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Thread: Help with complex numbers

  1. #1
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    Help with complex numbers

    Show that $\displaystyle
    \left| {a + b} \right|^2 + \left| {a - b} \right|^2 = 2\left( {\left| a \right|^2 + \left| b \right|^2 } \right)
    $

    where a and b are complex numbers. Interpret this results geometrically.
    $\displaystyle
    \begin{array}{l}
    \left| {a + b} \right|^2 + \left| {a - b} \right|^2 \\
    = \left| a \right|^2 + 2\left| a \right|\left| b \right| + \left| b \right|^2 + \left| a \right|^2 - 2\left| a \right|\left| b \right| + \left| b \right|^2 \\
    = 2\left| a \right|^2 + 2\left| b \right|^2 \\
    = RHS \\
    \end{array}
    $

    I don't know how to interpret this result geometrically though, any help would be appreciated.
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello,
    Quote Originally Posted by free_to_fly View Post
    $\displaystyle
    \begin{array}{l}
    \left| {a + b} \right|^2 + \left| {a - b} \right|^2 \\
    = \left| a \right|^2 + 2\left| a \right|\left| b \right| + \left| b \right|^2 + \left| a \right|^2 - 2\left| a \right|\left| b \right| + \left| b \right|^2 \\
    \end{array}
    $
    You're saying that $\displaystyle |a+b|^2=\left| a \right|^2 + 2\left| a \right|\left| b \right|+|b|^2$ which is wrong : if $\displaystyle b=-a$ the LHS equals 0 and the RHS equals $\displaystyle 4|b|^2$. I don't think there is a rule that tells us how to expand $\displaystyle |a+b|^2$.

    However, I'm sure you know that if $\displaystyle z$ is a complex number then $\displaystyle |z|^2=z\overline{z}$ so

    $\displaystyle \begin{aligned}
    \left| {a + b} \right|^2 + \left| {a - b} \right|^2&=(a + b)\overline{(a + b)} + (a - b)\overline{(a-b)}\\
    &=\ldots
    \end{aligned}$

    I don't know how to interpret this result geometrically though, any help would be appreciated.
    Take a look at equation #10 of Parallelogram -- from Wolfram MathWorld. It looks like the one you're asked to show, doesn't it ?
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