suppose $|a|<1$, show that $\frac{z-a}{1-\overline{a}z}$ is a mobius transformation that sends $B(0,1)$ to itself.
If $|z|=1$, then $\overline z=1/z$. So,
$$\left|{z-a\over 1-\overline az}\right|=\left|{z-a\over1-\overline{a/z}}\right|=\left|{\overline z(z-a)\over\overline z-\overline a}\right|=|\overline z|\left|{z-a\over\overline{z-a}}\right|=|\overline z|=1.$$
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: $(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2$.