Show that a Möbius transformation maps a straight line or circle onto a straight line or circle.
Solution
A generalized circle in \(\widehat{\mathbb C}\) means either an ordinary circle or a straight line together with \(\infty\). Such a set has an equation \[ A|z|^2+Bz+\overline B\,\overline z+C=0, \] where \(A,C\in\mathbb R\), \(B\in\mathbb C\), and \(|B|^2-AC>0\). If \(A=0\), this is a line; if \(A\ne0\), completing the square gives a circle.
Every Möbius map is a composition of translations, nonzero complex dilations, and inversion \(z\mapsto1/z\). Indeed, for \[ T(z)=\frac{az+b}{cz+d},\qquad ad-bc\ne0, \] there is nothing to prove when \(c=0\), while for \(c\ne0\), \[ T(z)=\frac ac+\frac{bc-ad}{c^2}\frac1{z+d/c}. \] Translations and nonzero complex dilations plainly preserve generalized circles. For inversion, substitute \(z=1/w\) into the equation and multiply by \(|w|^2\): \[ A+B\overline w+\overline B w+C|w|^2=0. \] This is again of the same form, now with coefficients \(A'=C\), \(B'=\overline B\), \(C'=A\), and \[ |B'|^2-A'C'=|B|^2-AC>0. \] Thus inversion also preserves generalized circles. Therefore every Möbius transformation maps every line or circle onto another line or circle.