Laurent Series of rational function about $z=0$ Find the Laurent expansion for $$\frac{\exp(\frac 1{z^2})}{z-1}$$ about $z=0$. I know that $\exp( \frac 1{z^2}) = \sum_{n=0}^\infty \frac{z^{-2n}}{n!}$ 
and $ \frac 1{z-1}=-\sum_{n=0}^\infty z^n$
 A: Write $$\exp(1/z^2) = \sum_{n=-\infty}^\infty a_n z^n \qquad\text{and}\qquad \frac{1}{z-1} = \sum_{n=-\infty}^\infty b_n z^n,$$
i.e. $a_{-2n} = 1/n!$ and $a_k = 0$ otherwise, and $b_n = -1$ for $n \ge 0$ and $b_n = 0$ for $n < 0$. Multiply the two series:
$$ \left(\sum_{n=-\infty}^\infty a_n z^n\right)\left(\sum_{n=-\infty}^\infty b_n z^n\right) = \left(\sum_{n=-\infty}^\infty c_n z^n\right)$$
and start identifying the $c_n$:s. We get
$$ c_0 = a_0 b_0 + a_{-2}b_2 + a_{-4}b_4 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$
$$ c_1 = a_0 b_1 + a_{-2}b_3 + a_{-4}b_5 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$
etc. In other words, $c_n = -e$ for all $n \ge 0$. For the negative indices you get
$$ c_{-1} = a_{-2} b_1 + a_{-4}b_3 + a_{-6}b_5 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$
$$ c_{-2} = a_{-2} b_0 + a_{-4}b_2 + a_{-6}b_4 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$
$$ c_{-3} = a_{-4} b_1 + a_{-6}b_3 + a_{-8}b_5 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1),$$
$$ c_{-4} = a_{-4} b_0 + a_{-6}b_2 + a_{-8}b_4 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1).$$
By now, the pattern should be obvious, if $n = -2k$,
$$ c_{-2k} = c_{-2k+1} = - \left(e - 1 - \frac1{1!} - \frac1{2!} - \cdots \frac1{k!}\right) = -e + \sum_{j=0}^k \frac{1}{j!}.$$
A: I have not read the previous anwer in detail. The short cut way is to notice the relationship that $$f^{n}=\frac{n!}{2\pi i}\oint \frac{f(z)dz}{(z-c)^{n+1}}$$
Notice that for Laurent series we have $$a_{n}=f^{n}/n!=\frac{1}{2\pi i}\oint \frac{f(z)dz}{(z-c)^{n+1}}$$
Thus the Laurent series coefficient for $f=\frac{e^{z^{-2}}}{z-1}$ is $$a_{n}=\frac{1}{2\pi i}\oint \frac{e^{z^{-2}}dz}{z^{n+1}(z-1)}$$
Consider the Cauchy integral formula which says $$f(c)=\frac{1}{2\pi i}\oint \frac{f(z)dz}{z-c}$$
Thus let $$g=\frac{e^{z^{-2}}}{z^{n+1}}, a_{n}=g(1)=e$$ should suffice. I might be wrong somewhere though. 
