If $h'(0)=0$, let $a_kz^k$ be the first nonzero Taylor term. Then $k\ge2$ and $a_k$ is real. But $\operatorname{Im}(a_kr^ke^{ik\theta})$ changes sign as $0<\theta<\pi$, and for small $r$ the leading term controls the sign. This contradicts positivity in the upper half-plane. Hence $h'(0)\ne0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 12 / 12.2A / b / ii / Solution

 Solution (source code)