# How to understand the residue $\text{Res}(\frac{e^{iz}}{\sinh(z)},0) = \frac{e^{iz}}{\cosh(z)}\bigg|_{z=0}$

I am reading a solution of a complex integral using residue, in one step it syas

$$\text{Res}(\frac{e^{iz}}{\sinh(z)},0) = \frac{e^{iz}}{\cosh(z)}\bigg|_{z=0}$$

I cannot understand how to get the RHS from the formula for polynomial case.
Can anyone please let me know how to get the RHS? Thanks!

Suppose $$z_0$$ is a simple pole of $$f(z)=\frac{P(z)}{Q(z)}$$. We have
$$\text{Res}(z_0, f) = \frac{P(z)}{Q'(z)} \bigg|_{z = z_0}.$$
$$\text{Res}(z_0, f) = \lim_{z \rightarrow z_0} (z-z_0) \frac{P(z)}{Q(z)} = \lim_{z \rightarrow z_0} \frac{P(z)}{\tfrac{Q(z)-Q(z_0)}{z-z_0}} = \frac{P(z_0)}{Q'(z_0)}.$$
Observe that this onluy works when $$z_0$$ is a simple pole, otherwise the expression above would also have a singularity.
• It is also useful to observe that if $P(z_0) \ne 0$ ,$Q'(z_0) \ne 0$ and $Q(z_0) = 0$ then $z_0$ is a simple pole of $f(z)$ and residue at $z_0$ is nothing but $\frac{P(z_0)}{Q'(z_0)}$ where $P$ and $Q$ are analytic at $z_0$ Dec 14, 2020 at 13:46