Proof

**Nichelle14** · June 11th 2006, 03:49 PM

If x, y are complex, prove that

||x| - |y|| < = |x - y|

**nath_quam** · June 13th 2006, 11:34 PM

absolute value or modulus signs

**TD!** · June 14th 2006, 06:53 AM

I'll be using the triangle inequality, which also holds for complex numbers:

$\left| {x + y} \right| \le \left| x \right| + \left| y \right|$

We can write:

$<br /> \left| x \right| = \left| {x - y + y} \right| \le \left| {x - y} \right| + \left| y \right| \Rightarrow \left| x \right| - \left| y \right| \le \left| {x - y} \right|<br />$

Doing the same in reverse (changing roles of y and x), yields:

$\left| y \right| - \left| x \right| \le \left| {x - y} \right|$

The result follows immediately now.

**ThePerfectHacker** · June 14th 2006, 02:28 PM

For all, $a_1,a_2,b_1,b_2\in \mathbb{R}$ ,

$(a_2b_1-a_1b_2)^2\geq 0$
Thus,
$a_2^2b_1^2-2a_1b_1a_2b_2+a_1^2b_2^2\geq 0$
Thus,
$a_2^2b_1^2+a_1^2b_2^2\geq 2a_1b_1a_2b_2$
Thus, adding same quality to both sides,
$a_1^2b_1^2+a_2^2b_1^2+a_1^2b_2^2+a_2^2b_2^2\geq$ $a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2$
Thus,
$(a_1^2+a_2^2)(b_1^2+b_2^2)\geq (a_1b_1+a_2b_2)^2$
Thus, take square root (note terms are non-negative),
$\sqrt{a_1^2+a_2^2}\cdot \sqrt{b_1^2+b_2^2}\geq |a_1b_1+a_2b_2|$ --> Cauchy-Swartchz
--------------
Assume, $|a_1b_1+a_2b_2|\geq 0$ then,
$\sqrt{a_1^2+a_2^2}\cdot \sqrt{b_1^2+b_2^2}\geq a_1b_1+a_2b_2$
Thus,
$-2\sqrt{a_1^2+a_2^2}\cdot \sqrt{b_1^2+b_2^2} \leq -2a_1b_1-2a_2b_2$
Thus, adding same quality to both sides,
$a_1^2+a_2^2+b_1^2+b_2^2-2\sqrt{a_1^2+a_2^2}\cdot \sqrt{b_1^2+b_2^2}$ $\leq a_1^2-2a_1b_1+b_1^2+a_2^2-2a_2b_2+b_2^2$
Thus,
$\left( \sqrt{a_1^2+a_2^2}-\sqrt{b_1^2+b_2^2} \right)^2 \leq (a_1-b_1)^2+(a_2-b_2)^2$
Take square roots, (note terms are non-negative),
$\left| \sqrt{a_1^2+a_2^2}-\sqrt{b_1^2+b_2^2} \right| \leq \sqrt{(a_1-b_1)^2+(a_2-b_2)^2}$
This is,
$||u|-|v||\leq |u-v|$
if, $u=a_1+ia_2,v=b_1+ib_2$

I did not yet complete the proof when,
$|a_1b_1-a_2b_2|<0$
I do not see how

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