# Cube roots of -8

• Aug 31st 2011, 06:11 PM
tarheelborn
Cube roots of -8
I am totally stuck on this problem. I need to find the cube roots of -8 in polar coordinates. I know that -8 lies on the real axis, so the angle of reference is $\displaystyle \pi$. I also know that $\displaystyle cos \pi=-1$. I know that the formula I need to use is cube root of $\displaystyle |z|= cis \frac{\theta}{3}$ and I know that $\displaystyle cos \frac{\theta}{3}=Sqrt(\frac{1+cos\theta}{3})$, $\displaystyle sin\theta=Sqrt(\frac{1-cos\theta}{3})$ and the cube root $\displaystyle =2(cos\frac{\theta}{3}+i sin\frac{\theta}3}$. I must be tired and doing something wrong with the signs or something. I keep getting a zero for the first term and I am pretty sure the answer is $\displaystyle 1+i \sqrt{3}$. Please help! Thanks!
• Aug 31st 2011, 06:25 PM
TKHunny
Re: Complex Analysis - Cube roots of -8
First of all, you'll need a 2, somewhere.

|z| = cis(x/3) really doesn't make quite enough sense. It is as if you ALMOST know where you are going.

Start with $\displaystyle z^{3} = -8 = 8\cdot cis(\pi)$

Just apply the theorem from DeMoivre.

$\displaystyle z = 8^{\frac{1}{3}}\cdot cis\left(\frac{\pi +2k\pi}{3}\right)$ for k = 0, 1, 2

k = 0 leads to $\displaystyle 2\cdot cis(\pi/3)$

k = 1 leads to $\displaystyle 2\cdot cis(\pi)$

k = 2 leads to $\displaystyle 2\cdot cis(5\pi/3)$

• Aug 31st 2011, 06:52 PM
tarheelborn
Re: Complex Analysis - Cube roots of -8
I think you're right; I ALMOST know where I am going! OK... I began with $\displaystyle 2 cis \pi=0-i\frac {\sqrt{3}}{2}$ which is where I thought I was off track because of the zero in the real part. However, I should have known that was ok because -8 has nothing in the imaginary part... At any rate, moving on, $\displaystyle 2 cis \frac{\pi}{3}=\frac{1}{2} + i \frac{\sqrt{3}}{2}$ and $\displaystyle 2 cis \frac{5\pi}{3} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$. But you say one for free and I should show you the next one. Aren't these three roots all of the roots of -8? Thank you so much for your help and your kind attitude.
• Aug 31st 2011, 07:06 PM
TKHunny
Re: Complex Analysis - Cube roots of -8
That was one problem for free, not one cube root of -8.

$\displaystyle 2\cdot cis(\pi) = -2 + i\cdot 0$

How are you managing square roots and imaginary values?
• Sep 1st 2011, 04:02 AM
HallsofIvy
Re: Complex Analysis - Cube roots of -8
I think you need to know that $\displaystyle cos(\pi/3)= \frac{1}{2}$ and $\displaystyle sin(\pi/3}= \frac{\sqrt{3}}{2}$.
• Sep 1st 2011, 05:17 AM
tarheelborn
Re: Complex Analysis - Cube roots of -8
Thank you; I thought that was the end of it! I managed the square roots with imaginary values with a graph of a right triangle and de Moivre's. This one threw me because it was just a line and there was no imaginary value, but you helped me see that it was basically the same as the other; I just had trouble with the angles. Thanks!