Trying to compute first few terms of the Laurent series for: $$ \frac{e^z}{z^2(z^2+1)} =\sum_{n=-2}^\infty c_n z^n = \quad ?$$
I know the expansion of $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ and I could have $\frac{e^z}{z^2}$ expanded. But how do I fit all together and be able to see when it converges.
From the comment I have:
$$ \frac{e^z}{1+z^2} = \sum_{n=0}^\infty \frac{z^n}{n!} \sum_{n=0}^\infty (-1)^n z^{2n} $$
And using Caucy product: $$ c_n =\sum_{k=0}^n \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)} $$
It follows that:
$$ \frac{e^z}{z^2(z^2+1)} = \frac{1}{z^2}\sum_{n=0}^\infty \sum_{k=0}^n \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)} $$
Via the Cauchy product, \begin{align} \frac{e^z}{z^2+1} &= e^z\cdot\frac{1}{1-(-z^2)} \\ &= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (-z^2)^n\right) \\ &= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (iz)^{2n}\right) \\ &= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{1+(-1)^n}{2}(iz)^n\right) \\ &= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{i^n+(-i)^n}{2}z^n\right) \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1}{k!} \cdot\frac{i^{n-k}+(-i)^{n-k}}{2}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^{n-k}+(-i)^{n-k}}{2(k!)}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^k+(-i)^k}{2((n-k)!)}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1+(-1)^k}{2}\cdot\frac{i^k}{(n-k)!}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{i^{2k}}{(n-2k)!}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n. \tag1 \end{align} So the desired Laurent series is \begin{align} \frac{e^z}{z^2(z^2+1)} &= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^{n-2} \\ &= \sum_{n=-2}^\infty \left(\sum_{k=0}^{(n+2)/2} \frac{(-1)^k}{(n+2-2k)!}\right) z^n. \end{align}
In hindsight, I took the long way to obtain $(1)$. Here's a shorter approach: \begin{align} \frac{e^z}{z^2+1} &= \frac{1}{1-(-z^2)}\cdot e^z \\ &= \left(\sum_{n=0}^\infty (-z^2)^n\right) \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right) \\ &= \left(\sum_{n=0}^\infty (iz)^{2n}\right) \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right) \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} i^{2k}\cdot \frac{1}{(n-2k)!}\right) z^n \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n. \end{align}