Differentiate $T(z)=(az+b)/(cz+d)$. 
Find the derivative of the function $T(z)=\frac{az+b}{cz+d}$, where
  $a, b, c, d\in\mathbb{C}$ with $ad - bc \neq 0$. When is $T'(z) = 0$?

Just want to check that this problem is analogous to what we do in real analysis. The quotient rule gives
$$T'(z)=\frac{a(cz+d)-c(az+b)}{(cz+d)^2}=\frac{ad-bc}{(cz+d)^2},$$
Thus the derivative of $T$ is zero iff $ad-bc=0.$ However this is excluded in the problem. So what do they want me to realise here?

If $ad-bc=0$ then $(a,b)$ is a multiple $(c,d)$, hence $T$ is constant. If $ad-bc\neq 0$ then $T'(z)$ is never zero as you proved. It folows that $T$ is a local biholomorphism by the (holomorphic) inverse function theorem. In fact, it is a global biholomorphism since the injectivity can be easily checked. 
