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

If $f$ is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula iswith positive orientation. On the circle the integrand has modulus at most $M/r^{k+1}$ and the contour has length $2\pi r$. Thus

Take a circle of radius $R\ge S$ about zero. The Cauchy estimate gives, for each integer $k>n$,Therefore all these derivatives vanish. The Taylor series of an entire function converges to it everywhere, and henceAs usual $n$ is a nonnegative integer; if a real exponent is allowed, the degree is at most its integer part. A negative growth exponent forces $f=0$.

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.