Show that if $|a|=1$ or $|b|=1$, then $\left|\dfrac{a-b}{1-\overline{a}b}\right|\leq 1$

Question

Let $a,b\in \mathbb{C}$, $|a|,|b|\leq 1$, $a\neq b$. Show that if $|a|=1$ or $|b|=1$, then $$\left|\dfrac{a-b}{1-\overline{a}b}\right|\leq 1$$

Proof: Let $a,b\in \mathbb{C}$, $|a|,|b|\leq 1$, $a\neq b$. Note that $|a|=|\overline{a}|$.

If $|a|=1$, then $|\overline{a}|=1$. This implies $$\left|\dfrac{a-b}{1-\overline{a}b}\right|=\dfrac{|a-b|}{|1-\overline{a}b|}\leq \dfrac{|a|+|b|}{|1|+|\overline{a}|\cdot |b|} = \dfrac{1+|b|}{1+|b|} = 1.$$

If $|b|=1$, then $$\left|\dfrac{a-b}{1-\overline{a}b}\right|=\dfrac{|a-b|}{|1-\overline{a}b|}\leq \dfrac{|a|+|b|}{|1|+|\overline{a}|\cdot |b|} = \dfrac{|a|+1}{1+|a|} = 1.$$

I was wondering if what I did is correct? I am not confident with my answer because I treated the inequalities of the numerator and denominator like normal and then just put one over the other.

Answer 1

If you want the fraction the increase, then you must decrease your denominator to achieve that. Thus, your solution is not correct but there are many ways to solve it. For instance, make observations like $a(1-\bar{a}b) = a - |a|^2b = a-b$ if $|a| = 1$

Answer 2

I am not confident with my answer because I treated the inequalities of the numerator and denominator like normal and then just put one over the other

You are right not to be confident since, indeed, you can't divide inequalities. In other words, if $a\le b, c\le d$ it does not follow that $a/c \le b/d\,$. Try for example $a=b=c=1$ and $d=2\,$.

Instead, you can show that if $|a|=1 \iff |a|^2 = 1 \iff a \bar a = 1 \iff \bar a = \dfrac{1}{a}$, then:

$$\require{cancel} \left|\dfrac{a-b}{1-\overline{a}b}\right|=\left|\dfrac{a-b}{1-\dfrac{1}{a}\,b}\right|= \left|\dfrac{\;\;\cancel{a-b}\;\;}{\dfrac{\cancel{a-b}}{a}}\right| = |a| = 1 $$

Since the problem is symmetric in $a,b$ the same applies if $|b|=1$.