Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2005 / Analysis / 4 / iii / Solution

For a bounded entire function, the preceding result with $n=0$ gives Liouville's theorem. If a nonconstant polynomial $p$ had no complex root, $1/p$ would be entire. The leading term gives $|p(z)|\to \infty$ at infinity, so $1/p$ is bounded outside a large disc and is also bounded on the remaining compact disc. Liouville's theorem would make $1/p$, and hence $p$, constant. This proves the fundamental theorem of algebra.

Finally, away from the zeros of $\cos z$, the holomorphic quotient $h=f/\cos z$ has modulus at most $C$. 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 $c$ with $|c|\le C$, andConversely every such multiple satisfies the assumed bound.