Proving $T'(p)T'(q)=1$ for a mobius transformation with fixed points p,q

Question

Let T be a mobius transformation with two fixed points p and q. I want to prove that $T'(p)T'(q)=1$.

I tried using S such that $STS^{-1}(z)=kz$ and deriving but got nowhere. I thougt maybe I had to assume that $ad-bc= 1$ but I don't know what to do

Answer 1

You are solving $$z=T(z)=\frac{az+b}{cz+d}.$$ You get a quadratic $$cz^2+(d-a)z-b=0.$$ So $p+q=(a-d)/c$ and $pq=-b/c$. One has $$T'(z)=\frac{ad-bc}{(cz+d)^2}.$$ Therefore $$T'(p)T'(q)=\frac{(ad-bc)^2}{[(cp+d)(cq+d)]^2}.$$ But $$(cp+d)(cq+d)=c^2pq+cd(p+q)+d^2=-bc+d(a-d)+d^2=ad-bc.$$