Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 12 (Complex Analysis OR Complex Methods)

Apply the residue theorem to $f(z)/(z-w{)}^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's derivative formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's derivative formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local uniform convergence therefore passes to every derivative.

On $|z-a|=\epsilon$, $g$ has a positive minimum modulus. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

If $f$ had a pole at one of the finitely many candidate points, choose a polynomial vanishing to sufficiently high order at all the other candidates. Hermite interpolation lets its remaining Taylor coefficients be chosen so that the residue of $p^{2}f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero integral by the residue theorem, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

Schwarz reflection shows that the Taylor coefficients along the real axis are real. Since $v(x,0)=0$ and $v(x,y)>0$ for $y>0$, one has $v_y(x,0)\ge 0$. The Cauchy–Riemann equations give $h^{\prime }(x)=u_x(x,0)=v_y(x,0)\ge 0$.

If $h^{\prime }(0)=0$, let $a_k{z}^k$ be the first nonzero Taylor term. Then $k\ge 2$ and $a_k$ is real. But $\mathrm{Im}(a_k{r}^k{e}^{ik\theta })$ 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 $h^{\prime }(0)\ne 0$.