How to show that $\frac{|u - \bar{v} z|}{|\bar{u} z - v|} \leq 1 \Leftrightarrow |z|\leq 1$

Question

I can't solve this math problem and I'm stuck in it! I think it must have a trick and I don't know that. the problem is this:

if u and v and z are complex numbers and $|u|<1$ and $|v|=1$ show that: $$ \frac{|u - \bar{v} z|}{|\bar{u} z - v|} \leq 1 \Leftrightarrow |z|\leq 1 $$ any help or tip to solve it would be appreciated.

Answer 1

HINT

We have that

$$\frac{|u - \bar{v} z|}{|\bar{u} z - v|} \leq 1 \iff \frac{|u - \bar{v} z|^2}{|\bar{u} z - v|^2} \leq 1 \iff |u - \bar{v} z|^2\le |\bar{u} z - v|^2$$

$$(u - \bar{v} z)(\bar u - v \bar z)\le(\bar{u} z - v)(u\bar{z} - \bar v)$$

$$|u|^2-uv\bar z-\bar u \bar v z+|v|^2|z|^2\le|u|^2|z|^2-\bar u \bar v z-uv\bar z+|v|^2$$