Calculating $Res_{z=w} {f'(z)\over f(z)} $?

Question

I have $$f(z) = \sum_{n=m}^\infty a_n (z-w)^n $$

where $0 < | z-w | < r$ and $a_m \neq 0$ and am asked to calculate

$$Res_{z=w} {f'(z)\over f(z)} $$

I have differentiated $f(z)$ to get $\sum_{n=m}^\infty na_n(z-w)^{n-1}$ and tried dividing, to get $\sum_{n=m}^\infty {n\over (z-w)}$ but this seems wrong. Can anyone help?

Answer 1

We have

$$f(z) = \sum_{n=m}^\infty a_n(z-w)^m = (z-w)^m \underbrace{\sum_{n=m}^\infty a_n (z-w)^{n-m}}_{g(z)}.$$

The assumption $a_m \neq 0$ guarantees that $g$ is holomorphic in $w$ with $g(w) = a_m \neq 0$. Then we have $f'(z) = m(z-w)^{m-1}g(z) + (z-w)^m g'(z)$ by the product rule, and hence

$$\frac{f'(z)}{f(z)} = \frac{m(z-w)^{m-1}g(z) + (z-w)^m g'(z)}{(z-w)^m g(z)} = \frac{m}{z-w} + \frac{g'(z)}{g(z)}.$$

The last term is holomorphic in $w$.