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The definition of a bounded function is:

$$\exists M\in\mathbb{R} \quad st \quad |f(x)| \leq M \quad\forall x\in Domain $$

So consider the complex entire function $f(z)$ such that $Re(f(z))<0$ for all $z\in \mathbb{C}$.

I need to use this to somehow show that $f(z)$ is bounded so that I can use Louiville's Theorem to show that $f(z)$ is constant.

My attempt.

$|f(z)-1| > 1$. Then $g(z) = \frac{1}{f(z)-1} < 1$. So $g(z)$ is bounded and entire and therefore constant and then $f(z)$ must as well be constant. Does this make sense?

Quibbly
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  • That makes sense, the same solution is given in http://math.stackexchange.com/questions/229312/an-entire-function-whose-real-part-is-bounded-must-be-constant. – Martin R Dec 08 '15 at 13:47
  • http://math.stackexchange.com/questions/561410/entire-function-with-positive-real-part-is-constant-no-picard –  Dec 08 '15 at 13:50

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