Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 13F (Complex Analysis)

The hypothesis says that $|f|$ attains a global, hence local, maximum at the interior point $a$. By the maximum modulus principle, a nonconstant holomorphic function cannot do this. Therefore $f$ is constant.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.

No. The nonvanishing holomorphic functions $f_n(z)=e^{nz}$ satisfy $f_n(0)=1$, butAvoiding zero alone permits an unbounded holomorphic logarithm's real part; the omitted ray in part (c) was what constrained the image after taking a square root.