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what did i do wrong this time? FP2 'de moivre theorem' watch

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    Use de Moivre's theorem to evaluate (-1+i)^8

    my working:

    r = sqrt(2)
    arg(-1+i) = -pi/4

    therefore (-1+i)^8 = [sqrt(2)[cos(pi/4) - isin(pi/4)]]^8

    using this dudes theorem: z^n = r^n (cos(nx) + isin(nx)

    16(cos2-isin2) is what i got. expanded to get 16cos2 - 16isin2.

    the answer is just 16 apparently. anyone know where i went wrong?

    thanks in advance
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    You forgot to include the pi value in the cosine and sine function
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    (Original post by Reml)
    You forgot to include the pi value in the cosine and sine function
    oh...

    thx
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    There's an easier way to do this.
    Any complex number can be written as Re^(i(theta)) where R is magnitude (in this case sqrt(2)) and theta is the argument (angle here is -pi/4) so that to the power 8 is 16e^(-2pi) and since 2 pi is one complete revolution its the same as an argument of 0 and e^0 = 1
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    (Original post by disco1000)
    There's an easier way to do this.
    Any complex number can be written as Re^(i(theta)) where R is magnitude (in this case sqrt(2)) and theta is the argument (angle here is -pi/4) so that to the power 8 is 16e^(-2pi) and since 2 pi is one complete revolution its the same as an argument of 0 and e^0 = 1
    oh yeah, nice..

    speaking of Re^(ix), can you help me out with this question?

    (a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

    4sqrt(2) [rcos(pi/4) - isin(pi/4)]

    (b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

    mind giving me a tutorial on how to do this? would be much appreciated
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    (Original post by racist mathematici)
    oh yeah, nice..

    speaking of Re^(ix), can you help me out with this question?

    (a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

    4sqrt(2) [rcos(pi/4) - isin(pi/4)]

    (b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

    mind giving me a tutorial on how to do this? would be much appreciated
    It's basically finding the nth root of a complex number, using de Moivre's theorem.
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    (Original post by racist mathematici)
    oh yeah, nice..

    speaking of Re^(ix), can you help me out with this question?

    (a) Express 4-4i in the form r(cosx + isinx) where r>0, -pi < x < pi, and r and x are exact values.

    4sqrt(2) [rcos(pi/4) - isin(pi/4)]

    (b) Hence, or otherwise solve the equation z^5 = 4 -4i leaving your answers in the form z = Re^ik(pi), where R is the modulus of z and k is a rational number such that -1 =< k =< 1.

    mind giving me a tutorial on how to do this? would be much appreciated
    Practically the same question, this time it becomes 4sqrt(2)*e^(-pi/4) which is 2^(5/2)*e^(-pi/4) now you take the 5th root which is just sqrt 2 * e (-pi/20) where k is - 1/20 + 2n(pi) for any value of n

    hope that helps, its kind of hard to explain it all just written down, check the proof for why e^i.theta = cos theta + isin theta, it will help.
 
 
 
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Updated: April 10, 2011
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