How to find the Laurent series of the following function in $z=i$.
$$f(z)=e^{\frac{1}{(z+i)(z-i)}}$$
I know that it is a essential singularity. There is a way to find the series rapresention without calculate the following integral for every single term
$$\sum a_{k}(z-z_{0})^{k}$$
$$a_{k}=\frac{1}{2\pi i}\oint_{\gamma } \frac{f(z)}{(z-z_{0})^{k+1}}dz$$
Your absolutely correct that the series expansion can be obtained using Cauchy's integral formula.
But there is an easier way.
Hint 1: Use the series expansion for exponential functions: $$ \exp \left( \frac 1 {(z + i)(z - i)}\right) = 1 + \frac 1 {(z + i)(z - i)} + \frac 1 {2!} \times \frac 1 { (z + i)^2(z - i)^2} + \dots $$
Hint 2: Use partial fractions: $$ \frac 1 {(z + i)(z - i)} = \frac i {2(z + i)} - \frac i {2(z - i)}$$
Hint 3: Use the geometric series formula to expand $1/(z + i)$ around $z = i$: \begin{align*} \frac 1 {z + i} & = \frac 1 {2i + (z-i)} \\\ & = (2i)^{-1} \times \frac 1 {1 + (2i)^{-1}(z-i)} \\\ & = (2i)^{-1} \times \left(1 - (2i)^{-1}(z-i) + (2i)^{-2}(z-i)^{2} - (2i)^{-3}(z-i)^{3} + \dots \right)\end{align*} which converges for $| z - i | < 2$.