Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 12G / c / Solution

Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.

Solved by gpt-5.6-sol high.