# Liouville's theorem (complex analysis)

The theorem is considerably improved by Picard's little theorem, which says that every entire function whose image omits two or more complex numbers must be constant.

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:

and *C _{r}* 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

where in the second inequality we have used the fact that |*z*| = *r* on the circle *C*_{r}. 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 *a*_{k} = 0 for all *k* ≥ 1. Thus *f*(*z*) = *a*_{0} and this proves the theorem.

There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.^{[2]}

Suppose that *f* is entire and |*f*(*z*)| is less than or equal to *M*|*z*|, for *M* a positive real number. We can apply Cauchy's integral formula; we have that

where *I* is the value of the remaining integral. This shows that *f′* is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that *f* is affine and then, by referring back to the original inequality, we have that the constant term is zero.

Any holomorphic function on a compact Riemann surface is necessarily constant.^{[6]}

Similarly, if an entire function has a pole of order n at ∞—that is, it grows in magnitude comparably to *z*^{n} in some neighborhood of ∞—then f is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if |*f*(*z*)| ≤ *M*|*z*^{n}| for |*z*| sufficiently large, then f is a polynomial of degree at most n. This can be proved as follows. Again take the Taylor series representation of f,

The argument used during the proof using Cauchy estimates shows that for all *k* ≥ 0,

Liouville's theorem does not extend to the generalizations of complex numbers known as double numbers and dual numbers.^{[7]}