Find the Laurent series for the given function about the indicated point, and the residue of the function at the point:
$$\frac{az+b}{cz+d}\quad \text{at}\quad z_0 = -\frac{d}{c}.$$
I'm not sure how to approach this problem. Any clue would be appreciated.
Since the point is clearly a simple pole:
$$\lim_{z\to -\frac{d}{c}}\left(z+\frac{d}{c}\right)\frac{az+b}{cz+d}\stackrel{\text{l'Hospital}}=\frac{a\left(-\frac{d}{c}\right)+b}{c}=\frac{bc-ad}{c^2}=-\frac{\Delta}{c^2}$$
with $\,\Delta=$ the given Moebius transformation's determinant.
About the Laurent series:
$$\frac{az+b}{cz+d}=\frac{a\left(z+\frac{d}{c}\right)+b-\frac{ad}{c}}{c\left(z+\frac{d}{c}\right)}=\frac{1}{c^2}\frac{ac\left(z+\frac{d}{c}\right)+bc-ad}{z+\frac{d}{c}}=$$
$$\frac{1}{c^2}\left(\frac{bc-ad}{z+\frac{d}{c}}+ac\right)$$
and if there was a doubt about the residue we can see it again in the above development.