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Likurg_2 [28]
3 years ago
14

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations.

Mathematics
1 answer:
Nostrana [21]3 years ago
4 0
x^3-27i=0
x^3=27i
x^3=27e^{i\pi/2}
x=\left(27e^{i\pi/2}\right)^{1/3}
x=3e^{i(\pi/2+2\pi k)/3}
x=3e^{i\pi(4k+1)/6}

where k\in\{0,1,2\}. This means you have

x=3e^{i\pi/6}=\dfrac32(\sqrt3+i)
x=3e^{i5\pi/6}=\dfrac32(-\sqrt3+i)
x=3e^{i9\pi/6}=-3i

as the solutions to the original equation.
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