Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$

Question

Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$.

I know how to do this for $|z|>1$ but I don't understand how I should set it up for $|z|>2$.

Here is how I would do it for $|z|>1$.

$\frac{1}{z-1}= \frac{1}{z}\frac{1}{1-1/z}=\frac{1}{z} \sum_{n=0}^\infty (\frac{1}{z})^n$

And similarly $\frac{1}{z+1}=\frac{1}{z} \frac{1}{1-(-1/z)} = \frac{1}{z} \sum_{n=0}^\infty \frac{(-1)^n}{z^n}$

Then subtract the two sums. Is it the same answer for |z|>2?

Answer 1

HINTS:

$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{b+c}\right)\,\left(\frac{1}{1+\left(\frac{z-c}{b+c}\right)}\right)$$

and

$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{z-c}\right)\,\left(\frac{1}{1+\left(\frac{b+c}{z-c}\right)}\right)$$