Proof of Mobius transformation

[Win_odd Dhamnekar]

What is the proof that the trajectory of the function z(t) is a circle passing through the origin?

 

[blamocur]

[imath]u(t) = \alpha t + \beta[/imath] is a straight line in [imath]\mathbf C[/imath]. [imath]z(t) = \frac{1}{u(t)}[/imath] is an inversion of [imath]u(t)[/imath] 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: [imath]\frac{1}{\alpha t+\beta} = \frac{1}{\alpha} \frac{1}{t + \gamma}[/imath], where [imath]\gamma = \frac{\beta}{\alpha}[/imath]. Since multiplication by [imath]\frac{1}{\alpha}[/imath] is just scaling and rotation it is enough to prove that [imath]z(t)=\frac{1}{t + \gamma}[/imath] represents a circle. The center of such circle is easy to guess: [imath]c = \frac{1}{2bi}[/imath], so what remains to show is that [imath]\left|\frac{1}{t+bi} - \frac{1}{2bi}\right| = \frac{1}{2b}[/imath]

 

[blamocur]

[imath]u(t) = \alpha t + \beta[/imath] is a straight line in [imath]\mathbf C[/imath]. [imath]z(t) = \frac{1}{u(t)}[/imath] is an inversion of [imath]u(t)[/imath] 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: [imath]\frac{1}{\alpha t+\beta} = \frac{1}{\alpha} \frac{1}{t + \gamma}[/imath], where [imath]\gamma = \frac{\beta}{\alpha}[/imath]. Since multiplication by [imath]\frac{1}{\alpha}[/imath] is just scaling and rotation it is enough to prove that [imath]z(t)=\frac{1}{t + \gamma}[/imath] represents a circle. The center of such circle is easy to guess: [imath]c = \frac{1}{2bi}[/imath], so what remains to show is that [imath]\left|\frac{1}{t+bi} - \frac{1}{2bi}\right| = \frac{1}{2b}[/imath]

Just realized that one step is missing: after [imath]z(t) = \frac{1}{t+\gamma}[/imath] represent [imath]\gamma[/imath] as [imath]a+bi[/imath]. Since [imath]t[/imath] ranges from [imath]-\infty[/imath] to [imath]+\infty[/imath] replacing [imath]t[/imath] with [imath]t+a[/imath] does not change the result. Thus [imath]\frac{1}{t+\gamma}\Rightarrow \frac{1}{t+bi}[/imath].