1. ## Laurent Series

Hi, I have to compute the laurent series of this function $\frac{z^2+1}{(z^3+1)^2}$ in $|z|>1$.
Sigularities are: $z=-1$ and $z=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$.
Now I do not know how to continue, I wrote partial fractions $\frac{z^2}{(z^3+1)^2}+\frac{1}{(z^3+1)^2}$ and I thought for the first fraction to use the derivative method, but there is $z^{3}$, so can't I use that method? While for the second fraction what I have to do... some ideas?

2. ## Re: Laurent Series

Originally Posted by quidh
Now I do not know how to continue, I wrote partial fractions $\frac{z^2}{(z^3+1)^2}+\frac{1}{(z^3+1)^2}$ and I thought for the first fraction to use the derivative method, but there is $z^{3}$, so can't I use that method? While for the second fraction what I have to do... some ideas?
Using the derivative method you'll obtain $1/(u+1)^2}=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)u^{-n-2}\;\;(|u|>1)$ . If $u=z^3$ then, $|u|>1\Leftrightarrow |z|>1$ so $1/(z^3+1)^2=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)z^{-3n-6}\;\;(|z|>1)$ and $z^2/(z^3+1)^2=\ldots=\sum_{n=0}^{+\infty} (-1)^n(n+1)u^{-3n-4}\;\;(|z|>1)$

3. ## Re: Laurent Series

Thanks for help!

4. ## Re: Laurent Series

gives you the exact expansion
$\frac{1}{4}+\frac{x^2}{4}-\frac{x^3}{4}+\frac{5 x^4}{36 \left((-2)^{1/3}-x\right)^2}-\frac{7 (-1)^{1/3} x^4}{36\ 2^{2/3} \left((-2)^{1/3}-x\right)^2}+\frac{x^5}{12 \left((-2)^{1/3}-x\right)^2}+\frac{(-1)^{2/3} x^5}{9\ 2^{1/3} \left((-2)^{1/3}-x\right)^2}+\frac{5 x^4}{36 \left(2^{1/3}+x\right)^2}+\frac{7 x^4}{36\ 2^{2/3} \left(2^{1/3}+x\right)^2}+\frac{x^5}{12 \left(2^{1/3}+x\right)^2}+\frac{x^5}{9\ 2^{1/3} \left(2^{1/3}+x\right)^2}+\frac{5 x^4}{36 \left((-1)^{2/3} 2^{1/3}+x\right)^2}+\frac{7 \left(-\frac{1}{2}\right)^{2/3} x^4}{36 \left((-1)^{2/3} 2^{1/3}+x\right)^2}+\frac{x^5}{12 \left((-1)^{2/3} 2^{1/3}+x\right)^2}-\frac{\left(-\frac{1}{2}\right)^{1/3} x^5}{9 \left((-1)^{2/3} 2^{1/3}+x\right)^2}$
bye