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

This is the question I'm stumbling with:

When $|\alpha| < 1$ and $|\beta| < 1$, show that:

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

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.

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.

• Interesting, this hint gets me to $|\alpha|^2 + |\beta|^2 < 1 + |\alpha|^2 |\beta|^2$. Any hint from there? Sep 26, 2013 at 18:07
• @Andomar What is $(1-|\alpha|^2)(1-|\beta|^2)$?
– mrf
Sep 26, 2013 at 18:33
• Very helpful, I like the partial answers! Thanks. Sep 26, 2013 at 18:44
• We have $|\alpha|^2 + |\beta|^2 < (1-|\alpha|^2)(1-|\beta|^2)$. Where can we go from here? Aug 29, 2019 at 1:37
• @AlJebr that’s not what you get. Check your algebra.
– mrf
Aug 29, 2019 at 7:20

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.

• Can down -voter explain us, why did you do it? Aug 16, 2019 at 19:48

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.

Using $$z \cdot \bar{z}=|z|^{2}$$ yields \begin{aligned} |1-\bar{\alpha} \beta|^{2}-|\alpha-\beta|^{2} &\left.=1+|\alpha|^{2}|\beta|^{2}-|\alpha|^{2}-\mid \beta\right)^{2} \\ &=\left(1-|\alpha|^{2}\right)\left(1-|\beta|^{2}\right) \\ &>0 \\ \therefore \quad |1-\bar{\alpha} \beta|^{2}&>|\alpha-\beta|^{2} \\ |1-\bar{\alpha} \beta|&>| \alpha-\beta \mid \end{aligned} Hence we may conclude that

$$\left|\frac{\alpha-\beta}{\mid-\bar{\alpha} \beta}\right|<1.$$