Find the Laurent series for $\frac1{(z-a)(z-b)}$ for $0<|z-a|<|a+b|$ where $0<|a|<|b|$
$$\frac1{(z-a)(z-b)}=\frac1{b-a}\left(\frac1{z-b}-\frac1{z-a}\right)$$
Here $\frac1{z-b}=\frac1{z-a+a-b}=\frac1{z-a-(b-a)}$
For binomial expansion of this term we write it as $-\frac1{b-a}\left(1-\frac{z-a}{b-a}\right)$
In this case $\left|\frac{z-a}{b-a}\right|<1 \implies |z-a|<|b-a|$ since $|b-a|<|a+b|$ it may not hold true $\forall|z-a|<|a+b|$
and if we expand it in $\frac1z$ form then for that $\frac1{z-a-(b-a)}=\frac1{z-a}\left(1-\frac{b-a}{z-a}\right)$
In this case
$$\left|\frac{b-a}{z-a}\right|<1 \implies |z-a|>|b-a|$$ but $|z-a|>0$ and it may or may not be greater than $|b-a|$
How it can be expanded in Laurent series