Mapping behavior of imaginary axis via $v=\frac{z-a}{z+a}$

Question

I would like to know what the bilinear transform $v=\frac{z-a}{z+a}$ does to the imaginary axis, where $a$ is a real number.

I substituted $z=yi$ and calculated $|v|$ giving me $|v| =1$. Is this enough proof to say that $v$ maps the imaginary axis to on the unit circle?

Answer 1

It's enough to show that it maps the imaginary axis to the unit circle, but not quite enough to show that it maps the imaginary axis onto the unit circle. To draw that conclusion, we need to know that linear fractional transforms map circles to circles (in the Riemann sphere).

Alternately, you must show explicitly that it maps onto the unit circle (with one point deleted, if you don't count the "point at $\infty$" as being on the unit circle).