Computing $\int_{|z-i|=\frac{3}{2}}\frac{e^{\frac{1}{z^2}}}{z^2+1}$ 
Compute the integral using residues: $\int_{|z-i|=\frac{3}{2}}\frac{e^{\frac{1}{z^2}}}{z^2+1}$

Inside the circumference there are the following singular points $-i$ which is a pole of order 1 and $0$ which is essential.
 So:
$\int_{|z-i|=\frac{3}{2}}
\frac{e^{\frac{1}{z^2}}}{z^2+1}=res_{z_0=i}\frac{e^{\frac{1}{z^2}}}{z^2+1}+res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}=\frac{\pi}{e}+res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}$
$res_{z_0=0}\frac{e^{\frac{1}{z^2}}}{z^2+1}=c_{-1}$ which is the first negative coefficient of the Laurent series. So developing the Laurent series:
$\frac{e^{\frac{1}{z^2}}}{z^2+1}=\sum_\limits{n=0}^{\infty}\frac{1}{n!z^{2n}}\sum_\limits{m=0}^{\infty}(-1)^m z^{2m}$
However I am not seeing the coefficient $c_{-1}$
Question:
How should I compute the coefficient $c_{-1}$?
Thanks in advance!

Here is a way to calculate the integral without Laurent series to find the residue at $z = 0$. It uses that the sum of all residues is $0$ including the residue at infinity:


*

*$f(z)= \frac{e^{\frac{1}{z^2}}}{z^2+1} \Rightarrow$
$$ \operatorname{Res}_{z=0}f(z)  + \operatorname{Res}_{z=i}f(z)  = - \left( \operatorname{Res}_{z=-i}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) $$
Calculating the residues:
$$\operatorname{Res}_{z=-i}f(z) =\lim_{z\to -i}\frac{(z+i) e^{\frac{1}{z^2}}}{(z+i)(z-i)} = -\frac{e^{-1}}{2i}$$
$$\operatorname{Res}_{z=\infty}f(z) = -\operatorname{Res}_{w=0}\frac{1}{w^2}f\left(\frac{1}{w} \right)=-\operatorname{Res}_{w=0}\frac{1}{w^2}\frac{ e^{w^2}}{\frac{1}{w^2}+1} = -\operatorname{Res}_{w=0}\frac{ e^{w^2}}{1+w^2} = 0$$
The last residue is $0$ since $\frac{ e^{w^2}}{1+w^2}$ is holomorphic in a neighbourhood of $0$.
So, all together
$\int_{|z-i|=\frac{3}{2}}
\frac{e^{\frac{1}{z^2}}}{z^2+1}dz = -2\pi i\left(\operatorname{Res}_{z=-i}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) = -2\pi i\left( -\frac{e^{-1}}{2i}+0 \right) = \frac{\pi}{e}$
