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Suppose a, b,c three complex numbers as: $|a|=|b|=|c|= 1$.

How can I prove that: $\left|\frac{a-b}{1-a\overline{b}}\right| = 1$ and

$|ab+bc+ca| = |a+b+c|$

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3  
As with your previous questions, start with $|w|^2 = w\bar{w}$. –  Arkamis Feb 15 '13 at 17:04
    
you can try to take the complex numbers in the form as $e^{i\theta}$, then it will really straight forward. –  Yimin Feb 15 '13 at 17:30

2 Answers 2

up vote 3 down vote accepted

From $|x|=1$ we can conclude $\bar x=1/x\ $ (because $1=|x|^2=x\bar x$).

Then, $|a-b|=|a|\cdot |1- a/b|=1\cdot |1-a\bar b|$, and for the second one: $$|ab+bc+ca|=\left|\frac{ab+bc+ca}{abc}\right|=|\bar c+\bar a+\bar b|=|a+b+c|. $$

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(1) As $|b|=1,|1-a\bar b|=\frac{|b||1-a\bar b|}{| b|}=|b(1-a\bar b)|=|b-ab^2|=|b-a|=|a-b|$

Alternatively, let $a=e^{i\alpha}=cis\alpha,b=cis\beta,c=cis\gamma $

So, \begin{align*} |a-b|&=|\cos \alpha-\cos \beta+i(\sin \alpha-\sin\beta) | =\sqrt{(\cos \alpha-\cos \beta)^2+(\sin \alpha-\sin\beta)^2}\\ &=\sqrt{2-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)}=\sqrt{2-2\cos(\alpha-\beta)} \end{align*}

$$|1-a\bar b|=|1-e^{i\alpha}e^{-i\beta}|=|1-e^{i(\alpha-\beta)}|=|1-\cos(\alpha-\beta)-i\sin(\alpha-\beta)|=\sqrt{(1-\cos(\alpha-\beta))^2+(-\sin(\alpha-\beta))^2}=\sqrt{2-2\cos(\alpha-\beta)}$$

(2) $|ab+bc+ca|=|abc(a^{-1}+b^{-1}+c^{-1})|=|abc||a^{-1}+b^{-1}+c^{-1}|=|\bar a+\bar b+\bar c|$ as $|abc|=|a||b||c|=1$

So, $|ab+bc+ca|=|\bar a+\bar b+\bar c|=|\overline{a+b+c}|=|a+b+c|$

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Sorry about accidentally losing part of your answer in my edit! Not sure how that happened. I was just editing for formatting, as I couldn't actually read parts of your answer, since they were going over into the side-bar. –  Tara B Feb 15 '13 at 17:37
    
@TaraB, I had to restore it resulting in one more version. –  lab bhattacharjee Feb 15 '13 at 17:38
    
The $\\bar$ does not to seem to work properly with $\{\}$ resulting in odd value in the last but one expression. –  lab bhattacharjee Feb 15 '13 at 17:40
    
I tend to use \overline instead if I want the line to go over several things. –  Tara B Feb 15 '13 at 17:43
    
@TaraB, thanks for your input –  lab bhattacharjee Feb 15 '13 at 18:08

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