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Math Help - prove of equation with complex numbers

  1. #1
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    prove of equation with complex numbers

    Here is the equation to prove:
    |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2
    I tried with something like this:
    |z|^2=z \hat{z}
    z+\hat{z}=2rez
    but i still can not get the right result.

    a,b are different complex numbers.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Vermax View Post
    Here is the equation to prove:
    |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2
    I tried with something like this:
    |z|^2=z \hat{z}
    z+\hat{z}=2rez
    but i still can not get the right result.

    a,b are different complex numbers.
    Multiply out (a - b)(\hat a - \hat b) then note that \hat a b = \hat {a \hat b}.
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  3. #3
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    Thank you, now almost everything is clear for me except one thing. How do you know that:
    |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2=(a - b)(\hat a - \hat b)?

    I am sure you, as an expert, presented me it correctly, but i am just curious where it comes from. How do you know these identities? (they are not the common ones i think...)
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Vermax View Post
    Thank you, now almost everything is clear for me except one thing. How do you know that:
    |a-b|^2=|a|^2-2 re \hat{a} b+|b|^2=(a - b)(\hat a - \hat b)?

    I am sure you, as an expert, presented me it correctly, but i am just curious where it comes from. How do you know these identities? (they are not the common ones i think...)
    I'm going to use \overline z for conjugate, it's easier to use (see below).

    You already have that |z|^2 = z \overline z.

    So |a - b|^2 = (a-b) \overline{(a-b)}.

    But you also have that \overline{(a-b)} = \overline a - \overline b (it's a standard result, but simple to prove).

    So |a - b|^2 = (a-b) (\overline a - \overline b).

    Multiply this out, you get

    a \overline a + b \overline b - a \overline b - \overline a b.

    But as \overline a b = \overline {a \overline b} (which you should be able to prove, it's very straightforward), this works out as

    a \overline a + b \overline b - (\overline a b + \overline {a \overline b}).

    The result follows from z + \overline z = 2 Re z.
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  5. #5
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    Thank you for that post. I did not think that complex numbers can be that complicated
    I proved these two identities that I was supposed to, but still I think I should work on beeing more "fluent" in it again
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