Let $$ f(z) = \frac{1}{(z - 4)(z + 8i)} $$
a) Find the domains where f(z) is valid
b) Find its power series at such domains
Considering three singularities, I believe the domains are:
$$ D_{1} = |z| < 4 \\ D_{2} = 4 < |z| < 8 \\ D_{3} = |z| > 8 $$
But I'm not sure if I should use a Laurent or a Taylor series. Since D1 deals with negative infinity, maybe a Laurent is needed there.
Thanks for your time.
We can decompose $f(z)$ as $$ f(z) = \frac{1-2i}{20(z-4)} - \frac{1-2i}{20(z+8i)} $$ For the first region, $\lvert z\rvert < 4\Rightarrow \frac{\lvert z\rvert}{4}<1$. Since $\frac{\lvert z\rvert}{4}<1$, $\frac{\lvert z\rvert}{8}<\frac{1}{2}<1$ \begin{align} \frac{1-2i}{20(z-4)} - \frac{1-2i}{20(z+8i)} &= \frac{1-2i}{-80}\frac{1}{1-\frac{\lvert z\rvert}{4}}-\frac{1-2i}{160i}\frac{1}{1+\frac{\lvert z\rvert}{8i}}\\ &= \frac{2i-1}{80}\sum_{n=0}^{\infty}\Bigl(\frac{z}{4}\Bigr)^n+\frac{2+i}{160}\sum_{n=0}^{\infty}i^n\Big(\frac{z}{8}\Bigr)^n\tag{1} \end{align} Additionally, you can simplify $(1)$ however you see fit. You then would do the same analysis for the second and third regions.