Question

x^3- 27i = 0. Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations.

Answer 1

\[x^3=27i~~\iff~~x=\sqrt[3]{27i}=3\sqrt[3]{i}\]
Do you know how to find all the n-th roots of a complex number?

Answer 2

No

Answer 3

Okay, well first you have to convert the given complex number (\(i\) in this case) to polar:
\[\large z=i=0+1i~~\Rightarrow~~z=e^{i\pi/2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\]
Then take the cube root:
\[z^{1/3}=\left(e^{i\pi/2}\right)^{1/3}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\]
That's one root.

The others will have the form
\[z^{1/3}=\cos\left(\frac{\pi}{6}+\frac{2\pi}{3}\right)+i\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}\right)\]
and
\[z^{1/3}=\cos\left(\frac{\pi}{6}+\frac{4\pi}{3}\right)+i\sin\left(\frac{\pi}{6}+\frac{4\pi}{3}\right)\]

Answer 4

You can find more information here: http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx