# Proof using Liouville's Theorem

Here is my current problem and I can't find any help for it anywhere.

Let the function $f:\mathbb{C}\rightarrow\mathbb{C}$ be defined and analytic on all of $\mathbb{C}$, and suppose that there exists $a\in\mathbb{C}$ and a real number $\delta\gt 0$ such that

$$\left|f(z)-a\right|\gt\delta\quad\forall z\gt\mathbb{C}$$

Use Liouville's Theorem to prove that $f$ is a constant.

I was given one hint which was to consider the function $\frac{1}{f(z)-a}$. Where do I even begin to approach this that hasn't be practised once in class.

• @Don ohh yeah, I fixed that – MRT Mar 15 '17 at 20:20

$\left|f(z)-a\right|\gt\delta\Rightarrow\frac{1}{\delta}>\left | \frac{1}{f(z)-a}\right |$. So, $g(z)=\frac{1}{f(z)-a}$ is bounded in $\mathbb{C}$.

Hints:

1) For all $\;z\in\Bbb C\;,\;\;f(z)\neq a\;$

2) $\;\cfrac1{f(z)-a}\;$ is analytic and bounded on the complex plane, and thus it is constant there.