Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 12B (Complex Analysis OR Complex Methods)

Let $|f(z)|\le K$ on $\mathrm{Re}z>0$. If $\mathrm{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The Cauchy derivative formula givesThe line segment joining $z_1$ to $z_2$ remains in $\mathrm{Re}z>c$, soThus one may take $M=K/c$, as in the lipschitz bound inside a bounded analytic half-plane.

The Liouville theorem states that every bounded entire function is constant.

Suppose $g(\mathbb{C})$ is not dense. Then some open disc $D(w,r)$ is disjoint from the image, sois entire and satisfies $|F(z)|\le 1/r$. Liouville's theorem makes $F$, and hence $g$, constant, a contradiction. Therefore every nonconstant entire $g$ has dense image, as asserted by dense image of a nonconstant entire function.

Write $h(z)={\sum }_{n=0}^{\infty }{a}_n{z}^n$. Cauchy's coefficient formula on $|z|=R$ givesExcept at the two measure-zero points where $\cos \theta =0$, the hypothesis givesSince $|\cos \theta {|}^{-1/2}$ is integrable,for a constant $C$ independent of $R$. Letting $R\to \infty$ shows that every $a_n=0$. Hence $h\equiv 0$, in particular $h$ is constant. This proves the entire function under a horizontal inverse-square-root bound result.