Question

Prove if \(|a| \), \(|b|\) \(< 1\), then \(|\frac{a-b}{1-b \overline{a}}|\)\(<1 \)

\(\overline{a} \) is the conjugate of \(a\)

Answer 1

\[
\begin{align*}
&\phantom{{}={}}\left|\frac{a-b}{1-b\bar{a}}\right|^2\\
&=\frac{a-b}{1-b\bar{a}}\frac{\bar{a}-\bar{b}}{1-\bar{b}a}\\
&=\frac{|a|^2+|b|^2-a\bar{b}-\bar{a}b}{1+|ab|^2-a\bar{b}-\bar{a}b}\\
\end{align*}
\]

Answer 2

Is this even correct? That looks wrong.

Answer 3

If I did the algebra right and the thing we are required to prove is true, then \(|a|^2+|b|^2<1+|ab|^2\) which seems unlikely to me.

Answer 4

\[
\begin{align*}
|a|^2+|b|^2&<1+|ab|^2\\
|a|^2-1&<|ab|^2-|b|^2\\
|a|^2-1&<|b|^2(|a|^2-1)\\
1&>|b|^2&|a|^2-1<1
\end{align*}
\]

Answer 5

Well, it comes from a conformal mapping. It's a function that will map the unit circle to itself, but move a point a which is inside the unit circle to the origin. So it should definitely come out somehow. I just don't know how to manipulate my inequalities or get the absolute value whatevers right to show it.

Answer 6

It does not move a point inside the unit circle to the origin. It does move it closer to the origin though.

Answer 7

Well, its definitely similar to a mapping that would do that. We've had to use it before for that purpose, move some sort of source point to the origin. Either way, though, are you thinking its incorrect?