Liouville's Theorem (Complex Analysis)

Theorem
Let $f: \C \to \C$ be a bounded entire function.

Then $f$ is constant.

Proof
By assumption, there is $M \ge 0$ such that $\left \vert {f \left({z}\right)} \right \vert \le M$ for all $z \in \C$.

For any $R \in \R: R > 0$, consider the function:
 * $f_R \left({z}\right) := f \left({R z}\right)$

Using the Cauchy Integral Formula, we see that:
 * $\displaystyle \left \vert {f_R' \left({z}\right)} \right \vert = \frac 1 {2 \pi} \left \vert {\int_{C_1 \left({z}\right)} \frac{f \left({w}\right)} {\left({w - z}\right)^2} dw} \right \vert \le \frac 1 {2 \pi} \int_{C_1 \left({z}\right)} M d w = M$

where $C_1 \left({z}\right)$ denotes the circle of radius $1$ around $z$.

Hence:
 * $\displaystyle \left \vert {f' \left({z}\right)} \right \vert = \left \vert {f_R' \left({z}\right)} \right \vert / R \le M / R$

Since $R$ was arbitrary, it follows that $\left \vert {f' \left({z}\right)} \right \vert = 0$ for all $z \in \C$.

Thus $f$ is constant.

Remark
In fact, the proof shows that, for a nonconstant entire function $f$, the maximum modulus $\displaystyle M \left({r, f}\right) := \max_{\left \vert {z}\right \vert = r} \left \vert {f \left({z}\right)}\right \vert$ grows at least linearly in $r$.