# 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.
• Thanks for the quick answer. But in this case, I don't understand the utility of the Laurent series, because we still don't have any value for $f(6i)$ – Sihtam Jan 11 '15 at 19:14
• The function has a pole at $6i$. There is no value there because $6i$ is not in the domain of the function. It's often easiest to obtain the Laurent series as a sort of trick or manipulation of another series or expression, instead of direct computation. – MPW Jan 12 '15 at 12:20