problem says :

function $\displaystyle \displaystyle f(z) = \frac {z^5}{(z^2-4)^2 } $ develop in too Laurent series in region $\displaystyle |z|>2$

i really don't have much trouble with series but (perhaps this is more thread for Pre-university math ) i have problem with this one, especially how to do partial fraction of this function ...

i have perhaps 10 - 15 pages of the work and jet i didn't get the correct answer i'll post here some of the work where I think that if i do this correctly i solve rest myself

because up (can't be sure how to say it on English ) in fraction have bigger power than down to use partial fractions i need to divide it first so sown should have "z" with higher power (or am I mistaking ? because if not than i really don't know how to do it)

$\displaystyle \dfrac {z^5}{(z^2-4)^2 } $

when dividing $\displaystyle z^5$ with $\displaystyle z^4-8z^2+16$ i get

$\displaystyle \dfrac {z^5}{(z^2-4)^2 } = z + \dfrac {8z^3-16z} {z^4-8z^2+16}$

and now i do partial fractions :

$\displaystyle \displaystyle \frac {8z^3-16z} {z^4-8z^2+16}= \frac {A}{z-2} + \frac {B}{(z-2)^2}+\frac {C}{z+2} + \frac {D}{(z+2)^2} $

am I on the good path so far ? if not how can i solve this (even if I am on the good path, is there some more simple way to do this ? )

sorry if this is wrong sub forum for this type of the question

thanks in advance