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I am a bit confused with the way of finding the pole of a function. I understand that to determine the type of isolated singularity, you make a Laurent series expansion of $f(z)$ such that you get $f(z) = \sum_k a_k (z-z_0)^k$. It is a pole if there is a finite number of $a_k$ such that $k<0$, a removable singularity if there are no $a_k$ such that $k<0$ and essential singularity if there is an infinite number of $a_k$ where $k<0$.

My question regards the example $f(z) = \frac{z^2}{(z^2-1)^2}$. This expression, according to the solution, has poles at $z = 1$ and $z =-1$. I can see why they are isolated singularities. However, when I do some work to see whether they are removable or poles, by doing the Laurent series expansion, I get $$f(z) =\frac{z}{(z^2-1)^2}=\frac{1}{4}(\frac{-1}{(z+1)^2}+\frac{1}{(z-1)^2})=\frac{1}{4}(-(1-z+z^2-z^3+...)^2+(1+z+z^2+z^3+...)^2)$$ In this expansion I don't get any term where $z$ has negative coefficient, so I don't see why it is a pole. Could anyone explain this to me?

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  • $\begingroup$ The Laurent series needs to be around the singularity, not around $0$. $\endgroup$
    – John B
    Apr 2 '16 at 23:30
  • $\begingroup$ So in this case I wouldn't need to do a Laurent series expansion, since I already have it of the form $f(z) = \frac{1}{4}(\frac{-1}{(z+1)^2}+\frac{1}{(z-1)^2}$? $\endgroup$
    – urpi
    Apr 2 '16 at 23:33
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For example, around $1$ you get $$ \begin{split} \frac z{(z^2-1)^2} &=\frac{(z-1)+1}{(z-1)^2(z+1)^2}\\ &=\frac{1}{z-1}\cdot \frac{1}{(z+1)^2}+\frac{1}{(z-1)^2}\cdot\frac{1}{(z+1)^2}\\ &=\left(\frac{1}{z-1}+\frac{1}{(z-1)^2}\right)\frac{1}{(z+1)^2}. \end{split} $$ The sum inside parentheses is already a sum of powers of $z-1$ and so you only need to expand $1/(z+1)^2$. Keep in mind that in order to determine the order of a singularity you only need to look at the negative powers and so you only need to go up to a certain order when developing $1/(z+1)^2$.

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