properties of complex modulus question If $|a| < 1$, prove that $|z| < 1$ is equivalent to
$$\frac{|z - a|}{|1-\bar{a}z|} \leq 1.$$
Where $a$ and $z$ are complex and $\bar{a}$ denotes the conjugate of $a$. 
I thought this was the easiest problem I would have to do tonight, but I have been bumbling around with the properties of the modulus ($|z + a| \leq |z| + |a|$, etc.) all night and I can't get the desired result. Is this a bad question?
 A: Some hints.


*

*Define $\phi_a$ by $\phi_a(z)=\frac{a-z}{1-\overline{a} z}$.
Prove that $\phi_a(\phi_a(z))=z$ for every $z$ with $|z|<1$.

*Thanks to part 1, it is sufficient to prove only one part:
if $|z|<1$, then $|\varphi_a(z)|<1$.

*Recall a formula for $w+\overline{w}$.
Simplify $z\overline{a} + a\overline{z}$.

*Recall that $|w|^2=w\overline{w}$. Calculate $|z-a|^2$ and $|1-\overline{a} z|^2$.

*Write the expression $1+|a|^2|z|^2-|z|^2-|a|^2$ as a product of two factors.
Assuming $|a|<1$ and $|z|<1$ deduce that $|z|^2+|a|^2\le 1+|a|^2|z|^2$.
A: Hint.  First rewrite the inequality as
$$|z-a|\le|1-\overline a z|\ ,$$
then use the relation
$$|w|^2=w\overline w\ .$$
This is often a good way to go because conjugates have "nicer" algebraic properties than magnitudes.  Squaring the left hand side,
$$|z-a|^2=(z-a)\overline{(z-a)}=(z-a)(\overline z-\overline a)=z\overline z-a\overline z-z\overline a+a\overline a\ .$$
Do something similar for the right hand side and see if you can take it from there.
Also don't forget that the question said "equivalent".
