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Calculate complex number to nth power with de Moivre's formula

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\left( \frac{2 \sqrt{3}+2i}{1-i} \right)^{50}[/tex]

    2. Relevant equations

    [tex]z = a+ib[/tex]

    1. [tex]z = r(cos \phi + isin \phi)[/tex]

      [tex]r = \sqrt{a^2+b^2}[/tex]

      [tex]cos \phi = \frac{a}{r}[/tex]

      [tex]sin \phi = \frac{b}{r}[/tex]

      [tex]\frac{z_1}{z_2} = \frac{r_1}{r_2}\left( cos(\phi_1-\phi_2) + isin(\phi_1 - \phi_2)\right)[/tex]

      [tex]z^{n} = r^{n}(cos n \phi + isin n \phi)[/tex]

    3. The attempt at a solution

    First I assign the numerator to [tex]z_1 = 2 \sqrt{3} + 2i[/tex] and convert it to trigonometric form:

    [tex]a=2\sqrt{3} , b=2[/tex]
    [tex]r = \sqrt{(2\sqrt{3})^2+2^2} = \sqrt{12+4} = 4[/tex]
    [tex]cos \phi_1 = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}[/tex]
    [tex]sin \phi_1 = \frac{2}{4} = \frac{1}{2}[/tex]

    From the [tex]sin \phi_1[/tex] and [tex]cos \phi_1[/tex] can be concluded that that [tex]\phi_1 = \frac{\pi}{6}[/tex]. Using the conversion formula we get the trigonometric form of [tex]z_1[/tex]: [tex]z_1 = 4(cos \frac{\pi}{6} + isin \frac{\pi}{6})[/tex]

    Second I assign the denominator to [tex]z_2 = 1-i[/tex]

    [tex]a = 1, b = -1, r = \sqrt{2}[/tex]
    [tex]cos \phi_2 = \frac{1}{\sqrt{2}}[/tex]
    [tex]sin \phi_2 = -\frac{1}{\sqrt{2}}[/tex]

    So [tex]\phi_2 = \frac{7\pi}{4}[/tex] and now we can derive the trigonometric form of [tex]z_2 = \sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})[/tex]

    What I get is:[tex]\left( \frac{4(cos \frac{\pi}{6} + isin \frac{\pi}{6})}{\sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})} \right)^{50}[/tex]

    Using the division formula I get the trigonometric version of the main equation:
    [tex]\left( \frac{4}{\sqrt{2}}(cos(\frac{\pi}{6}-\frac{7\pi}{4}) + isin (\frac{\pi}{6}-\frac{7\pi}{4}) \right)^{50}[/tex]

    How do I continue? I can simplify it further and I get [tex]cos (\frac{19\pi}{12})[/tex] and the same for sin but with the - (minus) sign in front.

    I can apply the de Moivre's formula but I can't simplify it to the answer given below.

    Can some one give me any guidance (step by step).

    Maybe I have an error somewhere or I just don't know how to arrive to the right answer which is: [tex]2^{74}(-\sqrt{3}+i)[/tex]

    Thank you in advance.
    Last edited: Sep 6, 2010
  2. jcsd
  3. Sep 6, 2010 #2


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    Homework Helper

    The first thing I would do is this:
    \frac{2 \sqrt{3}+2i}{1-i}\cdot\frac{1+i}{1+i}=\sqrt{3}-1+(1+\sqrt{3})i
    The the application of de Movire's theorem is relatively straight forward.

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