De Moivre's Theorem

**masoug** · December 18th 2010, 12:18 PM

Hi,
I was reading a textbook when I came upon their discussion on De Moivre's Theorem. The book talked about the difference between $(-8)^{\frac{1}{3}}$ and $(-8)^{\frac{2}{6}}$ , and they said something about how De Moivre's Theorem let us conclude that there are 3 cube roots of a number and 6 sixth roots. Can anyone decipher what this means?

Thanks!

-Masoug

**dwsmith** · December 18th 2010, 12:33 PM

$\displaystyle w^n=z$

It is obvious that $w=-2$ is a solution.

$r=\sqrt{(-8)^2}=8$

The angle of -8 is $\pi$ since it lies on the negative x axis.

$\displaystyle w_k=r^{1/n}\left(cos\left(\frac{\theta +2\pi k}{n}\right)+\mathbf{i}sin\left(\frac{\theta +2\pi k}{n}\right)\right)$

Where k goes from 0 to 1 less than the n so k = 0, 1, 2 and n = 3.

Principle root: $\displaystyle w_0=8^{1/3}\left(cos\left(\frac{\pi+2\pi*0}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*0}{3}\right)\right)=\cdots$

$\displaystyle w_1=8^{1/3}\left(cos\left(\frac{\pi+2\pi*1}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*1}{3}\right)\right)=\cdots$

$\displaystyle w_2=8^{1/3}\left(cos\left(\frac{\pi+2\pi*2}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*2}{3}\right)\right)=\cdots$

**HallsofIvy** · December 18th 2010, 12:43 PM

-8 has one real cube root (-2 as dwsmith says) and two complex roots. -8 has six complex (non-real) roots but if you square those, you will find that there are three sets of pairs that have the same square-giving you again the 3 cube roots of -8. For example, two of the sixth roots of -8 are $i\sqrt{2}$ and $-i\sqrt{2}$ both of which have square equal to -2.

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