= Solution
For a fixed <pole> away from the <saddle point>, the smooth amplitude there is $(i-z_0)^{-1}$. The <simple-saddle contribution in steepest descent> is
$$
J_{\rm sd}(z_0)=\frac{2\sqrt{\pi/k}\,e^{-k}}{i-z_0}\bigl[1+O(k^{-1})\bigr].
$$
The <residue theorem> supplies an additional contribution precisely when the <contour deformation> crosses $z_0$. The region swept above the original contour has positive orientation: the original contour from left to right followed by the reversed saddle contour encloses it counterclockwise. Consequently the original integral equals the saddle integral plus the pole contribution:
$$
\boxed{J(z_0)=J_{\rm sd}(z_0)+2\pi i\chi\,e^{k\phi(z_0)}.}
$$
For $|z_0|<1$, an upper-half-plane pole outside the indentation has $\chi=1$, while a lower-half-plane pole has $\chi=0$. The entire upper unit half-disk lies below the saddle <parabola>, so there is no further case inside that half-disk. With a fixed indentation radius $\delta$, a pole inside the upper semicircle, $|z_0|<\delta$, is also below the original contour and has $\chi=0$; real points in the indentation gap are excluded as well. Taking $\delta<|z_0|$ for a fixed nonzero pole gives the usual upper/lower classification. The pole contribution need not always dominate exponentially; its size depends on $\operatorname{Re}\phi(z_0)$, so keeping both terms makes that dependence explicit.
A pole on the original contour requires a stated <Cauchy principal value> or an indentation prescription. A pole parameter on the negative imaginary <branch cut> still defines the integral: the contour stays on its fixed sheet, and only the denominator uses $z_0$. Such a pole is not crossed, so only $J_{\rm sd}$ is needed and no value of $\sqrt{z_0}$ must be chosen. The residue exponential above is evaluated only when $\chi=1$. At $z_0=0$ the original indentation excludes the singular point, so no additional residue is crossed. These conventions matter before taking any limiting pole position.
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