\begin{align*}
f(z)&=\frac{\frac{a}{c}(cz+d)+b-\frac{ad}{c}}{cz+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cz+d}\\
&=\frac{a}{c}+(b-\frac{ad}{c})\cdot\frac{1}{cz+d}=\frac{a}{c}+(\frac{bc-ad}{c})\frac{1}{c}\cdot\frac{1}{z+\frac{d}{c}}\\
&=\frac{a}{c}+\frac{bc-ad}{c^2}(z+\frac{d}{c})^{-1}
\end{align*}
residue at $z_0=-\frac{d}{c}$ is $\frac{bc-ad}{c^2}$ which is coefficient of $(z-z_0)^{-1}$.
