Complex Analysis: Show |z| \leq 1 iff \frac{z-a}{1-a(bar)z} \leq 1

[FanofAFan]

Homework Statement

|a| < 1 a is arbitrary, then show that |z| \leq 1 iff \frac{z-a}{1-a(bar)z} \leq 1

Homework Equations

possible the triangle inequality

The Attempt at a Solution

\frac{z-a}{1-a(bar)z} is analytic everywhere except at 1/a(bar)
|z - a|² \leq |1-a(bar)z|²

|z|²-2|z||a| + |a|² =|1| -2|z||a(bar)| +|a|²|z|²

 

[praharmitra]

Surely you mean
<br /> |\frac{z-a}{1-\bar{a}z}| \leq 1<br />
since it makes no sense to use inequalities with complex numbers.

Now your first point (the function is analytic everywhere except at z = 1/\bar{a}. However, this point does not lie in our region of interest, i.e. |z|\leq1,\ |a|\leq1.

Now can you show that

|1-\bar{a}z|^2-|z-a|^2 \geq0

______________________________

EDIT: I read the question wrong. I thought we had to prove the last inequality. In any case it doesn't change the mathematics too much. Can you factorize the LHS of |1-\bar{a}z|^2-|z-a|^2 \geq0. Then use |a|\leq1 and you should be done!

 

[FanofAFan]

So \bar{a} just -a and would the factored out left side be...

|1| -2|\bar{a}||z| +|\bar{a}|²|z|² - |z|² -2|z||a| +|a|²

 

[praharmitra]

So \bar{a} just -a and would the factored out left side be...

|1| -2|\bar{a}||z| +|\bar{a}|²|z|² - |z|² -2|z||a| +|a|²

You forgot the bracket.
The LHS is
|1| - 2|\bar{a}||z| +|\bar{a}|^2|z|^2 - (|z|^2 - 2|z||a| + |a|^2)
Further note that
<br /> |\bar{a}|^2 = \bar{a}\bar{\bar{a}} = \bar{a}a = |a|^2<br />

Now simplify the expression and factorize.

 

[FanofAFan]

Ok thanks! also any ideas on this... https://www.physicsforums.com/showthread.php?t=487828