Hello to everyone! Could someone explain to me why the residue of the complex function $\displaystyle f(z)= \frac{e^{1/z}}{z^{2}+1}$ is $\displaystyle a_{-1}=Res(f,0)=1-\frac{1}{3!}+\frac{1}{5!}-...= \sin{1}$ Thanks in advance!
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$\displaystyle f(z)=e^{1/z}\cdot \dfrac{1}{1+z^2}=\sum_{n=0}^{+\infty}\dfrac{1}{z^n n!}\cdot \sum_{n=0}^{+\infty}(-1)^nz^{2n}$ for $\displaystyle 0<|z|<1$ . Now, find the coefficient of $\displaystyle 1/z$ .
So it was so easy... I should have thought it! Thanks for the help!