# Finding a 0 centered Laurent Series

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

• What is the first term of the Taylor series for $e^z$? – Zircht Apr 24 at 21:42

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: $$\frac{e^z - 1}{z^2} = \frac{z + \frac{z^2}2 + \frac{z^3}6 + \frac{z^4}{24} + \cdots}{z^2}\\ = \frac1z + \frac12 + \frac z6 + \frac{z^2}{24} + \cdots$$
$$e^z$$ 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 $$\frac{e^z-1}{z^2}=\frac{1}{z}+ \frac{1}{2}+ \frac{z}{3!}+ \cdot\cdot\cdot= \sum_{n=1}^\infty \frac{z^{n-2}}{n!}$$.