<s>[INST] Laurent series of $\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$ I want to expand the function following into Laurent series:
$$\frac{z^2 + iz - 1}{(z - 2)(z + 3)}$$
for $2 < |z| < 3$
My solution
First thing to apply is partial function decomposition:
$$\frac{1}{(z - 2)(z  +3)} = \frac{\frac 1 5}{z - 2} + \frac{-\frac 1 5}{z + 3}$$
From which we have that:
$$\frac{z^2 + iz - 1}{(z - 2)(z  +3)} = \frac{\frac 1 5 (z^2 + iz - 1)}{z - 2} + \frac{-\frac 1 5 (z^2 + iz - 1)}{z + 3}$$
Now we expand those two series:
$$\frac{1}{z - 2} = \frac{1}{z}\frac{1}{1 - \frac 2 z} = \sum_{n = 0}^\infty \frac{2^n}{z^{n + 1}}$$
$$\frac{1}{z + 3} = \frac{1}{3(\frac z 3 + 1)} = \frac 1 3 \frac{1}{1 - ( - \frac z 3)} = \sum_{n = 0}^\infty (-1)^n \frac{z^n}{3^{n + 1}}$$
So final result is:
$$\frac{z^2 + iz - 1}{(z - 2)(z + 3)} = \sum_{n = 0}^\infty \frac{z^2 + iz - 1}{5}[\frac{2^n}{z^{n + 1}} - \frac{(-1)^nz^n}{3^{n + 1}}]$$
Does it make any sense to you?
 [/INST] An idea for you: using Taylor's polynomial, or whatever, write the numerator as power of $\;z-2\;$ (first...), so you get:
$$f(z)=z^2+iz-1=f(2)+f'(2)(z-2)+\frac{f''(2)(z-2)^2}{2!}$$
and now multiply termwise the above by the development around $\;z=2\;$ to get an actual (Laurent) power series:
$$\left(f(2)+f'(2)(z-2)+\frac{f''(2)(z-2)^2}{2!}\right)\sum_{n=0}^\infty\frac{2^n}{z^{n+1}}=\frac{f''(2)(z-2)}{2}+f'(2)+\ldots$$
(you can write the above more detailed), and then do the same with $\;z=3\;$ ...
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