# Calculus of residues

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?

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 $0$.
• is it not $\frac{(b-a)z}{1-bz}+o(z)$? – vidyarthi Sep 20 '17 at 8:18
• so is the residue $0$? – vidyarthi Sep 20 '17 at 8:22