# Laurent Expansion of $\frac{(z+1)^2}{z(z^3+1)}$

I’m stuck on finding the Laurent expansion this function about $$z=0$$:

$$\frac{(z+1)^2}{z(z^3+1)}$$

What I tried was to compute the Binomial Expansion for the top bit and then expand:

$$\frac{1}{1+z^3}=\frac{1}{z^3}\frac{1}{1-(-\frac{1}{z^3})}$$

I ended up with the following:

$$\sum_{n=0}^{\infty}\begin{pmatrix}2\\n\end{pmatrix}z^{n-4}\sum_{n=0}^{\infty} (-1)^nz^{4-3n}$$

Now I don’t know if I’m right what I have and if I am how can I continue. I don’t know if I’m allowed to do a Cauchy Product here.

Many thanks !

• $z$ is small, so you don't want to manipulate $\frac{1}{1+z^3}$ like that.
– J.G.
Jan 11, 2023 at 15:29
• @J.G. Oh yeah you are right, but what about the rest? Jan 11, 2023 at 15:30
• The maximum $n$ with $\binom{2}{n}\ne0$ is $n=2$. It's easiest to start from $\frac1z(1+2z+z^2)(1-z^3+z^6-\cdots)$.
– J.G.
Jan 11, 2023 at 15:31
• Ok it makes perfect sense now, cheers! Jan 11, 2023 at 15:33

You have : $$\begin{array}{rcl} \displaystyle\frac{(z+1)^2}{z(z^3+1)} &=&\displaystyle \left(z+2+\frac{1}{z}\right) \sum_{k=0}^\infty (-1)^kz^{3k} \\ &=&\displaystyle \sum_{k=0}^\infty (-1)^kz^{3k+1} + 2\sum_{k=0}^\infty (-1)^kz^{3k} + \sum_{k=0}^\infty (-1)^kz^{3k-1} \\ &=&\displaystyle \frac{1}{z} + \sum_{k=0}^\infty(-1)^k\left(2z^{3k}+z^{3k+1}-z^{3k+2}\right) \end{array}$$ where the first term of $$\sum_{k=0}^\infty z^{3k-1}$$ (corresponding to $$k=0$$) has been extracted from the sum, before re-indexing it.
• Not quite: the last factor should give $z^{3k}$ a $(-1)^k$ coefficient.
• How? I found the second term so the sum just with $z^{3k-1}$ at the end Jan 11, 2023 at 15:38