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!


Yes, you can compute the Laurent series directly from the integral definition.

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 $z_0$.

  • $\begingroup$ 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 :) $\endgroup$ – Vim Jun 29 '16 at 0:01

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