Develop into Laurent series around $0$: $\frac{e^{1/z}}{z - 3i} .$ Develop into Laurent series around $0$:
$$\frac{e^{\frac{1}{z}}}{z - 3i} .$$
I was thinking of developing $e^{\frac{1}{z}}$ first and then $\frac{1}{z - 3i}$, but I got stuck while writing its multiple as one sum. Is that a good way of solving this?
I'm new to developing to Laurent series so any hint helps!

If you have two functions $f$ and $g$ holomorphic on an annulus $$\mathcal{A}=\{z\in\mathbb{C}: r_0<|z-z_0|<r_1\},$$ then given Laurent series' $$f(z)=\sum\limits_{n=-\infty}^\infty a_n (z-z_0)^n$$ and $$g(z)=\sum\limits_{n=-\infty}^\infty b_n (z-z_0)^n$$  valid on $\mathcal{A}$, we will have $fg$ hololomorphic on $\mathcal{A}$, as well, with Laurent series $$f(z)g(z)=\sum\limits_{n=-\infty}^\infty c_n (z-z_0)^n, $$ where $$c_n=\sum\limits_{k=-\infty}^\infty a_kb_{n-k},$$ valid on $\mathcal{A}.$ This is analogous to taking the product of two power series' with the same radius of convergence.
