Show that $|z-c|\leq|1-\bar{c}z|$ for $c,z\in\mathbb{C}$, $|c|\leq1, |z|\leq1$

Question

"Show that $|z-c|\leq|1-\bar{c}z|$ for $c,z\in\mathbb{C}$, $|c|\leq1, |z|\leq1$"

I tried to solve this by just defining $z$ and $c$ like this:

$c:=a+bi$
$z:=x+yi$

Because of $|c|\leq1, |z|\leq1$, we know this:

$\sqrt{a^2+b^2}\leq1 \implies a^2+b^2\leq1$
$\sqrt{x^2+y^2}\leq1 \implies x^2+y^2\leq1$

Now my idea was to just plug in the definitions of $c$ and $z$ and do lots of rearranging and use many definitions for complex numbers, which takes a while and is very messy. Sadly I got stuck at some point, namely here:

$\sqrt{2-2(ax+by)}$

But what now? I really don't know what to do next. Also, I think that this can be done so much quicker and way more elegant than just plugging in the numbers and somehow trying to get to the end. Does somebody have some advice for me? Thanks in advance!

Answer 1

This is true $$|z-c|\leq|1-\bar{c}z|$$

iff this is true $$|z-c|^2\leq|1-\bar{c}z|^2$$

and that is true if $$(z-c)(\bar{z}-\bar{c})\leq(1-\bar{c}z)(1-\bar{z}c) $$

is true, which is the same as:

$$z\bar{z}+c\bar{c}\leq 1+c\bar{c}\cdot z\bar{z} $$

and this is the same as $$(|z|^2-1)(|c|^2-1)\geq 0$$