$

$\implies f'(z)=0 $ in $\mathbb C$,since $e^{f(z)}\neq 0$ in $\mathbb C$

$\implies f(z)$ is constant

  • $\begingroup$ What is wrong with this answer for which it is downvoted? $\endgroup$
    – Styles
    Oct 21, 2017 at 5:46
  • 2
    $\begingroup$ Might be that it's extremely similar to the other answer given here. $\endgroup$
    – JKEG
    May 2, 2019 at 11:23

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