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 $$M\geq0$$ such that $$|f(z)|\leq M$$ for all $$z\in\C$$.

For any $$R>0$$, consider the function
 * $$f_R(z) := f(Rz).$$

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

(where $$C_1(z)$$ denotes the circle of radius $$1$$ around $$z$$).

Hence $$|f'(z)| = |f_R'(z)|/R \leq M/R$$.

Since $$R$$ was arbitrary, it follows that $$|f'(z)| = 0$$ 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 $$M(r,f):=\max_{|z|=r} |f(z)|$$ grows at least linearly in $$r$$.