Inequalty with complex numbers

Question

i'm being fighting with this for a long, it shouldn't be that hard, can you provide some ideas?

Let a,b on $\mathbb{C}$ if |a|<1 and |b|<1 prove:

$$\left|\frac{a-b}{1-\bar{a}b}\right| < 1$$

Thanks!

Answer 1

The inequality is equivalent to

$$|a-b|<|1-\bar{a}b|$$

thus to

$$(a-b)(\bar{a}-\bar{b})<(1-\bar{a}b)(1-a\bar{b})$$

which you can expand to

$$a\bar{a}-a\bar{b}-\bar{a}b+b\bar{b}<1-\bar{a}b-a\bar{b}+a\bar{a}b\bar{b}$$

so it reduces to

$$|a|^2 + |b|^2 < 1 + |a|^2\,|b|^2$$

that is

$$(1-|a|^2)(1-|b|^2)>0$$

which is true by hypothesis.

Notice that $\bar{a}b\ne1$, because that would imply

$$|a||b|=1$$

which can't happen under your hypotheses.