So, I have a homework question about Laurent series expansion - I feel comfortable with these question but this one is quite a bit more complicated and I wanted to ask whether what I've done makes sense for a function like this? So, I need to find a Laurent expansion about $0$, valid in the annulus $1 < \vert z \vert < 2$ for
$$ f(z) = \frac{z+1}{(z+2)(z-i)} $$
- Partial fraction decomposition:
$$ f(z) = \frac{z+1}{(z+2)(z-i)} = \frac{\frac{2}{5}-\frac{1}{5}i}{z+2} + \frac{\frac{3}{5}+\frac{1}{5}i}{z-i} $$
- Find series valid for $\vert z \vert > 1$
$$ \begin{align} \frac{\frac{3}{5}+\frac{1}{5}i}{z-i} &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\left(\frac{1}{1 - zi}\right)\\ \\ &= i\left(\frac{3}{5}+\frac{1}{5}i\right)\sum_{n = -\infty}^{-1}(-zi)^{n}, \; \vert zi\vert > 1\\ \\ &= \sum_{n = -\infty}^{-1}\left(\frac{3}{5}+\frac{1}{5}i\right)(-1)^{n}z^{n}i^{n+1}, \; \vert z\vert > 1 \end{align} $$
- Find series valid for $\vert z \vert < 2$
$$ \begin{align} \frac{\frac{2}{5}-\frac{1}{5}i}{z+2} &= 2\left(\frac{2}{5}-\frac{1}{5}i\right)\left(\frac{1}{1 - (-z/2)}\right)\\ \\ &= 2\left(\frac{2}{5}-\frac{1}{5}i\right)\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^{n}, \; \left| -\frac{z}{2}\right| < 1\\ \\ &= \sum_{n = 1}^{\infty}\left(\frac{2}{5}-\frac{1}{5}i\right)(-1)^{n -1}\frac{z^{n-1}}{2^{n}}, \; \left| z\right| < 2 \end{align} $$
- Thus, $$ \begin{align} f(z) = \frac{z+1}{(z+2)(z-i)} = \sum_{n = -\infty}^{-1}\left(\frac{3}{5}+\frac{1}{5}i\right)(-1)^{n}z^{n}i^{n+1} + \sum_{n = 1}^{\infty}\left(\frac{2}{5}-\frac{1}{5}i\right)(-1)^{n-1}\frac{z^{n-1}}{2^{n}}\\ \end{align} $$
Valid in the annulus $1 < \vert z \vert < 2$
Does this work?