I was recently attempting the question:
Express the Mobius tranformation $f(z) = (2z + 3)/(z − 4)$ as the composition of maps of the form $z → az, z → z + b$ and $z → 1/z$. Hence show that f maps the circle $|z − 2i| = 2$ onto the circle $|8z + (6 + 11i)| = 11$.
So for the first part, let $a(z)=z-4 \ , b(z)= 1/z , \ c(z) = 11z, \ d(z) = z+2 $
Then $dcba(z)=dcb(z-4)=dc(1/(z-4)) = d(11/(z-4)) = 2+ 11/(z-4)= \frac{2z+3}{z-4}=f(z) $
But I'm not sure how this relates to the second part. I thought maybe looking at the transformation of the circle with each of these simpler mappings separately but I'm not sure it works: after applying $a(z) $ to the circle, we translate it $-4$ in direction of +ve real axis so the image is $ |z-4-2i|=2 $. But then what about after applying $b(z) $ ? I assume I have missed something obvious. Any hints/help is appreciated.