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Complex Numbers, Polar Form

  1. May 29, 2007 #1
    IS there a formula for:

    z=(Cosx +iSinx)^4 (Cosy + iSiny)^2 ??
  2. jcsd
  3. May 29, 2007 #2
    You mean a way to simplify it? Just change each expression in ( ) to its corresponding exponential form.
  4. May 29, 2007 #3
    This doesn't make any sense: one of x and y has to represent the real part scalar and the other the imaginary part scalar. The expression you have there implies that neither x nor y are scalars and hence aren't axial values.
    Last edited: May 29, 2007
  5. May 29, 2007 #4
    Well I just put them in instead of the actual values....I'm just looking for the general way.

    So do I simplify each and then multiply them or something?
  6. May 29, 2007 #5
    If so, have you tried, as daveb suggested

    [tex] z=(\cos x +i\sin x)^{4} * (\cos y + i\sin y)^{2} [/tex]

    [tex] z = e^{4xi} * e^{2yi}[/tex]

    [tex] z = e^{(4x + 2y)i}[/tex]

    [tex] z = \cos (4x + 2y) + i \sin (4x + 2y) [/tex]
    Last edited: May 29, 2007
  7. May 29, 2007 #6
    I have only ever seen the last line there before, but thats actually what I thought it was. It's one of those rules.

    Cheers guys!
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