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Math Help - DeMoivre's Theorem

  1. #1
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    DeMoivre's Theorem

    I need to find the real and imaginary values of z^5, when z = 2-2i

    I got r = 2 sqrt(2) and theta = -pi/4

    However now I am stuck. I tried to put it into the form r^k e^(i k theta)

    and got (2 rt(2)) ^5 * e^(-5pi/4) but then I don't know what to do to find the real and imaginary parts.

    What should I do?

    Thank you
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  2. #2
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    \text{Re} \left( {\left[ {x + yi} \right]^5 } \right) = \sum\limits_{k = 0}^2 {\left( { - 1} \right)^k \dbinom{ 5}{2k} x^{5 - 2k} y^{2k} }
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  3. #3
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    Hello, ironz!

    \text{Find the real and imaginary components of }\,(2 - 2i)^5

    Write 2 - 2i in polar form.
    . . We have: . x = 2,\;y =2,\;r = 2\sqrt{2},\;\theta = \text{-}\dfrac{\pi}{4}


    Hence: . 2 - 2i \:=\:2\sqrt{2}\left(\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right)


    Then: . (2-2i)^5 \;=\;\left(2\sqrt{2}\right)^5\,\left(\cos\frac {5\pi}{4} - \sin\frac{5\pi}{4}\right)

    . . . . . . . . . . . . =\;2^5(\sqrt{2})^5\left(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right)

    . . . . . . . . . . . . =\;32(4\sqrt{2})\left(\dfrac{-1+i}{\sqrt{2}}\right)

    . . . . . . . . . . . . =\;128(-1 + i)

    . . . . . . . . . . . . =\;\underbrace{-128}_{\mathbb{R}}   + \underbrace{128}_{\mathbb{I}}i

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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by ironz View Post
    I need to find the real and imaginary values of z^5, when z = 2-2i
    An alternative: taking into account that (1-i)^2=1-1-2i=-2i:

    (2-2i)^5=2^5(-2i)^2(1-i)=-2^7(1-i)=-128+128i

    In general this is the shortest way for computing (x+yi)^n when y=x or y=-x .

    Fernando Revilla
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  5. #5
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    Quote Originally Posted by ironz View Post
    I need to find the real and imaginary values of z^5, when z = 2-2i

    I got r = 2 sqrt(2) and theta = -pi/4

    However now I am stuck. I tried to put it into the form r^k e^(i k theta)

    and got (2 rt(2)) ^5 * e^(-5pi/4) but then I don't know what to do to find the real and imaginary parts.

    that should read e^(-i5pi/4)

    What should I do?

    Thank you
    You reached

    \displaystyle\ z^5=\left[2\sqrt{2}\right]^5e^{-i\frac{5{\pi}}{4}}

    To regenerate the real and imaginary parts from this...

    \displaystyle\ e^{-i\frac{5{\pi}}{4}}=cos\left(-\frac{5{\pi}}{4}\right)+isin\left(-\frac{5{\pi}}{4}\right)

    \displaystyle\ -\frac{5{\pi}}{4}=2{\pi}-\frac{5{\pi}}{4}=\frac{3{\pi}}{4}

    \displaystyle\ cos\left(\frac{3{\pi}}{4}\right)=-cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}

    \displaystyle\ sin\left(\frac{3{\pi}}{4}\right)=sin\left(\frac{\p  i}{4}\right)=\frac{1}{\sqrt{2}}

    Then

    \displaystyle\ z^5=\frac{-2^5\left(\sqrt{2}^5\right)}{\sqrt{2}}+\frac{i2^5\l  eft(\sqrt{2}^5\right)}{\sqrt{2}}
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  6. #6
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    Don't let those square roots of 2 in Archie Meade's response fool you. When you actually do the calculation, they will cancel out and give the same answer FernandoRevilla gives.
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  7. #7
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    I understand. I think I read the question wrong and tried to solve for all the roots instead of just the z^5, also, I didn't know how to simplify it properly. Thank you for your help
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  8. #8
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    Quote Originally Posted by ironz View Post
    I understand. I think I read the question wrong and tried to solve for all the roots instead of just the z^5, also, I didn't know how to simplify it properly. Thank you for your help
    Having found z^5

    z=2-2i is one of the 5 "fifth roots" of z^5

    but that's a different question.
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