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Math Help - how to (-64i)^1/3?

  1. #1
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    how to (-64i)^1/3?

    Find (-64i)^1/3.
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    Re: how to (-64i)^1/3?

    (-64i)^{\frac{1}{3}}=4(-i)^{\frac{1}{3}}=4(e^{\frac{\pi}{2}(4k+3)})^{\frac  {1}{3}}=4e^{\frac{\pi}{6}(4k+3)}

    where k\in\{0,1,2\}

    Now, use Euler's formula to find the 3 roots.
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    Re: how to (-64i)^1/3?

    (-64 i) ^1/3 = - 2√(3) - 2i???
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    Re: how to (-64i)^1/3?

    Quote Originally Posted by mous99 View Post
    (-64 i) ^1/3 = - 2√(3) - 2i???
    That is one of he cube roots of -64i. There are two others - do you know how to find them? You can use de Moivre's tehorem:

      ( R \cos \theta + i \sin \theta)^n = R^n (\cos (n \theta) + \sin (n \theta))

    From this you can derive the fact that for  z = R(\cos \theta + i \sin \theta), z^{1/n} = R^{1/n}[\cos(\frac {\theta + 2 \pi k}n) + i \sin (\frac {\theta + 2 k \pi} n)]

    Here you have R = 64, \theta = \frac {3 \pi}2, and  n =  3 . Set k = 0, 1, and 2 to find the three cube roots.
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    Re: how to (-64i)^1/3?

    (-64i)^1/3 = (-64)^1/3 *(i)^1/3
    = -4 (i)^1/3
    = -4(cos π/6 + i sin π/6)


    why the 0= 3π/ 6, not π/6???
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    MHF Contributor ebaines's Avatar
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    Re: how to (-64i)^1/3?

    Quote Originally Posted by mous99 View Post
    (-64i)^1/3 = (-64)^1/3 *(i)^1/3
    Not quite right:  (-64 i)^{1/3} = 64^{1/3}(- i)^{1/3} = 4 [ \cos(\frac {\frac { 3\pi} 2 + 2k \pi} 3) + i \sin(\frac {\frac { 3\pi} 2 + 2k \pi} 3 ) ]
     = -4i, \ 2\sqrt 3 - 2i,\ -2 \sqrt 3 - 2i for  k = 0, \ 1,\ 2 respectively.
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    Re: how to (-64i)^1/3?

    Quote Originally Posted by mous99 View Post
    Find (-64i)^1/3.
    I don't understand why all the replies assume that the question is about cube roots.

    I would have thought that it is straightforward complex expatiation:
    z^w=\exp(w\log(z)).

    In this case the principal value:
    \left( { - 64i} \right)^{\frac{1}{3}}  = \exp \left( {\frac{{\ln (64) - \frac{{i\pi }}{2}}}{3}} \right) = \exp \left( {\ln (4) - \frac{{i\pi }}{6}} \right)
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  8. #8
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    Re: how to (-64i)^1/3?

    Just possibly all replies assume that the question is about cube roots because the question specifically [b]asked[/b about a cube root.
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    Re: how to (-64i)^1/3?

    Quote Originally Posted by HallsofIvy View Post
    Just possibly all replies assume that the question is about cube roots because the question specifically asked about a cube root.
    Where in
    Quote Originally Posted by mous99 View Post
    Find (-64i)^1/3.
    is there anything about cube roots?

    Is that not just complex exponentiation?
    Last edited by Plato; December 19th 2012 at 12:13 PM.
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    Re: how to (-64i)^1/3?

    how to find θ= 3π/2 +2kπ,

    because my answer for θ = π/2.

    its no same with u.

    (-64i)^1/3 = 4(-i)^1/3

    z = re^iθ

    z^3 = (r^3)(e^3θ )


    so, r^3 = 64
    r = 4

    how to find 3θ = 3π/2 +2kπ???
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    Re: how to (-64i)^1/3?

    z = (-64i)^1/3
    z^3 = (-64i)1/3

    r = √[(0)^2 + (-64)^2 ] =64

    cos 3
    π/2 = 0
    sin 3
    π/2 = -1

    z^3 = (r^3) ( e^(3
    π/2 + 2kπ)

    theta =
    π / 2 + (2kπ/3)


    when k = 0, theta =
    π/2
    z0 = 4i


    when k =1, theta = ??? (how to find this theta)

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  12. #12
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    Re: how to (-64i)^1/3?

    Did you consider putting k= 1 in the formula you are given?
    \theta= \frac{\pi}{2}+ \frac{2k\pi}{3}
    If k= 0, \theta= \frac{\pi}{2}+ 0= \frac{\pi}{2} as you say.

    If k= 1, \theta= \frac{\pi}{2}+ \frac{2\pi}{3}= \frac{3\pi}{6}+ \frac{4\pi}{6}= \frac{7\pi}{6}

    If k= 2, \theta= \frac{\pi}{2}+ \frac{4\pi}{3}= \frac{3\pi}{6}+ \frac{8\pi}{6}= \frac{11\pi}{6}

    If k= 3, \theta= \frac{\pi}{2}+ \frac{6\pi}{3}= \frac{3\pi}{6}+ \frac{12\pi}{6}= \frac{15\pi}{6} which will give the same complex number as with k= 0 because sine and cosine (and so e^{ix}) have period 2\pi.
    Last edited by HallsofIvy; December 22nd 2012 at 02:11 PM.
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