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Math Help - The exponential form of complex number

  1. #1
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    The exponential form of complex number

    z = (-i) ^1/3

    z^3 = -i

    let z = re ^i ϴ

    z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]

    pls help me to understand how to find "3ϴ"?

    what is the formula to find 3ϴ??
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  2. #2
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    Re: The exponential form of complex number

    Hey mous99.

    For this problem do you know how to use De-Moivre's formula for the angle? (Hint: its in the form of (ϴ + 2kπ)/n where n = 3 k = 0,1,2) [In this case theta = [3/2]*π]
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  3. #3
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    Re: The exponential form of complex number

    behold:

    e^{i (\theta + \phi )} = \cos(\theta + \phi ) + i \sin(\theta + \phi )

     = \cos(\theta)\cos(\phi ) - \sin(\theta)\sin(\phi ) + i (\sin(\theta)\cos(\phi ) + \cos(\theta)\sin(\phi ))

     = (\cos(\theta) + i \sin(\theta))(\cos(\phi ) + i \sin(\phi )) = e^{i \theta}e^{i \phi }

    so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles).
    Last edited by Deveno; December 21st 2012 at 02:52 PM.
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  4. #4
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    Re: The exponential form of complex number

    i cant see the invalid equation.

    can u show me again?

    thanks...
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  5. #5
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    Re: The exponential form of complex number

    Unfortunately, Latex does not seem to be working today. I have removed the Latex tags.

    Quote Originally Posted by Deveno View Post
    behold:

    e^{i (\theta + \phi )} = \cos(\theta + \phi ) + i \sin(\theta + \phi ) = \cos(\theta)\cos(\phi ) - \sin(\theta)\sin(\phi ) + i (\sin(\theta)\cos(\phi ) + \cos(\theta)\sin(\phi ))

    = (\cos(\theta) + i \sin(\theta))(\cos(\phi ) + i \sin(\phi )) = e^{i \theta}e^{i \phi }

    so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles).
    The exponential form of complex number



    z = (-i) ^1/3

    z^3 = -i

    let z = re ^i ϴ

    z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]

    pls help me to understand how to find "3ϴ"?

    what is the formula to find 3ϴ??
    Just what you wrote: 3ϴ= [(3π/2) + 2nπ]

    Now, solve for ϴ.
    Thanks from mous99
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