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 $0$.
Robert Z
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is it not $\frac{(b-a)z}{1-bz}+o(z)$? – vidyarthi Sep 20 '17 at 08:18
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@vidyarthi At the first order they are the same! – Robert Z Sep 20 '17 at 08:20
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so is the residue $0$? – vidyarthi Sep 20 '17 at 08:22
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@vidyarthi Yes, the residue at 0 is 0. BTW if you are new here take a few seconds for a tour https://math.stackexchange.com/tour – Robert Z Sep 23 '17 at 14:26