How to find the residue of
$$f(z)=\frac{1}{z(1-e^{2z})}$$
at simple poles $z=n\pi i $?
I know how to find the residue of a function on a simple pole with a formula about that, but here I can't use that…
I also know that the result should be $\frac{i}{2n\pi}$ , but I don't know how can I get this…
Can someone help me?
Thank you in advance!
Best Answer
The residues at $z = in\pi$, $n \in \Bbb{Z} \text{\\} \{0\}$ can be calculated directly
$$\lim_{z\to in\pi} \frac{z-in\pi}{z(1-e^{2z})} \equiv -\frac{1}{in\pi}\cdot \frac{1}{(e^{2z})'}\Biggr|_{z=in\pi} = \frac{i}{2n\pi}$$
The residue at $z = 0$ can be calculated by Taylor expanding the denominator
$$\frac{1}{z(1-e^{2z})} = \frac{1}{z(-2z-2z^2-\cdots)} = -\frac{1}{2z^2(1+z+\cdots)} = -\frac{1}{2z^2}(1-z+\cdots)$$
$$ = -\frac{1}{2z^2} + \frac{1}{2z} - \cdots$$
which gives a residue of $\frac{1}{2}$