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Thread: demoivre theorem

  1. #1
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    demoivre theorem

    write in rectangular form

    (square root 2-i) 4th power
    Last edited by SHORTY; Nov 15th 2006 at 11:17 PM. Reason: wrong symbol stated
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  2. #2
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    Quote Originally Posted by SHORTY View Post
    write in rectangular form

    (divied sign 2-i) 4th power
    Do you mean $\displaystyle \frac{1}{(2-i)^4}$?

    We need to rationalize this before we can work with it, so:
    $\displaystyle \frac{1}{(2-i)^4} = \frac{1}{(2-i)^4} \cdot \frac{(2+i)^4}{(2+i)^4}$

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

    = $\displaystyle \frac{(2+i)^4}{(4 - i^2)^4} = \frac{(2+i)^4}{(4+1)^5}$

    = $\displaystyle \frac{(2+i)^4}{5^4} = \frac{(2+i)^4}{625}$

    So we need to know what $\displaystyle (2+i)^4$ is.

    Note that $\displaystyle 2 + i = r cos \theta + ir sin \theta$.

    So
    $\displaystyle r cos \theta = 2$
    $\displaystyle r sin \theta = 1$

    Dividing the two equations gives:
    $\displaystyle tan \theta = \frac{1}{2}$
    So
    $\displaystyle \theta \approx 0.463648$ rad.

    Squaring the two equations and adding gives:
    $\displaystyle r^2 sin^2 \theta + r^2 cos^2 \theta = 1^2 + 2^2$

    $\displaystyle r^2(sin^2 \theta + cos^2 \theta ) = 5$

    $\displaystyle r^2 = 5$

    $\displaystyle r = \sqrt{5}$.

    Thus $\displaystyle 2 + i \approx \sqrt{5}cos(0.463648) + i \sqrt{5}sin(0.463648)$.

    By DeMoivre's theorem:
    $\displaystyle (cos\theta + i sin\theta)^n = cos(n\theta) + i sin(n\theta)$

    So
    $\displaystyle (2 + i)^4 \approx (\sqrt{5}cos(0.463648) + i \sqrt{5}sin(0.463648))^4$

    = $\displaystyle (\sqrt{5})^4 (cos(4 \cdot 0.463648) + i sin(4 \cdot 0.463648))$

    = $\displaystyle 25(cos(1.85459) + i sin(0.85459))$

    = $\displaystyle 25(-0.28 + 0.96i)$ <-- These are apparently exact, though I don't know how to prove it.

    So finally:
    $\displaystyle \frac{1}{(2-i)^4} = \frac{(2+i)^4}{625} = \frac{25(-0.28 + 0.96i)}{625}$

    = $\displaystyle \frac{-0.28 + 0.96i}{25}$

    -Dan
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