= Solution
One possible open contour begins at $t=0$ along a ray with $\operatorname{Re}(1/t)>0$, avoids the <essential singularity> at the origin, and ends at infinity along a ray with $\operatorname{Re}(zt)<0$. 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, $e^{-1/t}\to0$; at its final end, $e^{zt}\to0$. More generally, the terminal ray may be chosen with
$$
\arg t=\pi-\arg z,
$$
so that $zt$ is negative real. This contour satisfies the endpoint condition from part (a) and therefore produces a solution of the differential equation.
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