Let $C$ is positive oriented $|z|=1/2$ circle.
Calculate $\int_C \frac{e^{1/z^2}}{z-1}dz$
I think I can solve it via using residue. The only singular point is $0$ in there. Should I write Laurent series of $e^{1/z^2}$ in $1<|z-1|< \infty $ neighborhood?
Last, if the circle were $|z|=1$ what should I do in that situation?
Thanks in advance
On
The area of convergence for $e^\frac{1}{z^2}$ is always $\mathbb C\setminus\{0\}$, which is good, beacuse you want it to be convergent in the vincinity of $0$.
You have $$e^\frac{1}{z^2} = \sum_{n=0}^\infty \frac{1}{n!}z^{-2n}$$ and in the vincinity of $z=0$: $$ \frac{1}{1-z} = \sum_{m=0}^\infty z^m$$ so \begin{align} \frac{e^\frac{1}{z^2}}{1-z} &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{1}{n!}z^{m-2n} = \\ &= \sum_{n=0}^\infty \sum_{k=-2n}^\infty \frac{1}{n!}z^{k} = \\ &= \sum_{k=-\infty}^\infty\sum_{n\ge\max\{0,-\frac{k}{2}\}} \frac{1}{n!}z^{k} = \\ \end{align} Therefore the residdum at $0$, the coefficient at $z^{-1}$ is equal to $$ a_{-1} = \sum_{n\ge\max\{0,\frac{1}{2}\}} \frac{1}{n!} = \sum_{n=1}^\infty \frac{1}{n!} = e-1$$
If the circle were to be $|z|>1$ you need to specify how you surround the singularity at $z=1$, or whether you want to calculate the principal value of the integral.
For the circle $|z|=R>1$ you'd need to include the residuum for $z=1$, or expand function $\frac{1}{1-z}$ in the region $|z|>1$ ($\frac{1}{1-z} = -\sum_{k=1}^\infty z^{-k}$) and find the coefficient $a_{-1}$ in subsequently obtained series for $\frac{e^{1/z^2}}{1-z}$.
Since$$\frac{e^{\frac1{z^2}}}{1-z}=\left(1+\frac1{z^2}+\frac1{2!z^4}+\frac1{3!z^6}+\cdots\right)(1+z+z^2+z^3+\cdots)\tag1$$and since the residue of $\frac{e^{\frac1{z^2}}}{1-z}$ at $0$ is the coefficient of $\frac1z$ that you get multiplying the series from $(1)$, then that residue is equal to$$1+\frac1{2!}+\frac1{3!}+\cdots=e-1.$$Therefore, your integral is equal to $2\pi i(e-1)$.