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How can we find the residue at $ z=0 $ of $$f(z) = \log\left(\frac{1-az}{1-bz}\right)$$ where $a, b$ are complex constants?

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Note that as $z\to 0$, $$f(z) = \log\left(\frac{1-az}{1-bz}\right)=\log\left(1+\frac{(b-a)z}{1-bz}\right)=(b-a)z+ o(z)$$ What may we conclude? Recall that the residue is the coefficient of $z^{-1}$ in the expansion of $f$ at $0$.

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  • $\begingroup$ is it not $\frac{(b-a)z}{1-bz}+o(z)$? $\endgroup$ – vidyarthi Sep 20 '17 at 8:18
  • $\begingroup$ @vidyarthi At the first order they are the same! $\endgroup$ – Robert Z Sep 20 '17 at 8:20
  • $\begingroup$ so is the residue $0$? $\endgroup$ – vidyarthi Sep 20 '17 at 8:22
  • $\begingroup$ @vidyarthi Yes, the residue at 0 is 0. BTW if you are new here take a few seconds for a tour math.stackexchange.com/tour $\endgroup$ – Robert Z Sep 23 '17 at 14:26

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