Liouville's Theorem (complex Analysis) - Proof

Proof

The theorem follows from the fact that holomorphic functions are analytic. If f is an entire function, it can be represented by its Taylor series about 0:

where (by Cauchy's integral formula)

a_k = \frac{f^{(k)}(0)}{k!} = {1 \over 2 \pi i} \oint_{C_r} \frac{f( \zeta )}{\zeta^{k+1}}\,d\zeta

and Cr is the circle about 0 of radius r > 0. Suppose f is bounded: i.e. there exists a constant M such that |f(z)| ≤ M for all z. We can estimate directly

| a_k | \le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta| \le \frac{1}{2 \pi} \oint_{C_r} \frac{ M }{ r^{k+1} } \, |d\zeta| = \frac{M}{2 \pi r^{k+1}} \oint_{C_r} |d\zeta| = \frac{M}{2 \pi r^{k+1}} 2 \pi r = \frac{M}{r^k},

where in the second inequality we have used the fact that |z|=r on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.

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