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Prove that $$|\frac{a-b}{1-\bar ab}|=1$$ if $|a|=1$ or $|b|=1$

I assumed $|a|=1$. Then tried to show that our statement holds.

I wrote $a=a_1+ia_2$ and $b=b_1+ib_2$ and $\bar a=a_1-ia_2$

Also $$|a|=|\bar a|=a_1^2+a_2^2=1$$

However, after multiplying it all out it doesn't get me anywhere.

I figured maybe I have to use triangle inequality, in particular I know that:

$|a+b|\leq |a|+|b|$ and $|a+b| \geq |a|-|b|$

Should I take $-b$ instead of $b$? Then the inequality becomes:

$|a-b| \leq |a|+|-b|$ and $|a-b| \geq |a|-|-b|$ which looks more similar to what I need.

GRS
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3 Answers3

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In case modulus $|a|=1$ you can use the fact that $$|a|^2 = a\bar{a} =1$$ to express the ratio as follows. $$\frac{a-b}{1-\bar{a}b} = \frac{a-b}{a\bar{a}-\bar{a}b} = \frac{1}{\bar{a}}\cdot\frac{a-b}{a-b}\ .$$ Since the modulus $|\bar{a}| = 1$ this implies that $$\left|\frac{a-b}{1-\bar{a}b}\right| = \frac{1}{|\bar{a}|} = 1\ .$$

In case modulus $|b|=1$ you can use the fact that $$|b|^2 = b\bar{b} =1$$ to express the ratio as follows. $$\frac{a-b}{1-\bar{a}b} = \frac{a-b}{b\bar{b}-\bar{a}b} = \frac{1}{b}\cdot\frac{a-b}{\bar{b}-\bar{a}}\ .$$ Since the modulus $|\bar{b}-\bar{a}| = |a-b|$ this implies that $$\left|\frac{a-b}{1-\bar{a}b}\right| = \frac{1}{|b|}\cdot \frac{|a-b|}{|\bar{b}-\bar{a}|} = \frac{1}{|b|}\cdot 1 = 1\ .$$

Anders Muszta
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  • Thank you very much, I was looking at this question again, and what would happen if I set both |a| and |b| to 1? – GRS Jan 17 '16 at 18:53
  • @Epsilon. You are welcome. It is perfectly fine to set |a|=1 and |b|=1; the result will be the same. The premise does not say "either |a|=1 or |b|=1" but just "|a|=1 or |b|=1". – Anders Muszta Jan 17 '16 at 21:45
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Some hints: $a\bar{a} = 1$ and $a-b = a(1-b/a)$. And for $|b| = 1$, note that $1 = \bar{1}$.

Unit
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If $|a|=1$ then $$\left|\frac{a-b}{1-\bar ab}\right|= \left|\frac{a(1-b/a)}{1-(\bar a a)(b/a)}\right|=|a|\left|\frac{1-b/a}{1-b/a}\right|=1. $$

Intelligenti pauca
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