Statement of the Theorem

Let $f \colon \mathbb{C} \to \mathbb{C}$ be an entire function. If $f$ is bounded, then $f$ is constant.

Proof

Suppose $f$ is an entire bounded function. Say $|f| \le M$. Let $a, b \in \mathbb{C}$. We will show that $f(a) = f(b)$ by calculating the difference.

Let $R > 0$ be large enough so that $|a|, |b| < R$. Then since $f$ is entire, we have that $f(a) = \frac{1}{2 \pi i} \int_{|z| = R} \frac{f(z)}{z-a} dz$ and $f(b) = \frac{1}{2 \pi i} \int_{|z| = R} \frac{f(z)}{z-b} dz$.

Then we find that \[f(a)-f(b) = \frac{1}{2 \pi i} \int_{|z| = R} \frac{f(z) (a-b)}{(z-a)(z-b)} dz.\] We use the modulus bound for integrals: \[|f(a)-f(b)| \leq \frac{M R |a-b|}{(R-|a|)(R-|b|)}.\] And since we may take $R$ as large as we want, we find that $|f(a)-f(b)|=0$. Hence $f$ is constant.

$\square$