Show that $\left|\frac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$ when $|\alpha|,|\beta| < 1$

This is the question I'm stumbling with:

$$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$$

The chapter that contains this question contains (among others) the triangle inequalities:

$$\left||z_1| - |z_2|\right| \le |z_1 + z_2| \le |z_1| + |z_2| $$

I've tried to use the triangle inequalities to increase the dividend and/or decrease the divisor:

$$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < \cfrac{|\alpha| +|\beta|}{\left|1-|\bar{\alpha}\beta|\right|}$$

But it's not clear if or why that would be smaller than one. I've also tried to multiply the equation by the conjugated divisor $\cfrac{1-\alpha\bar{\beta}}{1-\alpha\bar{\beta}}$, which gives a real divisor, but the equation does not appear solvable.

Any hint would be much appreciated.

Solution 1:

Hint: $$ \left| \frac{\alpha-\beta}{1-\bar\alpha\beta}\right| < 1 \Leftrightarrow |\alpha-\beta|^2 < |1-\bar\alpha\beta|^2. $$

Expand both sides, remembering that $|z|^2 = z\bar z$ and simplify. That should get you where you want to be after some algebra.

Solution 2:

The hint of mrf gives a nicest proof, of course, but we can use here also the following algebra:

Let $\alpha=x+yi$ and $\beta=a+bi,$ where $\{x,y,a,b\}\subset\mathbb R.$

Thus, we need to prove that $$|x-a+(y-b)i|<|1-(x-yi)(a+bi)|$$ or $$(x-a)^2+(y-b)^2<(1-ax-by)^2+(ay-bx)^2$$ or $$x^2+y^2+a^2+b^2<1+a^2x^2+b^2y^2+a^2y^2+b^2x^2$$ or $$1-(x^2+y^2+a^2+b^2)+(x^2+y^2)(a^2+b^2)>0$$ or $$(1-x^2-y^2)(1-a^2-b^2)>0,$$ which is obvious.

Solution 3:

A silly answer, given the context - but:

Consider the function $$f(z)= {z -\beta\over 1- z\overline \beta},$$ where $|\beta|<1$, with domain the disc $|z| \le 1$. The denominator does not vanish: if $\beta\not = 0$, $$ 1-z\overline\beta = 0 \iff z = \beta/|\beta|^2,$$ requiring $|z|>1$. Hence $f$ is analytic on the disc.

Now, if $|z|=1$, then $z^{-1}= \overline z$, so that $$|f(z)|= \left|{z -\beta\over 1- z\overline \beta}\right| = \left|{z -\beta\over \overline z- \overline \beta}\right| = 1.$$ So, by the maximum modulus principle, $|f(\alpha)|< 1$, if $|\alpha| <1.$

As I recall, I saw this argument in Rudin's RCA.