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Absolute value of complex exponential equals 1

  1. Apr 16, 2013 #1
    Hello all,

    I'm having trouble showing that |e^it|=1, where i is the imaginary unit. I expanded this to |cos(t)+isin(t)| and then used the definition of the absolute value to square the inside and take the square root, but I keep getting stuck with √(cos(2t)+sin(2t)). Does anyone have any suggestions for how to show this equals 1?
     
  2. jcsd
  3. Apr 16, 2013 #2
    Why would you square cos(t)+isin(t)? To find the magnitude of a complex number you square the real and imaginary parts separately and take the square root of their sum (or alternatively, take the square root of the complex number multiplied by its conjugate). Absolute value is defined this way so that it still means the 'distance' from the origin.

    Edit: Oh, I see your mistake. A definition of absolute value for real numbers is [itex] |x|= \sqrt{x^2} [/itex]. This is no longer true for complex numbers. See http://en.wikipedia.org/wiki/Complex_number#Absolute_value_and_argument
     
    Last edited: Apr 16, 2013
  4. Apr 16, 2013 #3
    Make sure you are applying absolute value correctly. The absolute value of a complex number is not equivalent to the square root of the square unless the number has no imaginary part. It is generalized to adhere to being the positive distance of the complex number from the origin in the complex plane.
     
  5. Apr 16, 2013 #4

    rbj

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    so you got this right:

    [tex] e^{i t} = \cos(t) + i \sin(t) [/tex]

    but this is not:

    [tex] |\cos(t) + i \sin(t)| = \sqrt{ \cos(2 t) + \sin(2 t) } [/tex]

    the quantity inside the square root is not always non-negative. in the domain of real numbers, how do you deal with the square root when the argument is negative?
     
  6. Apr 17, 2013 #5

    HallsofIvy

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    [itex]|z|= \sqrt{zz^*}[/itex]
    where * indicates the complex conjugate. In particular
    [itex]|cos(t)+ i sin(t)|= \sqrt{(cos(t)+ i sin(t))(cos(t)- i sin(t))}[/itex]
    That should give what you want.
     
  7. Apr 17, 2013 #6

    Bacle2

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    Yet another way: in coordinates, cost+isint corresponds to the point (cost,sint) in the plane, and then apply a basic identity using squares of sine and cosine.
     
  8. Apr 18, 2013 #7
    e^(it) is in polar form already with r = 1

    If you want to put it in rectangular form

    [tex] e^{it} = cost + isint = a + ib [/tex]

    then

    [tex]|e^{it}| = \sqrt{a^2 + b^2} = \sqrt{cos^2t + sin^2t} = \sqrt{1} = 1[/tex]

    or consider...

    [tex] |e^{it}| = e^{it} \cdot e^{-it} = e^0 = 1 [/tex]
     
    Last edited: Apr 18, 2013
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