# What is the Laurent expansion of $\frac{e^{1/z^2}}{z-1}$?

As above, I do not know how to get the Laurent expansion of $$\frac{e^{1/z^2}}{z-1}$$ over 0. What I think I understand is splitting the denominator into $$\frac{1}{z}\sum_0^\infty \frac{1}{z}^n + \frac{1}{z}\sum_0^\infty z^n$$

(I am not sure this is correct)

But then, what do I do about $$e^{1/z^2}$$? Clear steps and explanation would be very helpful!

• The answer will depend on which annulus you are expanding over. – Lord Shark the Unknown May 17 at 1:53

By definition we have that $$e^{1/z^2}=\sum_{k=0}^\infty\frac{z^{-2k}}{k!}$$ for $$z\neq 0$$, and
$$\frac1{z-1}=-\sum_{k=0}^\infty z^k,\quad\text{ when } |z|<1\\ \frac1{z-1}=\frac1z\frac1{1-z^{-1}}=\frac1z\sum_{k=0}^\infty z^{-k}=\sum_{k=0}^\infty z^{-(k+1)},\quad\text{ when } |z|>1$$
$$\frac{e^{1/z^2}}{z-1}=\begin{cases}-\sum_{k=0}^\infty\sum_{j=0}^\infty \frac{z^{j-2k}}{k!}=\sum_{n\in\Bbb Z} c_n z^n,\quad 0<|z|<1\\ \sum_{k=0}^\infty\sum_{j=0}^\infty \frac{z^{-j-2k-1}}{k!}=\sum_{n\in\Bbb Z} b_n z^n,\quad |z|>1\end{cases}$$
$$c_n:=-\sum_{j,k\in\Bbb N:j-2k=n}\frac1{k!} \\b_n:=\sum_{j,k\in\Bbb N:-j-2k-1=n}\frac1{k!}$$