So the Mobius Transformation $S(z) = \dfrac{az+b}{cz+d}$ is analytic on its domain $\left\{z \in \mathbb{C} ~|~ z \neq -\dfrac{d}{c}\right\}$
And $$S^{'}(z) = \dfrac{ad-bc}{(cz+d)^2} \neq 0$$ because of the condition $ad-bc \neq 0$, such condition will imply $S^{'}(z)$ is never zero.
Hence $S(z)$ is a conformal map on the $\mathbb{C} \setminus -\left\{\dfrac{d}{c}\right\}$
However, I think the Professor mentioned something that Mobius Transformation is conformal on the whole of $\mathbb{C}$. I do not quite understand where I went wrong with the Theorem. I follow closely on this theorem:
If $D$ is an open subset of the complex plane $\mathbb{C}$, then a function
$f: D \rightarrow \mathbb{C}$ is a conformal mapping if and only if it is
holomorphic (analytic) on its domain and its derivative $f^{'}(z)$ is
everywhere non-zero on $D$.
Best Answer
The statement is false. Consider the exponential function on a horizontal strip like $D=\{|\Im z|<\pi+\epsilon\}$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$. The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.