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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!

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    $\begingroup$ The answer will depend on which annulus you are expanding over. $\endgroup$ – Lord Shark the Unknown May 17 at 1:53
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By definition we have that $e^{1/z^2}=\sum_{k=0}^\infty\frac{z^{-2k}}{k!}$ for $z\neq 0$, and

$$\frac1{z-1}=-\sum_{k=0}^\infty z^k,\quad\text{ when } |z|<1\\ \frac1{z-1}=\frac1z\frac1{1-z^{-1}}=\frac1z\sum_{k=0}^\infty z^{-k}=\sum_{k=0}^\infty z^{-(k+1)},\quad\text{ when } |z|>1$$

Hence

$$\frac{e^{1/z^2}}{z-1}=\begin{cases}-\sum_{k=0}^\infty\sum_{j=0}^\infty \frac{z^{j-2k}}{k!}=\sum_{n\in\Bbb Z} c_n z^n,\quad 0<|z|<1\\ \sum_{k=0}^\infty\sum_{j=0}^\infty \frac{z^{-j-2k-1}}{k!}=\sum_{n\in\Bbb Z} b_n z^n,\quad |z|>1\end{cases}$$

where

$$c_n:=-\sum_{j,k\in\Bbb N:j-2k=n}\frac1{k!} \\b_n:=\sum_{j,k\in\Bbb N:-j-2k-1=n}\frac1{k!}$$

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