Liouville's Theorem

Liouville's Theorem is as follows:

Liouville's Theorem. If $f$ is holomorphic and bounded on $\mathbb C$, then it is constant.

Proof. We have the following formula for the derivative of the function $f$: $$f'(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)d\zeta}{(\zeta-z)^2}$$ where $\mathrm{ind}(\gamma,z)=1$. If we suppose that $f$ is bounded in magnidue so that $|f(z)|0$, and we choose $\gamma$ to be a circle of radius $R$ centered at $z$, the integral identity can be transformed into the following bound: $$|f'(z)|=\bigg|\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)d\zeta}{(\zeta-z)^2}\bigg| \leq \frac{1}{2\pi}\int_\gamma \frac{M}{R^2}ds = \frac{M}{R}$$ but because $R$ can be chosen to be arbitrarily large, we have that $|f'(z)|0$, and hence $f'(z)$ vanishes for all $z$, implying that $f$ is constant. $\blacksquare$

complex-analysis

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