Let \(a,b,c,d\in\CC\) with \(ad-bc\neq0\). The map \[ T(z)\coloneqq\frac{az+b}{cz+d}, \] defined for \(z\in\CC\) with \(cz+d\neq0\), is a linear fractional transformation, or Möbius transformation.
Let \(T(z)=(az+b)/(cz+d)\) with \(ad-bc\neq0\).
\(T\) is a bijection from \(\{z\in\CC:cz+d\neq0\}\) onto \(\{w\in\CC:-cw+a\neq0\}\), with inverse \[ T^{-1}(w)=\frac{dw-b}{-cw+a}. \]
\(T\) is holomorphic on \(\{z\in\CC:cz+d\neq0\}\), with \[ T'(z)=\frac{ad-bc}{(cz+d)^2}\neq0, \] so \(T\) is conformal there.
For \(cz+d\neq0\), \(-c\,T(z)+a=(ad-bc)/(cz+d)\neq0\), and \[ \frac{d\,T(z)-b}{-c\,T(z)+a}=\frac{d(az+b)-b(cz+d)}{-c(az+b)+a(cz+d)}=\frac{(ad-bc)z}{ad-bc}=z. \] The map \(w\mapsto(dw-b)/(-cw+a)\) has coefficients with \(da-bc\neq0\), and the same computation shows that \(T\) inverts it on \(\{w:-cw+a\neq0\}\).
By the quotient rule, \[ T'(z)=\frac{a(cz+d)-c(az+b)}{(cz+d)^2}=\frac{ad-bc}{(cz+d)^2}, \] which is nonzero because \(ad-bc\neq0\).