Proof of Mobius transformation

Win_odd Dhamnekar

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What is the proof that the trajectory of the function z(t) is a circle passing through the origin?
 
u(t)=αt+βu(t) = \alpha t + \beta is a straight line in C\mathbf C. z(t)=1u(t)z(t) = \frac{1}{u(t)} is an inversion of u(t)u(t) combined with a reflection relative to the horizontal axis (a.k.a. complex conjugation). It is known that an inversion of a straight line not passing through 0 is a circle passing through 0 (https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle).
To prove more formally can do this variable transformation: 1αt+β=1α1t+γ\frac{1}{\alpha t+\beta} = \frac{1}{\alpha} \frac{1}{t + \gamma}, where γ=βα\gamma = \frac{\beta}{\alpha}. Since multiplication by 1α\frac{1}{\alpha} is just scaling and rotation it is enough to prove that z(t)=1t+γz(t)=\frac{1}{t + \gamma} represents a circle. The center of such circle is easy to guess: c=12bic = \frac{1}{2bi}, so what remains to show is that 1t+bi12bi=12b\left|\frac{1}{t+bi} - \frac{1}{2bi}\right| = \frac{1}{2b}
 
u(t)=αt+βu(t) = \alpha t + \beta is a straight line in C\mathbf C. z(t)=1u(t)z(t) = \frac{1}{u(t)} is an inversion of u(t)u(t) combined with a reflection relative to the horizontal axis (a.k.a. complex conjugation). It is known that an inversion of a straight line not passing through 0 is a circle passing through 0 (https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle).
To prove more formally can do this variable transformation: 1αt+β=1α1t+γ\frac{1}{\alpha t+\beta} = \frac{1}{\alpha} \frac{1}{t + \gamma}, where γ=βα\gamma = \frac{\beta}{\alpha}. Since multiplication by 1α\frac{1}{\alpha} is just scaling and rotation it is enough to prove that z(t)=1t+γz(t)=\frac{1}{t + \gamma} represents a circle. The center of such circle is easy to guess: c=12bic = \frac{1}{2bi}, so what remains to show is that 1t+bi12bi=12b\left|\frac{1}{t+bi} - \frac{1}{2bi}\right| = \frac{1}{2b}
Just realized that one step is missing: after z(t)=1t+γz(t) = \frac{1}{t+\gamma} represent γ\gamma as a+bia+bi. Since tt ranges from -\infty to ++\infty replacing tt with t+at+a does not change the result. Thus 1t+γ1t+bi\frac{1}{t+\gamma}\Rightarrow \frac{1}{t+bi}.
 
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