Find the Laurent series about $z=i$

Question

Let $g(z)=\cfrac{3z+1}{(z-i)^3}$. Find the Laurent expansion of $g$ about $z=i$.

My idea is first to find the Laurent series of $\cfrac{1}{z-i}$ about $z=i$, and then diferenciate, but I have problem with that. I want to use $$\displaystyle\sum_{n=0}^{+\infty}z^n=\cfrac{1}{1-z},\quad \|z\|<1$$ How can I do that with $\cfrac{1}{z-i}$ about $z=i$?

Thank you,

Answer 1

You have a direct way, upon writing $$ g(z)=\cfrac{3z+1}{(z-i)^3}=\cfrac{3(z-i)+1+3i}{(z-i)^3}=\cfrac{3}{(z-i)^2}+\cfrac{1+3i}{(z-i)^3}, $$ this is the Laurent series expansion for $g(z)$ on $0<|z-i|$.

Answer 2

Observe that $$ \frac{3z+1}{(z-i)^3}=\frac{3(z-i)}{(z-i)^3}+\frac{3i+1}{(z-i)^3}=\frac{3}{(z-i)^2}+\frac{3i+1}{(z-i)^3} $$ And this is the Laurent series for the function