Find Laurent series Let $$f(z):=\frac{e^{\frac{1}{z}}}{z^2+1}$$ and let $$\sum_{k}a_kz^k$$ with k in Z the Laurent series of $f(z)$ for $0<|z|<1$. I have to find a formula for $a_k$. I've tried a lot, but I'm stuck. Can somebody help me? I want to do it with ordening of absolute series.

We have $$e^\frac{1}{z}=1+\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+ \dots$$ as well as the geometric expansion$$\frac{1}{z^2+1}=1-z^2+z^4 -+ \dots $$
valid in the annulus $0<|z|<1$. Multiply the two and collect coefficients from terms of the same degree.
