Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 7A / b

One possible open contour begins at $t=0$ along a ray with $\mathrm{Re}(1/t)>0$, avoids the essential singularity at the origin, and ends at infinity along a ray with $\mathrm{Re}(zt)<0$. For example, when $z>0$ is real, it may leave the origin along the positive real axis, bend through the upper half-plane, and approach the negative real axis as $|t|\to \infty$. At its initial end, $e^{-1/t}\to 0$; at its final end, $e^{zt}\to 0$. More generally, the terminal ray may be chosen withso that $zt$ is negative real. This contour satisfies the endpoint condition from part (a) and therefore produces a solution of the differential equation.