Laurent series of $\frac{e^z}{z^2+1}$

Question

I cant figure out the laurent series of the following function.

Let $f(z)= \frac{e^z}{z^2+1} $ and $|z|\gt 1$

$$\frac{1}{z^2+1}=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}$$

and

$$e^z = \sum_{n=0}^{\infty}\frac{z^{n}}{n!}$$

$$e^z*\frac{1}{z^2+1} =\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^nz^{-2n-2}*\frac{z^{n-k}}{(n-k)!}$$

How can I go from here ?

Answer 1

Let

\begin{align*} f(z)&= \frac{e^z}{z^2+1} \end{align*}

We assume we need a Laurent expansion for all $|z|>1$ with center $z_0=0$.

We obtain \begin{align*} f(z)&=\frac{e^z}{z^2+1}\\ &=\frac{e^z}{z^2}\cdot\frac{1}{1+\frac{1}{z^2}}\\ &=\frac{e^z}{z^2}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k}}\\ &=\sum_{j=0}^\infty\frac{z^j}{j!}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k+2}}\\ &=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k-2=n}\atop{j\geq 0;k\geq 0}}\frac{(-1)^k}{j!}\right)z^n\tag{1}\\ &=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k=n}\atop{j\geq 0;k\geq 1}}\frac{(-1)^{k+1}}{j!}\right)z^n\tag{2}\\ &=\sum_{n=-\infty}^\infty\left(\sum_{k=K}^\infty\frac{(-1)^{k+1}}{(n+2k)!}\right)z^n\tag{3}\\ \end{align*}

Comment:

• In (1) we apply the Cauchy product formula

• In (2) we shift the index $k$ by one

• In (3) we replace $j$ with $n+2k$ and since $j=n+2k\geq 0$ and $k\geq 1$ we have to set \begin{align*} K=\max\left\{1,-\left\lfloor\frac{n}{2}\right\rfloor\right\} \end{align*}