Laurent series for $\frac{z}{1+z^3}$ for $|z|>1$

Question

I managed a decent start but got stuck:

$|z|>1$ implies $1/|z|<1$, so we can use $\sum_{k=0}^{\infty}z^{-k} = \frac{1}{1-1/z}$ to help find the Laurent series:

$$\frac{z}{1+z^3} =\frac{z-1}{1+z^3}\frac{1}{1-1/z} = \frac{z-1}{(z-1)(z^2-z-1)}\sum_{k=0}^{\infty}z^{-k} = \frac{1}{z^2-z-1}\sum_{k=0}^{\infty}z^{-k} $$

But now I have this nasty factor out in front, when I'm looking for an answer which is only powers of $z$. I thought about factorising the quadratic but it wouldn't help get an answer only in terms of powers of $z.$ Any hints would be greatly appreciated!

Answer 1

$\frac z {1+z^{3}}=\frac 1{z^{2}} \frac 1{1+z^{-3}}=\frac 1{z^{2}} \sum\limits_{n=1}^{\infty} (-1)^{n}z^{-3n}$.

Answer 2

Put $w = 1/z$. Then $$ \frac{z}{1 + z^3} = \frac{zw^3}{w^3 + z^3 w^3} = \frac{w^2}{1 + w^3}. $$ Hence $$ \frac{w^2}{1 + w^3} = \frac{w^2}{1 - (-w^3)} = w^2 \sum_{k = 0}^{\infty} {(-w^3)^k} = \sum_{k = 0}^{\infty} {(-1)^k w^{3k+2}} $$ for $0 < |w| < 1$. Hence $$ \frac{z}{1 + z^3} = \sum_{k = 0}^{\infty} {\frac{(-1)^k}{z^{3k + 2}}} $$ for $|z| > 1$.