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Complex numbers + proof

  1. Jan 28, 2005 #1
    Hi :smile:

    I just started to look at complex numbers.

    Prove the ``Parallellogram law''
    http://www.sosmath.com/complex/number/complexplane/img4.gif


    This is how I solved it:

    [tex]z=a+bi[/tex]
    [tex]w=c+di[/tex]

    [tex]|z+w|^2=\sqrt{(a+c)^2+(b+d)^2}=a^2+2ac+c^2+b^2+2bd+d^2[/tex]

    then we have

    [tex]|z-w|^2=\sqrt{(a-c)^2+(b-d)^2}=a^2-2ac+c^2+b^2-2bd+d^2[/tex]

    [tex]2(|z|^2+|w|^2)=2((a^2+b^2)+(c^2+d^2)) = 2a^2+2b^2+2c^2+2d^2[/tex]

    [tex]a^2+2ac+c^2+b^2+2bd+d^2+a^2-2ac+c^2+b^2-2bd+d^2=2a^2+2b^2+2c^2+2d^2[/tex]

    My question is:

    Was my calculation OK or was it a misscalculation (a lucky one which prooved the formula)
     
    Last edited: Jan 28, 2005
  2. jcsd
  3. Jan 28, 2005 #2

    dextercioby

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    Those 2 radicals should be squared.Or should be absent altogether.

    The rest is okay.

    Daniel.
     
  4. Jan 29, 2005 #3
    Oh....That was a typo :blushing:

    Thank you dextercioby :cool:
     
  5. Jan 29, 2005 #4

    mathwonk

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    vector methods are nice too. i.e. |z+w|^2 = (z+w).(z+w), (dot product),

    and |z-w|^2 = (z-w).(z-w).

    Expanding and adding, the cross terms cancel, leaving

    z.z + w.w + z.z + w.w = 2(|z|^2 + |w|^2).



    you can do this with complex numbers notation too, no vectors. i.e. let zbar be the conjugate of the complex number z. Then |z|^2 = z(zbar).

    Hence |z+w|^2 = (z+w)([z+w]bar). But bar commutes with sums and products, so this equals

    (z+w)(zbar + wbar) = z(zbar) + w(wbar) + wzbar + zwbar.

    Similarly |z-w|^2 = z(zbar) + w(wbar) - wzbar - zwbar.

    so the sum is 2 (z[zbar] + w[wbar]) = 2 (|z|^2 + |w|^2).
     
    Last edited: Jan 29, 2005
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