suppose $|a|<1$, show that $\frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself. Suppose $|a|<1$, show that $f(x) = \frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself.
To make such a mobius transformation i tried to send 3 points on the edge to 3 points of the edge. so filling $i,1,-1$ in $f(x)$ we should get on the edges of the unit ball. But i don't seem to know how to calculate these exactly:
$$f(1) = \frac{1-a}{1-\overline{a}1}$$
$$f(-1) = \frac{-1-a}{1+\overline{a}}$$
$$f(1) = \frac{i-a}{1-\overline{a}i}$$
I don't seem to get how i could write these formula's in such a way that i get into the edges of the circle.
Anyone can help me?
Kees

hint: prove that : $\left|\dfrac{z-a}{1-\bar{a}z}\right| < 1$ by using $z = x+iy, a = m+in$, and use the well-known CS inequality: $(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2$.
