Cube roots of a purely imaginary number
Cube roots of a purely imaginary number
IB Year 6 | Grade 12
Solve the equation \(z^3=-27i\), giving your answers in polar form.
Solution:
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Answer:\(3\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right),\ 3\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right),\ 3\left(\cos\left(-\frac{5\pi}{6}\right)+i\sin\left(-\frac{5\pi}{6}\right)\right)\)
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