Finding singularities of A Function

Question

I want to find singularities of $$f(z)=\frac{{z}^{2}}{e^z + {e}^{-z} - 2}$$ I solved this problem but I am not sure about it. Is it correct?

$${e^z + {e}^{-z} - 2}= 0$$ Then I divide by $$e^{-z}$$ to get $${{e^{2z}} - 2{e}^{z} +1 }= 0$$ Letting $$x=e^z$$ Then $$x^2-2x+1=0$$ $$(x-1)(x-1)=0$$ so $$e^z=1=e^{i2πk}$$ $$z= i2πk$$ Where k=0,1,-1,2,-2...

Is it correct?

Answer 1

Yes it's correct, but an easier way is to remark that $$f(z)=\frac{1}{2}\cdot \frac{z^2}{\cosh(z)-1}$$ and thus $$\cosh(z)=1\iff z=2i\pi n$$ with $n\in\mathbb Z$.