How to rigorously deduce the Laurent series of $\log\frac{z-p}{z-q}$?

Of course, the logarithm here is defined on the ring region $|z|>R\ge\max\{|p|,|q|\}$ as $$\log\frac{z-p}{z-q}=\int_{z_0}^z \left(\frac1{w-p}-\frac1{w-q}\right)\mathrm d w.$$ Here the integral is along an arbitrary curve connecting $z_0$, a fixed point, to $z$ in the ring region. It's noteworthy that this logarithm is actually well defined, although not appearing so at first glance.

All I know is $\log(1+z)=z-\frac12z^2+\frac13z^3+\cdots$ when $|z|<1$ and $\log$ is chosen to be the principal branch. With this idea I can work out a naive argument: for our fractional logarithm, first rewrite the expression as $\log\frac{1-p/z}{1-q/z}$ by dividing both the numerator and denominator simultaneously (I don't know how to justify this from the integral definition, though); then it all reduces to $\log(1-p/z)-\log(1-q/z)$, with the two logarithm both chosen as suitable branches, to which the canonical power expansion applies.

I believe I'm almost on the right track, but can't get over that confusion. Could you help me? Thanks!

For $\lvert w \rvert >\max\{\lvert p \rvert,\lvert q \rvert\}$, $$\frac{1}{w-p} - \frac{1}{w-q} = \frac{1}{w(1-p/w)} - \frac{1}{w(1-q/w)} = \sum_{n=1}^\infty \frac{p^n - q^n}{w^{n+1}}$$ The terms for $n=0$ cancel, and all remaining terms have an antiderivate in $\Bbb C$. Therefore $$\int_{z_0}^z \left(\frac1{w-p}-\frac1{w-q}\right)\, dw = \sum_{n=1}^\infty (p^n - q^n) \left( -\frac{1}{nz^n} + \frac{1}{nz_0^n} \right) = C + \sum_{n=1}^\infty \frac{q^n - p^n}{n z^n}$$ for some constant $C$ depending on $z_0$.
• Thank you! It's clear now. Actually this I have tried to but I didn't realise that the $n=0$ terms will cancel and was having quite a hard time finding their anti derivatives which of course don't exist in the first place :) – Vim Jun 29 '16 at 0:01