Inverse of a Mobius transformation
by Hidenori
Proposition
Show that if $f(z) = \frac{az + b}{cz + d}$ is a Mobius transformation then $f^{-1}(z) = \frac{dz - b}{-cz + a}$.
Solution
\[\begin{align*} f^{-1}(f(z)) &= \frac{d(az + b)/(cz + d) - b}{-c(az + b)/(cz + d) + a} \\ &= \frac{d(az + b) - b(cz + d)}{-c(az + b) + a(cz + d)} \\ &= \frac{adz - bcz}{ad - bc} \\ &= z. \end{align*}\] \[\begin{align*} f^{-1}(f(z)) &= \frac{d(az + b)/(cz + d) - b}{-c(az + b)/(cz + d) + a} \\ &= \frac{d(az + b) - b(cz + d)}{-c(az + b) + a(cz + d)} \\ &= \frac{adz - bcz}{-bc + ad} \\ &= z. \end{align*}\]Subscribe via RSS