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Math Help - A complex number problem

  1. #1
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    A complex number problem

    Hi guys I need help in proving this problem:


    thanks guys. really need your immediate reply.
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  2. #2
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    i've got the 'answer'

    We have to prove that |(a-b)| = |(1-ab*)| , where |a| = 1

    Hence, we need to show that the magnitude of the LHS = RHS

    |a| = aa* = 1 => a = 1/a*

    Hence,

    LHS = |(a-b)|
    RHS = |(1-b*/a*)| => |(a*-b*)|/|a*| = |(a*-b*)|/1 = |(a*-b*)|

    ============> LHS = |(a-b)| = |(a*-b*)| = RHS ==================> It is PROVEN.

    Since a & b are distinct complex numbers,

    a = b + ci ; a* = b - ci ; b = x +yi ; b* = x - yi

    LHS = |(a-b)| = |(b+ci-x-yi )| = |(b-x)+(c-y)i| = [(b-x)^2 + (c-y)^2]^(1/2)
    RHS = |(a*-b*)| = |(b-ci-x+yi)| = |(b-x)+(y-c)i| = [(b-x)^2 + (y-c)^2]^(1/2)

    Note: (c -y) = [-(c-y)]
    let (c - y) be h,

    h = -h => h^2 = (-h)^2 = |h|^2

    Hence,

    let (b-x) be p,

    LHS = (p^2 + h^2)^(1/2)
    RHS = (p^2 + h^2)^(1/2) = LHS <=========== PROVEN
    Last edited by werepurple; December 6th 2009 at 05:36 PM.
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  3. #3
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    Quote Originally Posted by werepurple View Post

    => |(a-b)| / |a| = |(a-b)/(a)| = |(1-b/a)| = |1-ab*|
    why is |(1-b/a)| = |1-ab*|?
    is there a property why this is true?

    thanks for the rep btw.
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  4. #4
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    hmmm.. you know what? I think the denominator should be 1-ba* not 1-ab*..

    maybe our teacher just printed it wrong.. but I hope you do have a solution..

    Thanks!!
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  5. #5
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    I don't think there's a mistake with the question, haha.

    Btw, its not exactly a property, but just in case you are not aware.

    First off, from the start, what we want to prove is this:

    |(a-b)| = |1-ab*|

    And, what you were asking is this: |(1-b/a)| = |(1-ab*)| (Why?)

    Please note that the RHS of both equations are similar, which then tells me what you really were asking is how |(1-b/a)| come about from |(a-b)/(a)|.

    What really happened happens within the absolute signs: (a-b)/(a) = ((a/a)-(b/a)) = (1-b/a).

    I changed my workings which my teachers always complained messy; I hope you would now find it clearer, and easier to understand!
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  6. #6
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    hehe.. i get it all actually except now for the first part..
    why did you substitute |a|= 1 to WTF?
    it's because WTF was the problem we wanted to prove so we can't assume that its equal to 1 unless we already proved it.
    so i tried starting from the LHS of WTF and started manipulating it so that it became 1. what i got was

    |(a-b)/(1-ab*)|=|a-b|/|1-ab*|

    = (|a-b|/1)/|1-ab*|

    since 1 = |a|,

    =(|a-b|/|a|)/|1-ab*|

    =|(a-b)/a|/|1-ab*|

    =|1-(b/a)|/|1-ab*| (1)

    but b/a = ba' (where a' is the inverse of a) and since a' = a*/(|a|^2) = a*/(1^2) = a*,
    b/a = ba*, therefore continuing from 1:

    =|1-ba*|/|1-ab*| should be equal to 1.

    which makes me suspect that our professor typed it wrong. hehe.
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  7. #7
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    I made a mistake, I've corrected it. Lol, I was reading my workings, then I realized I made a really really stupid mistake. Haha, joined this forum just yesterday, not really careful doing math using my keyboard lols.
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  8. #8
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    hey thanks!!
    actually I finally realized the missing step earlier in class...hehe.. I realized that

    |1-ba*| = |(1-ba*)*|
    = |1*-(ba*)*|
    = |1-ab*|

    anyways, you also proved it another way so thanks!!!

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