I have to find out what are the isolated singularities of function: $f(z)=\frac{1}{(z)(1-e^(2z)))}$ .
Firstly,I thought that $z=0$ is a simple pole of this function, but then I tried to find the residue in $z=0$ (with the formula for finding residues for simple poles) and I got that the $Res(f,0)=\infty$ , which made me doubt that 0 is a simple pole of this function... Can someone help me with this?
Any help for $z=\infty$ (what kind of isolated singularity it is) would be appreciated. EDIT: I'm not sure how to put the exponent (2z) in $e$
Zero is a pole of order two, because we can factor $z$ out of $1-e^{2z}=\sum(2z)^n/n!$.
$z=kπi$ are simple poles.
$\infty$ isn't a pole, because $\lim_{z\to\infty}f(z)=0$. Neither is it a zero.
$f$ isn't meromorphic at infinity.