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anonymous

  • 5 years ago

Find the three cube roots of the complex number 8i. Give your answers in the form x + iy

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  1. amistre64
    • 5 years ago
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    if I recall correctly, its the three spokes of a wheel; at 90degrees, 210degrees, and 330 degrees....

  2. anonymous
    • 5 years ago
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    how do I change it into x + iy form though?

  3. amistre64
    • 5 years ago
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    change it into polar corrdinate form first and use Demovers thrm onit...

  4. amistre64
    • 5 years ago
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    8i is: z = 8(cos(90) + i sin(90)) ; now take the 1/3 power of it...

  5. amistre64
    • 5 years ago
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    when we take this to a normal integer power of 3 we get: z^3 = 8^3(cos(90*3) + i sin(90*3)) right? its the same for rational exponents as well.

  6. amistre64
    • 5 years ago
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    z^(1/3) - 2(cos(90/3) + i sin(90/3))

  7. amistre64
    • 5 years ago
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    90/3 = 30; the cos(30)=sqrt(3)/2 and the sin(30) = 1/2 2(sqrt(3)/2 + i 1/2) = sqrt(3) + 1i should be one of the cube roots

  8. amistre64
    • 5 years ago
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    if I did it right ;)

  9. amistre64
    • 5 years ago
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    the others should be at 120 degree intervals from that one... at 150 and 270

  10. amistre64
    • 5 years ago
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    (sqrt(3) + i)^2 = 2 +2sqrt(3) i (sqrt(3)+i) (2 +2sqrt(3) i) 2sqrt(3) + 6i +2i -2sqrt(3) = 8i

  11. amistre64
    • 5 years ago
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    that was cool ..... I wonder if im right on the others... 2(cos(270) + i sin(270)) 2(0 + i (-1)) = -2i (-2i)^2 = -4 -4(2i) = -8i ..... Doh!!; well I got the first one right ;)

  12. amistre64
    • 5 years ago
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    2(cos(150) + i sin(150)) 2(-sqrt(3)/2 + i 1/2) -sqrt(3) + i; should be another one, (-sqrt(3) +i)^2 = 3 -1 -2i sqrt(3) (2 -2i sqrt(3))(-sqrt(3) + i) -2sqrt(3) +2i +6i +2sqrt(3) 8i ..... well that ones good ;)

  13. amistre64
    • 5 years ago
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    0 - (2i) is the last one; - (2i)^2 = -(4i^2) = -(-4)(2i) = -(-8i) = 8i

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