1. ## Residue calculation

Hello to everyone!
Could someone explain to me why the residue of the complex function $f(z)= \frac{e^{1/z}}{z^{2}+1}$ is $a_{-1}=Res(f,0)=1-\frac{1}{3!}+\frac{1}{5!}-...= \sin{1}$

2. ## Re: Residue calculation

$f(z)=e^{1/z}\cdot \dfrac{1}{1+z^2}=\sum_{n=0}^{+\infty}\dfrac{1}{z^n n!}\cdot \sum_{n=0}^{+\infty}(-1)^nz^{2n}$ for $0<|z|<1$ . Now, find the coefficient of $1/z$ .

3. ## Re: Residue calculation

So it was so easy... I should have thought it! Thanks for the help!