I am faced with the integral
$$I = \frac{1}{2\pi i}\int_{-\infty}^\infty \frac{(z-i)^{1/2}}{(z-k)(z-c)} \, dz$$
where $c \in \Bbb R$, $0<\text{Im}(k)<1$, and the multifunction is defined as
$$(z-i)^{1/2} = \sqrt{|z-i|} \, e^{i\theta/2} \qquad \theta = \arg(z-i) \in \bigg(-\frac{3\pi}{2}, \frac{\pi}{2}\bigg]$$
From what I understand, the integrand tends to $0$ as $z \rightarrow \infty$, and there is a branch cut at $[i,i\infty)$, so we must close off the contour in the lower half plane and make an indentation at $c$.
There are no singularities in the lower half plane, and the indentation goes half way around the pole at $c$. Hence,
$$0 = I + \frac 12 \text{res}\bigg(\frac{(z-i)^{1/2}}{(z-k)(z-c)} \, ; \, z=c\bigg) \\ \implies I = -\frac{(c-i)^{1/2}}{2(c-k)} = \frac{(c-i)^{1/2}}{2(k-c)}$$
However, I think that the answer should instead be
$$I = \frac{(c-i)^{1/2}}{k-c}$$
based on the problem that I am working on. Can anyone explain to me what the correct approach should be? Thanks!