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Complex Numbers - Forms and Parts

  1. Feb 11, 2012 #1
    Hi, I have a complex number and understand that the rectangular form of the number is represented by

    s = σ + jω, where σ is the real part and jω is imaginary.

    I am having trouble locating them in the number below:
    [​IMG]

    I know that "2" is a real number, and the numerator is imaginary along with j*2*pi*k. Since the numerator is dividing both the elements at the bottom, does this number have a real and imaginary part? (This is where I am a little confused).

    My guess would be that σ = 2 and the rest is imaginary.

    If I could figure out what parts are real and imaginary, I can go on to find the rectangular form and the polar form.

    Thanks
     
  2. jcsd
  3. Feb 11, 2012 #2
    Euler's relation e^ix = cos(x) + i*sin(x) vastly simplifies the exponential in the numerator. Cos and sin have period 2pi so the numerator is -1. Then multiply numerator and denominator by the conjugate of the denominator (assuming k is real). That leaves you with a real number in the denominator and a complex number in the numerator whose real and imaginary parts can be readily evaluated.

    By the way, multiplying numerator and denominator by the conjugate of the denominator is the standard thing to do with this kind of problem.
     
    Last edited: Feb 11, 2012
  4. Feb 12, 2012 #3

    HallsofIvy

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    No, the numerator is NOT imaginary. In fact it is real- it is [itex]e^{j3\pi}= cos(3\pi)+ jsin(3\pi)= -1[/itex].
     
    Last edited: Feb 12, 2012
  5. Feb 13, 2012 #4
    Just try a look into www.wolframalpha.com and enter

    Exp[3 Pi I] / (2 + 2 Pi k I)

    what is the Mathematica version of your formula
     
  6. Feb 13, 2012 #5

    Deveno

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    as others have suggested, evaluate the numerator at the specific angle (3pi), and then multiply the resulting fraction by:

    [tex]\frac{2 - j2\pi k}{2 - j2\pi k} (= 1)[/tex]

    to make the denominator real.
     
  7. Feb 21, 2012 #6
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