I have given $f(z)=\frac{z}{(z^2-1)^2}$ and want to classify the singularity at $\infty$.
My idea was the following:
I remark that $f(z)$ is analytic on $|z|>R$ for $R>\sqrt{2}$. Then by definition I need to consider $g(z)=f(1/z)$ which is then analytic on $0<|z|<1/R$. Hence $g$ has an isolated singularity at $0$ so $f$ has an isolated singularity at $\infty$. Now I remark that $$z\cdot g(z)=\frac{1}{\left(\frac{1}{z^2}-1\right)^2}\rightarrow 0$$ if $z\rightarrow 0$. Hence by the Riemann Theorem $g$ has a removable singularity at $0$ and thus $f$ has a removable singularity at $\infty$?
Riemann Theorem: Let $f$ be analytic on $0<|z-b|<\delta$. Assume $$\lim_{z\rightarrow b} (z-b)f(z)=0$$then $b$ is a removable singularity.
Is this correct so?
Thanks for your help.
Yes, it's a removable singularity. But, even easier, for $z\ne 0$, $$g(z)=f(1/z) = \frac{1/z}{(1/z^2-1)^2} = \frac{z^3}{(1-z^2)^2}$$ obviously extends to an analytic function on $|z|<1$ if we set $g(0)=0$. Indeed, we see that $f$ has a zero of order $3$ at $\infty$.