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Math Help - Proof

  1. #1
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    Proof

    If x, y are complex, prove that

    ||x| - |y|| < = |x - y|
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  2. #2
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    Complex signs??

    absolute value or modulus signs
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  3. #3
    TD!
    TD! is offline
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    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.
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  4. #4
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    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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