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Complex Number basic problem

  1. Jul 9, 2011 #1
    In a complex number sum, I have encountered a minute difficulty in understanding a step:

    [itex]\left|(cos\theta-1)+i.sin\theta\right|[/itex]
    = [itex]\sqrt{}(cos\theta-1)^2+sin^2\theta[/itex]


    Now my question is, how did the 'i' got eliminated from the second step? Now, i equals [itex]\sqrt{}-1[/itex], so when squared, there should be a minus sign in the second step. Can anyone help me clearing my basics?
     
  2. jcsd
  3. Jul 9, 2011 #2
    Are you remembering that the modulus of a complex number is the square root of the product of the number with its conjugate? That is:

    [tex]\left | a \right | = \sqrt{\overline{a}a},[/tex]

    where

    [tex]a =(\text{Re}(a)+i \, \text{Im}(a)),[/tex]

    [tex]\overline{a}=(\text{Re}(a)-i \, \text{Im}(a)),[/tex]

    and Re(a) is the real part of a, and Im(a) the imaginary part.
     
  4. Jul 9, 2011 #3
    Thanks for the help! Simply ignored this basic rule initially.
     
  5. Jul 9, 2011 #4

    micromass

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    The equality

    [tex]|a+bi|=\sqrt{a^2+b^2}[/tex]

    is just the definition of the absolute value! There is no reasoning behind it, it's just true by definition. Your OP was also true by definition.
     
  6. Jul 10, 2011 #5
    As often happens, we have two identities, and whichever is taken as the definition, the other pops out as a theorem.
     
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