Stuck on trying to find the Laurent series for $$\frac{e^z -1}{z^2}$$ centered at $z_0 = 0$. Still new to Laurent series, so not entirely sure how to get it. I know the Taylor series for $e^z$ but don't know how to deal with the -1 like that.
Any guidance would be very helpful
On
$e^z$ has Taylor series, about z= 0, $\sum_{n=0}^\infty \frac{z^n}{n!}= 1+ z+ \frac{z^2}{2}+ \frac{z^3}{3!}+ \cdot\cdot\cdot$. So $e^z- 1$ has series $\sum_{n=1}^\infty \frac{z^n}{n!}= z+ \frac{z^2}{2}+ \frac{z^3}{3!}+ \cdot\cdot\cdot$. Then $\frac{e^z-1}{z^2}=\frac{1}{z}+ \frac{1}{2}+ \frac{z}{3!}+ \cdot\cdot\cdot= \sum_{n=1}^\infty \frac{z^{n-2}}{n!}$.
So you know the series for $e^z$. Which is to say, you're familiar with $$ e^z = 1 + z + \frac{z^2}2 + \frac{z^3}6 + \frac{z^4}{24} + \cdots $$ Now, subtracting $1$ means just that: subtract $1$. It's as easy as can be. $$ e^z - 1 = 1 + z + \frac{z^2}2 + \frac{z^3}6 + \frac{z^4}{24} + \cdots - 1\\ = z + \frac{z^2}2 + \frac{z^3}6 + \frac{z^4}{24} + \cdots $$ Finally, dividing by $z^2$ is similarly straight-forward: $$ \frac{e^z - 1}{z^2} = \frac{z + \frac{z^2}2 + \frac{z^3}6 + \frac{z^4}{24} + \cdots}{z^2}\\ = \frac1z + \frac12 + \frac z6 + \frac{z^2}{24} + \cdots $$