1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...