I'm working through a question where I'm asked to find the series expansions of various functions around certain points. I've found the answer for most of the examples, but the last part of the question asks me to find the series expansion of $f(z) = \frac{1}{z-b}$ around some point $a$, and I can't work out how it's done.
I'd really appreciate any help you could offer me.
Just us the fact that$$\frac1{z-b}=\frac1{(z-a)-(b-a)}=-\frac1{b-a}\times\frac1{1-\frac{z-a}{b-a}}$$and expand $\dfrac1{1-\frac{z-a}{b-a}}$ using the geometric series.