Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 12 (Complex Analysis OR Complex Methods)

Cauchy’s formula is $f(w)=(2\pi i{)}^{-1}{\int }_{\gamma }f(z)/(z-w)dz$, proved by removing the removable singularity of $[f(z)-f(w)]/(z-w)$ and applying Cauchy’s theorem; similarly $f\prime (w)=(2\pi i{)}^{-1}\int f(z)/(z-w{)}^{2}dz$. The resulting estimates prove Liouville’s theorem. Finally ${\cos }^{2}x=(1+\cos 2x)/2$ and residues in the upper half-plane give ${\int }_{-\infty }^{\infty }{\cos }^{2}x/(x^2+1)dx=\frac{\pi }{2}(1+e^{-2})$.

Choose $\sqrt{z^2-1}\sim z$ at infinity. On $-1<x<1$ its upper/lower values are $\pm i\sqrt{1-x^2}$, divided by $x+c$. Under $z=1/\zeta$ the transformed integrand is the stated $g$; inversion reverses orientation, maps the slit exterior/interior accordingly, and encloses $0,-1/c$ negatively. Their residues give $J=-2\pi i(c-\sqrt{c^2-1})$. Comparing the two slit banks gives $J=-2iI$, hence $I=\pi (c-\sqrt{c^2-1})$.