Find the laurent series of the function $$f(z) = \frac{1}{(z+i)(z-5i)}$$ which is valid in the annulus where $1 < |z| < 5$
This is where I'm at currently. I'm not sure if this is the correct way of going about answering this:
$$\frac{1}{(z+i)(z-5i)}$$
$$= \frac{-1}{6i(z+i)} + \frac{1}{6i(z-5i)}$$
$$= \frac{i}{6(z+i)} + \frac{-i}{6(z-5i)}$$
Partial fraction 1:
$$\frac{i}{6(z+i)} = \frac{i}{6z+6i} = \frac{i}{6z(1+\frac{i}{z})} = \frac{i}{6z} \cdot \frac{1}{(1-(-\frac{i}{z}))}$$
$1 < |z|$, so: $$|-\frac{i}{z}| = \frac{1}{|z|} < 1$$
$$\frac{i}{6(z+i)} = \frac{i}{6z} \cdot \sum_{n=0}^{\infty}(-\frac{i}{z})^n = \frac{1}{6} \cdot \sum_{n=0}^{\infty}(-\frac{i}{z})^{n+1}$$
Partial fraction 2:
$$\frac{-i}{6(z-5i)} = \frac{-i}{6z-30i} = \frac{-i}{-30i(1-\frac{z}{5i})} = \frac{1}{30(1-\frac{z}{5i})} =\frac{1}{30} \cdot \frac{1}{1-\frac{z}{5i}}$$
$|z| < 5$, so: $$|\frac{z}{5i}| = \frac{|z|}{5} < 1$$
$$-\frac{i}{6(z-5i)} = \frac{1}{30i} \cdot \sum_{n=0}^{\infty}(\frac{z}{5i})^n = \sum_{n=0}^{\infty}(\frac{z}{5})^n \cdot (\frac{1}{i})^{n+1}$$
So
$$f(z) = \frac{1}{6} \cdot \sum_{n=0}^{\infty}(-\frac{i}{z})^{n+1} + \sum_{n=0}^{\infty}(\frac{z}{5})^n \cdot (\frac{1}{i})^{n+1}$$
$$ = \frac{1}{6} [ \sum_{n=-\infty}^{-1}-(\frac{z}{i})^{2n+1} + \sum_{n=1}^{\infty}(\frac{i}{z})^{2n}] + \sum_{n=0}^{\infty}(\frac{z}{5})^n \cdot (\frac{1}{i})^{n+1}$$
Valid in the annulus $1< |z| < 5$
(Apologies if there are any small mistakes)
If this is the final result, I imagine it could be simplified further.
This is mostly your argument, but cleaned up and expressed as an actual Laurent series.
It is probably easier to do this first for $g(w)=f(iw).$ Then we don't have to keep track of all those $i$ values just yet, staying in the real coefficients.
We get $$g(w)=\frac1{(1+w)(5-w)}=\frac16\left(\frac1{1+w}+\frac1{5-w}\right)$$ and you get, for $|w|>1,$
$$\frac1{1+w}=\sum_{n=-\infty}^{-1}(-1)^{n+1}w^n$$
and:
$$\frac1{5-w}=\sum_{n=0}^{\infty} \frac{w^n}{5^{n+1}},$$ for $|w|<5.$
So: $g(w)=\sum_{n=-\infty}^{\infty} b_nw^n$ where:
$$b_n=\begin{cases}\frac16(-1)^{n+1}&n<0\\ \frac16\frac{1}{5^{n+1}}&n\geq0 \end{cases}$$
Now $f(z)=g(-iz).$ So you get $f(z)=\sum_{n=-\infty}^\infty a_nz^n$ where $$a_n=(-i)^n b_n=\begin{cases}-\frac{i^n}6&n<0\\ \frac16\frac{(-i)^n}{5^{n+1}}&n\geq 0 \end{cases}$$