# How do I find the cube roots of this?

• Dec 3rd 2012, 06:14 PM
beastcoast515
How do I find the cube roots of this?
I had this question on a test I recently took and I am still totally unaware of how to solve it. If anyone could help I would greatly appreciate it. Thanks!

Find the cube roots of 3sqrt(3) - 3i
• Dec 3rd 2012, 06:32 PM
Prove It
Re: How do I find the cube roots of this?
\displaystyle \begin{align*} 3\sqrt{3} - 3i &= 3\left( \sqrt{3} - i \right) \\ &= 3 \left( 2 \, e^{ -\frac{\pi}{6} \, i } \right) \\ &= 6 \, e^{-\frac{\pi}{6} \, i } \end{align*}

So the cube roots can be found by taking this to the power of \displaystyle \begin{align*} \frac{1}{3} \end{align*} and remembering that there will be three roots evenly spaced around the circle, so each separated by an angle of \displaystyle \begin{align*} \frac{2\pi}{3} \end{align*}.

The first cube root is

\displaystyle \begin{align*} \left( 6\, e^{-\frac{\pi}{6}\, i }\right) ^{\frac{1}{3}} &= \sqrt[3]{6} \, e^{ -\frac{\pi}{18} \, i } \end{align*}

and the other two cube roots are \displaystyle \begin{align*} \sqrt[3]{6}\, e^{\left( -\frac{\pi}{18} - \frac{2\pi}{3} \right) i} = \sqrt[3]{6}\, e^{ -\frac{13\pi}{18} \, i } \end{align*} and \displaystyle \begin{align*} \sqrt[3]{6}\, e^{\left( -\frac{\pi}{18} + \frac{2\pi}{3} \right) i} = \sqrt[3]{6}\, e^{\frac{11\pi}{18} \, i } \end{align*}
• Dec 3rd 2012, 06:47 PM
beastcoast515
Re: How do I find the cube roots of this?
I can't believe I forgot about e! Thank you very much, Prove it.
• Dec 3rd 2012, 07:34 PM
Prove It
Re: How do I find the cube roots of this?
Just something to note, since you were given the original complex number in Cartesians, you should really give your answers in Cartesians as well.