Question

Let $a$ be a fixed non-zero complex number with $|a|<1$ and
$w=\left(\frac{z-a}{1-\overline{a}z}\right)$.
where $z$ is a complex number. Then

• A. there exists a complex number $z$ with $|z|<1$ such that $|w|>1$
• B. $|w|>1$ for all z such that $|z|<1$
• C. $|w|<1$ for all z such that $|z|<1$
• D. there exists z such with $|z|<1$ and $|w|=1$

Answer 1

The correct option is C $|w|<1$ for all z such that $|z|<1$

$w=\left(\frac{z-a}{1-\overline{a}z}\right)\Rightarrow w-w\overline{a}z=z-a\Rightarrow z=\frac{w+a}{1+\overline{a}w}$

If $|z|<1$,
$\left|\frac{w+a}{1+\overline{a}w}\right|<1\Rightarrow |w+a{|}^2<|1+\overline{a}w{|}^2\Rightarrow (w+a)(\overline{w}+\overline{a})<(1+\overline{a}w)(1+a\overline{w})\Rightarrow |w{|}^2+|a{|}^2+\overline{a}w+a\overline{w}<1+\overline{w}a+w\overline{a}+|w{|}^2|a{|}^2\Rightarrow |w{|}^2+|a{|}^2-1-|w{|}^2|a{|}^2<0\Rightarrow |w{|}^2(1-|a{|}^2)-1(1-|a{|}^2)<0\Rightarrow (|w{|}^2-1)(1-|a{|}^2)<0$

We know that $|a|<1$
So $|w{|}^2<1\Rightarrow |w|<1$