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
$$|\frac{z-a}{1-\bar{a}z}| \leq 1$$
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

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

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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
$$|\bar{a}|^2 = \bar{a}\bar{\bar{a}} = \bar{a}a = |a|^2$$

Now simplify the expression and factorize.

 

[FanofAFan]

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