Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$1-|w| \le |1+w| \le 1+|w| \ \forall w \in \mathbb{C}$

The book gives a geometric but not an algebraic proof. Does anybody see a way? Please do tell

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Suppose you can assume the triangle inequality $$|a+b|\le |a|+|b|\quad \text{for}\ a,\,b\in\mathbb{C}.$$

By letting $a=-w$ and $b=1+w$ we get $|1|\le |-w|+|1+w|$, hence $1-|w|\le |1+w|$;

By letting $a=1$ and $b=w$ we get $|1+w|\le |1|+|w|$, hence $|1+w|\le 1+|w|$.

Edit. "Algebraic proof" of the triangle inequality. We show that $(|a|+|b|)^2-(|a+b|)^2\ge 0$.

$$\begin{align*} (|a|+|b|)^2-(|a+b|)^2 &=|a|^2+2|a||b|+|b|^2-(a+b)(\overline{a+b})\\ &=a\overline{a}+2|a||b|+b\overline{b}-a\overline{a}-b\overline{b} -a\overline{b}-\overline{a}b\\ &=2|a||\overline{b}|-(a\overline{b}+\overline{a\overline{b}})\\ &=2|a\overline{b}|-2\text{Re}\,(a\overline{b})\ge 0. \end{align*}$$

share|improve this answer
    
pharmine, Thank you. –  VVV Nov 13 '11 at 22:00
    
You're welcome. –  pharmine Nov 15 '11 at 6:23

You don't need geometry, this is just basic properties of norms : The triangle inequality $|x+y|\leq |x|+|y|$ yields your upper bound by taking $x=1$, and you obtain your lower bound by using $||x|-|y||\leq |x-y|$ and the trivial fact that $|x|-|y|\leq||x|-|y||$ specifying $x$ to be $1$. But maybe I didn't understand what you mean by "algebraic proof" ?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.