Laurent series of $z/(z-6i)^9$ I'm really confused by the Laurent series, and need a little help with this one:
$ f(z) = \frac z{(z-6i)^9} $ around $z = 6i$
The partial fraction method doesn't seems adapted for this problem.
I wrote 
$$ f(z) = \sum_{n=-\infty}^{\infty} a_{n} (z-6i)^{n} $$
With $$a_{n} = \frac1 {2\pi i} \oint_C \frac {f(z)}{(z-6i)^{n+1}} dz$$
$$ f(z) = \sum_{n=-\infty}^{\infty} \frac1 {2\pi i} \oint_C \frac z{(z-6i)^{n+10}} dz . (z-6i)^{n} $$
I don't know what to do next. I'm also supposed to tell in which region the expansion is valid. As there's only one pole in $z=6i$ I was thinking that the expansion would be valid in the whole complex plane?
Thanks

Hint: You can write $z$ as $z-6i + 6i$. Then
$$\frac{z}{(z-6i)^9} = \frac{z-6i}{(z-6i)^9} +\frac{6i}{(z-6i)^9}
$$
$$= \frac{6i}{(z-6i)^9}+\frac{1}{(z-6i)^8}
$$
This looks valid for all $z\neq 6i$ to me. The series is finite.
