# What are the isolation singularities of this function?

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 $$e$$

• Did u try with z^2 (for pole at z=0?) – user166305 May 16 at 19:53
• I'm not sure I understood what you meant to ask :( – Tota May 16 at 20:01
• 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. – user166305 May 16 at 20:09
• 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. – Tota May 16 at 20:32

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.
• 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...:) – Tota May 17 at 8:15