I am given $f(z) = \frac{1}{z^2 + 2}$ and asked to provide all Laurent series around the point $z_{0} = 0$.
I've decomposed this into a partial fraction by having, $\frac{1}{2}(\frac{1}{(z-i\sqrt2)} - \frac{1}{(z+i\sqrt2)})$.
Basically from here I'm lost. I know I have a break at $i\sqrt2$. So $0<|z|<i\sqrt2$ and $i\sqrt2<|z|<\infty$?
Apologies, I am not familiar with formatting on here.
For $\|z\| > \sqrt{2}$, $$ \begin{align*} \frac{1}{z^2+2}&= \frac{1}{z^2}\frac{1}{1+\dfrac{2}{z^2}}\\ &= \frac{1}{z^2}\sum_{n=0}^{\infty}\left(-\frac{2}{z^2}\right)^n. \end{align*} $$ For $\|z\| < \sqrt{2}$, $f(z)$ is analytic and has a Taylor's power series representation, $$ \begin{align*} \frac{1}{z^2+2}&= \frac{1}{2}\left(\frac{1}{1+\dfrac{z^2}{2}}\right)\\ &=\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{z^2}{2}\right)^n. \end{align*} $$