Residue of a a complex quotient

Question

I have the following Laurent expansion corresponding to the function: $$f(z)=\frac{z+2}{z^2-4iz-3}$$ $$f(z)=\left(-1+\frac{1}{2}i\right)\sum_{n=1}^\infty \left(\frac{i}{z}\right)^n+\left(\frac{1}{3}+\frac{1}{2}i\right)\sum_{n=0}^\infty \left(\frac{z}{3i}\right)^n$$

The residue of $f(z)$ at $3i$ is: $Res(f;3i)=\frac{3}{2}-i$. Where can I see it in the Laurent expansion written above?

Answer 1

Hint: Rather than the Laurent expansion at $z=0$, use partial fractions: $$ \frac{z+2}{z^2-4iz-3}=\frac{\frac32-i}{z-3i}+\frac{-\frac12+i}{z-i} $$

Comment: Note that your Laurent expansion diverges at $z=3i$, so it is going to be very difficult to draw any conclusion about the behavior of $f(z)$ at $z=3i$ from it.

Answer 2

In order to see it use use to expand $f$ as a Laurent series around $3i$, i.e., $$ f(z)=\sum_{k=-\infty}^{\infty}a_k(z-3i)^k, $$ and your question is related to the coefficient $a_{-1}$.