I need to find the Laurent series that represents the function: $f(z) = \dfrac{1}{z({1+z^{2}})}$, when $1<|z|<\infty$.
What I've tried thus far is:
Write f as the product $\dfrac{1}{z}\dfrac{1}{1+z^2}$ so that I could rewrite $\dfrac{1}{1+z^2}$ as $\displaystyle\sum_{i=0}^{\infty}(-1)^nz^{2n}$. I realise that this isn't correct though, because I need $|z|<1$ for this representation. I then tried to break $\dfrac{1}{1+z^2}$ up into $\dfrac{1}{z-i}\dfrac{1}{z+i}$ but I don't know how to proceed from here.
Q: How do I represent the function $f(z) = \dfrac{1}{z({1+z^{2}})}$ as a Laurent series, when $1<|z|<\infty$.
$$f(z)=\frac{1}{z^3}\,\frac{1}{1+\frac{1}{z^2}}$$ Does that help?