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  1. #1
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    demoivre theorem

    write in rectangular form

    (square root 2-i) 4th power
    Last edited by SHORTY; November 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 \frac{1}{(2-i)^4}?

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

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

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

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

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

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

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

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

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

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

    r^2 = 5

    r = \sqrt{5}.

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

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

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

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

    = 25(cos(1.85459) + i sin(0.85459))

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

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

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

    -Dan
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