Laurent series: how to join the 2 sums for $f(z)= \frac{1}{(z-1)(z+1)}$ about z = 1 for $0 < |z − 1| < 2$

Question

We are to find the Laurent series for f(z) about $z = 1$ for $0 < |z − 1| < 2$:

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

Assumptions: $|\frac{z−2}{1}| < 1 ⇔ |z − 1| < 2$

For $\frac{1}{(z-1)}$ we have the standard geometric series expansion as follows:

As for the second part we can expand $\frac{1}{(z+1)}$ in the following way:

Now my question is how to join these two sums, should they multiplied togheter like this?

$\sum_{n=1}^\infty\frac{(-1)^n (z-1)^n}{2^{n}}\sum_{n=1}^\infty{z^n} = \sum_{n=1}^\infty\frac{(-1)^n (z-1)^n}{2^{n}} z^n$

But the actual answer looks different, and is supposed to be the below sum. How exactly is this achieved?:

Answer 1

First of all, note that you cannot use the expansion $$ \frac{1}{1 - z} = 1 + z + z^2 + … \text{,}$$ because the condition $|z| < 1$ does not necessarily hold. In fact, it would be undesirable to produce any power series expansion in $z$, because you are supposed to find the Laurent series centered at $1$. Merely take your correct equation $$ \frac{1}{1+z} = \frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^n(z-1)^n $$ and multiply both sides by $(z-1)^{-1}$.