Hey guys I'm looking to find the Laurent Series Expansion of $$f(z)=e^{\frac {-3} {z^2}}, ann(0,0,\infty)$$
I know that $$e^{\frac 1 z} = \sum_{n=0}^\infty \frac{(\frac{1} {z})^n}{n!}$$
Can I use this in this case or is there an alternative way? Thanks
Well for any fixed $z_0\ne 0$, $w_0=\frac{-3}{z_0^2}$ is a well defined complex number, so we can plug it in and get
$$e^{w_0}=\sum_0^\infty \frac{w_0^n}{n!}=\sum_0^\infty\frac{(\frac{-3}{z_0^2})^n}{n!}=\sum_0^\infty\frac{(-3)^n}{z_0^{2n}n!}$$
Since $z_0$ was arbitrary, this must be valid in the entire punctured plane. So we have
$$e^{\frac{-3}{z^2}}=\sum_0^\infty\frac{(-3)^n}{z^{2n}n!}$$
This is a power series centered at the origin and valid in some (any) punctured disk at the origin, so by uniqueness it must be the Laurent series
Keep in mind that a function can have different Laurent series in different annuluses centered at the same point, but for a given annulus there is only one Laurent series. The power series above is valid in every annulus so it is the only Laurent series at this point.