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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 $e$

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    $\begingroup$ Did u try with z^2 (for pole at z=0?) $\endgroup$ – user166305 May 16 at 19:53
  • $\begingroup$ I'm not sure I understood what you meant to ask :( $\endgroup$ – Tota May 16 at 20:01
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    $\begingroup$ limit z->0 zf(z)=infinity suggests it has a higher order. As I understand, your denominator is -z(e^{2z}-1). e^{2z} has a power series expansion around zero, right? Use it and see if you can take z^2 from the denominator out and use the definition of pole to conclude what the order of pole at zero is. $\endgroup$ – user166305 May 16 at 20:09
  • $\begingroup$ Got it.Thanks:) Would you give me any advice for other singularities? For example I know that $z=n\pi i$ , for $n$ = whole number, are also singularities, but I'm not sure of what type.Neither for $z=\infty$ ... Any advice is appreciated!Thanks a lot. $\endgroup$ – Tota May 16 at 20:32
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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.

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  • $\begingroup$ Thanks a lot! For $z=\infty$ I knew it wasn't a pole, I asked what kind of singularity it is... I have problem with identifying of what kind a singularity is...:) $\endgroup$ – Tota May 17 at 8:15

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