Liouville's Theorem (Complex Analysis)/Proof 2

Proof
By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.

Let $r > 0$.

consider:
 * $D_r = \set {z \in \C: \cmod z \le r}$

Then, consider the parameterization $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ given by:
 * $\map {\gamma_r} t := r e^{2 \pi i t}$

For all $z \in \C$, we have:

where:
 * $\map {g_z} w := \map f {w + z}$

Hence, for all $z \in \C$:

Thus it follows that $\map {f'} z = 0$ for all $z \in \C$.

By Zero Derivative implies Constant Complex Function, $f$ is constant.