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Math Help - Find all 3 cube roots of i

  1. #1
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    Find all 3 cube roots of i

    1. z(1/3) = r^(*1/3) e^(i*pi/2/3)
    = cos(pi/6) + i*sin(pi/6)
    = squareroot(3/4) + i/2

    2. z(1/3) = r^(*1/3) e^(i*(pi/2 + 2*pi)/3)
    = cos(5*pi/6) + i*sin(5*pi/6)
    = -squareroot(3/4) + i/2

    3. z(1/3) = r^(*1/3) e^(i*(pi/2 + 4*pi)/3)
    = cos(3*pi/2) + i*sin(3*pi/2)
    = 0 - i

    Is that correct?

    A friend of mine says its supposed to be

    k = 0 ==> i^(1/3) = cos(π/6) + i sin(π/6) = (√3 + i)/2
    k = 1 ==> i^(1/3) = cos(5π/6) + i sin(5π/6) = (-√3 + i)/2
    k = 2 ==> i^(1/3) = cos(3π/2) + i sin(3π/2) = -i.

    How do i figure this out =/
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  2. #2
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    Quote Originally Posted by FailingTrig View Post
    1. z(1/3) = r^(*1/3) e^(i*pi/2/3)
    = cos(pi/6) + i*sin(pi/6)
    = squareroot(3/4) + i/2

    2. z(1/3) = r^(*1/3) e^(i*(pi/2 + 2*pi)/3)
    = cos(5*pi/6) + i*sin(5*pi/6)
    = -squareroot(3/4) + i/2

    3. z(1/3) = r^(*1/3) e^(i*(pi/2 + 4*pi)/3)
    = cos(3*pi/2) + i*sin(3*pi/2)
    = 0 - i

    Is that correct?

    A friend of mine says its supposed to be

    k = 0 ==> i^(1/3) = cos(π/6) + i sin(π/6) = (√3 + i)/2
    k = 1 ==> i^(1/3) = cos(5π/6) + i sin(5π/6) = (-√3 + i)/2
    k = 2 ==> i^(1/3) = cos(3π/2) + i sin(3π/2) = -i.

    How do i figure this out =/
    Both sets of answers are correct and equivalent. Surely you can see that all your friend has done is simplified what you got.
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