Find the Laurent series expansion for
$$f(z)=\frac{1}{z+1}+\frac{1}{z-3},$$
in the domains
(a) $0<\vert z \vert <1$,
(b) $1<\vert z \vert<3$,
and
(c) $3<\vert z \vert<\infty.$
Attempt:
(a) We have
\begin{align} \frac{1}{z+1}+\frac{1}{z-3}&=\frac{1}{1+z}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}\\ &=\sum_{n=0}^\infty (-1)^nz^n-\sum_{n=0}^\infty \frac{z^n}{3^{n+1}}. \end{align}
For (b) I get
\begin{align} \frac{1}{z+1}+\frac{1}{z-3}&=\frac{1}{z}\frac{1}{1+\frac{1}{z}}-\frac{1}{3}\frac{1}{1-\frac{z}{3}}\\ &=\frac{1}{z}\sum_{n=0}^\infty (-1)^n\bigg(\frac{1}{z}\bigg)^n-\frac{1}{3}\sum_{n=0}^\infty \bigg(\frac{z}{3}\bigg)^n\\ &=\sum_{n=-1}^{-\infty} (-1)^{n+1}z^n - \sum_{n=0}^\infty \frac{z^n}{3^{n+1}}. \end{align}
For (c) I have
\begin{align} \frac{1}{z+1}+\frac{1}{z-3}&=\frac{1}{z}\frac{1}{1+\frac{1}{z}}+\frac{1}{z}\frac{1}{1-\frac{3}{z}}\\ &=\frac{1}{z}\sum_{n=0}^\infty \bigg(\frac{1}{z}\bigg)^n+\frac{1}{z}\sum_{n=0}^\infty \bigg(\frac{3}{z}\bigg)^n\\ &=\sum_{n=-1}^{-\infty}z^n+\sum_{n=-1}^{-\infty}\frac{z^n}{3^{n+1}}. \end{align}
Am I correct? Thanks for any tips or hints if I am incorrect anywhere. Ill take an upvote as in, I am doing things correctly..