Laurent Series of $\frac{1}{z(1-z)}$ in neighborhood of $z=1$ and $z=0$

Question

So, the question is: Laurent Series of $\frac{1}{z(1-z)}$ in neighborhood of $z=1$ and $z=0$.
I know I can find Laurent series' all over MSE, but in an effort to build my own intuition, and to see the entire process, I'm just trying to show all the details.
So, here is my attempt: First, we need to break $\frac{1}{z(1-z)}=\frac{1}{z}+\frac{1}{1-z}$ using partial fraction decomposition.

First question: "In a neighborhood of $z=0$ and $z=1$" does this mean we consider $|z|<1$ for "neighborhood of $z=0$" and $|z-1|<1$ for "neighborhood of $z=1$"?

Supposing that I am right about my first question, then $\frac{1}{1-z}=\sum_{n=0}^\infty z^n$, but we still have to play with the $\frac{1}{z}$ part.

Second question: Is there an obvious way to break down $\frac{1}{z}$?

I was thinking I could answer my second question this way: $\frac{1}{z}=\frac{1}{1+z-1}=\frac{1}{1-(-(z-1))}=\sum_{n=0}^\infty(-1)^n(z-1)^n$.

So, getting back to the problem, we would have, in a neighborhood of $z=0$, the Laurent series of $\frac{1}{z(1-z)}$ is $\sum_{n=0}^\infty(-1)^n(z-1)^n+\sum_{n=0}^\infty z^n=\sum_{n=0}^\infty(-1)^nz^n(z-1)^n$..... right?

Now, for the "in a neighborhood of $z=1$" part, I (assuming I am thinking of my questions $1$ correctly) consider $|z-1|<1$, and so $\frac{1}{1-z}=-\sum_{n=1}^\infty\frac{1}{z^n}$ and $\frac{1}{z}=-\sum_{n=1}^\infty\frac{(-1)^n}{(z-1)^n}$ (then add), right?

Any insight, tips, ideas, etc. would be greatly appreciated! Thank you.

Answer 1

It is much simpler than what you did:

• Near $z=0$, just write $$\frac1{z(1-z)}=\frac1z(1+z+z^2+z^3+\dots)=\frac1z+1+z+z^2+\dotsm.$$
• Near $z=1$, use substitution first: set $u=z-1$, so \begin{align}\frac1{z(1-z)}&=-\frac1{u(1+u)}=-\frac1u(1-u+u^2-u^3+\dotsm)\\ &=-\frac1u+1-u+u^2-\dotsm \end{align} and eventually, substitute $u$ with $z-1$.