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Complex Conjugates

  1. Jul 20, 2009 #1
    Statement:
    [tex]VV* = (V_r + jV_i)(V_r - jV_i) = V_{r}^{2} + V_{i}^{2} = |V|^{2}[/tex]


    Question:
    I am not sure why the second equality isn't written as, [tex]V_{r}^{2} + V_{i}^{2} = V^{2}?[/tex]
     
  2. jcsd
  3. Jul 20, 2009 #2
    Well no. Let's try V = a + ib and square it to see what happens.

    V^2 = VV = (a + ib)(a + ib) = a^2 + aib + iba + (ib)^2 = a^2 + 2iab - b^2

    Notice the extra 2iab term.

    The modulus of a complex number is the length of the "vector" in the 2-d complex plane. Such vectors have x-component equal to the real part and y-component equal to the imaginary part. So, according to the Pythagorean theorem, they have modulus sqrt(a^2 + b^2). Squaring this gives the desired a^2 + b^2.
     
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