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Thread: De Moivre's Theorem

  1. #1
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    De Moivre's Theorem

    I have the problem: (1+i)/\5 (One plus i, raised to the 5th). I know r= square root of 2 and arctan= pi/4. Then we have square root of 2 raised to the 5th. This is where I run into a problem. How exactly does this work out. For example i know how to get the cube 8 by 8/\(1/3). I just don't know how to raise an nth power to something that is squared.
    Thanks.
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  2. #2
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    $\displaystyle (\sqrt{2})^5 = (2^{1/2})^5 = 2^{5/2}$

    Or you can write this as $\displaystyle \sqrt{2^5} = \sqrt{32}$

    And simplifying you can get $\displaystyle = 4\sqrt{2}$

    But as far as I know it doesn't really matter which way you write it...
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  3. #3
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    Hello Struggle_Through
    Quote Originally Posted by Struggle_Through View Post
    I have the problem: (1+i)/\5 (One plus i, raised to the 5th). I know r= square root of 2 and arctan= pi/4. Then we have square root of 2 raised to the 5th. This is where I run into a problem. How exactly does this work out. For example i know how to get the cube 8 by 8/\(1/3). I just don't know how to raise an nth power to something that is squared.
    Thanks.
    You don't actually say what it is you have to do with $\displaystyle (1+i)^5$, but I hope you may find this helpful:

    $\displaystyle (1+i)^2 = 1 +2i + i^2 = 2i$

    $\displaystyle \Rightarrow (1+i)^4 = (2i)^2 = -4$

    $\displaystyle \Rightarrow (1+i)^5 = -4(1+i)$

    Grandad
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  4. #4
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    If you have to use Moivre's Formula:

    $\displaystyle
    (1+i)^5=\sqrt{2^5} ( \frac {\sqrt{2}} {2}+i \frac {\sqrt{2}} {2} )^5=
    $

    $\displaystyle
    =\sqrt{2^4\cdot2}(cos\frac{\pi}{4}+isin\frac{\pi}{ 4})^5=4\sqrt{2}(cos\frac{5\pi}{4}+isin\frac{5\pi}{ 4})
    $

    $\displaystyle
    =4\sqrt{2}(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=-4-4i=-4(1+i)
    $

    Moivre's Formula is:
    $\displaystyle
    (cos(x)+isin(x))^n=cos(nx)+isin(nx)
    $
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