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 !
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.