Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2005/analysis/4/iii/solution

For a bounded entire function, the preceding result with gives Liouville's theorem. If a nonconstant polynomial had no complex root, would be entire. The leading term gives at infinity, so is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make , and hence , constant. This proves the fundamental theorem of algebra.
Finally, away from the zeros of , the holomorphic quotient has modulus at most . At each zero it therefore has a removable singularity. These isolated points have no finite accumulation, so removing all of them gives a bounded entire function. Liouville's theorem gives a constant with , and
Conversely every such multiple satisfies the assumed bound.

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