Let $$f(z) = \frac{1}{(z+1)(z-1)(z-3)} = \frac{1}{8(z+1)}-\frac{1}{4(z-1)}+\frac{1}{8(z-3)}$$
If we want to find all possible Laurent series at center $z=0$, there will be three domains.
- $|z|<1$
- $1<|z|<3$
- $|z|>3$
The first and last domain can be found by expressing the function in analytic and principal form respectively and using the geometric series expansion. But how do we express the function under the second domain in Laurent series? By factoring out 3 in the third term of $f(z)$ it resembles the third domain $|\frac{z}{3}|<1$ and we can express it in the analytic form. Does the first and second term get expressed in principal part and therefore we should factor out the $z$ instead?