solution 1
solution 2
solution 3 Found 2 solutions by stanbon, lynnlo:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! I can't really tell what your problem is.
The z^3 = 3-3i seems to indicate you want to solve for "z"
which would be the three cube roots of 3-3i.
If that is true:
z^3=3-3i
module argument exponential representation
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r = sqrt(3^2+3^2) = 3sqrt(2)
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theta = tan^-1(-3/3) = tan^-1(-1) = (3/4)pi
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3 -3i = 3sqrt(2)(cis[(3/4)pi]
z = (3sqrt(2))^(1/3)*(cis(3/4)pi+(2npi))
Let n = 0 to get z = (3sqrt(2))^(1/3)cis((1/4)pi)
If n = 1 get z = (3sqrt(2))^(1/3)cis((1/4)pi+(2/3)pi)
If n = 2 get z = (3sqrt(2))^(1/3)cis((1/4)pi + (4/3)pi)
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Cheers,
Stan H.
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