How would one find the residue of the following function:
$$ f(z) = \frac{e^{iz}}{e^z+e^{-z}}$$
I tried finding the poles of the function by setting $e^z+e^{-z} = 0 $ and then deducing that the poles are:
$$z_0 = \frac{\pi i}{2}(1+2k)$$ for integers k.
However, I am unsure how to find the residue given these poles.
Any help would be appreciated.
If $g(z)=e^z+e^{-z}$ then $g(z_0)=0$ and we can compute the limit:
$$\lim_{z\to z_0} \frac{g(z)-g(z_0)}{z-z_0}\tag{1}$$
and invert the result to get the limit:
$$\lim_{z\to z_0} \frac{z-z_0}{g(z)-g(z_0)}$$
But then you can just multiply by $e^{iz_0}$ to get the residue.
Finally, (1) is just the derivative of $g(z)$ at $z_0,$ by definition.
So the residue at $z_0=\frac{\pi i }2(2k+1)$ is:
$$\frac{e^{iz_0}}{e^{z_0}-e^{-z_0}}=\frac{e^{-\frac \pi2(2k+1)}}{(-1)^k2i}$$
This wouldn’t work if $g$ had a double zero at $z_0$ - that is, if $g(z_0)=g’(z_0)=0.$