# Complex Analysis: Liouville's Theorem

I am stuck on the following question.

Suppose $f = u+iv$ is entire and there exists $M > 0$ such that $|u(z)| \leq M$ for all $z\in C$. Show that $f$ is constant.

I would figure we would have to use Liouville's Theorem to show this is true because the theorem states if "A bounded entire function is constant."

How would I continue to approach this problem?

I want to simply say because the function is entire and $|u(z)|$ is bounded that $f$ is constant. The only thing that makes me uneasy is $v(z)$. We do not know if it is bounded or not. Hence why I am stuck.

Thank you for your time and Thanks in advanced for any feedback.

-

I would try using $|e^{f}|=|e^{u+iv}|=|e^{u}e^{iv}|=|e^{u}||e^{iv}|=|e^{u}|\le e^{|u|}\le e^{M}$.

Thus, we can say $e^f$ is constant by Liouville's Theorem... Now show this implies $f$ is constant.

HINT: Think of how $f$ could not be constant even though $e^{f}$ is constant, and show this condition contradicts the assumption that $f$ is entire or $u$ is bounded.

-
Seems close, but $\left|e^u\right| \ne e^\left|u\right|$. (But $e^u \le e^{\left|u\right|}$) – Michael Anderson May 12 '14 at 6:41
The edited version is above. – user2154420 May 13 '14 at 18:44

Hint: Consider the function $g(z) = f(z) + M + 1$. Can you show that $1/g(z)$ is a bounded, entire function, using the fact that $g$ has real part which can be bounded below?

-