
De Moivre's Theorem
Hi,
I was reading a textbook when I came upon their discussion on De Moivre's Theorem. The book talked about the difference between $\displaystyle (8)^{\frac{1}{3}}$ and $\displaystyle (8)^{\frac{2}{6}}$, 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! (Bow)
Masoug

$\displaystyle \displaystyle w^n=z$
It is obvious that $\displaystyle w=2$ is a solution.
$\displaystyle r=\sqrt{(8)^2}=8$
The angle of 8 is $\displaystyle \pi$ since it lies on the negative x axis.
$\displaystyle \displaystyle w_k=r^{1/n}\left(cos\left(\frac{\theta +2\pi k}{n}\right)+\mathbf{i}sin\left(\frac{\theta +2\pi k}{n}\right)\right)$
Where k goes from 0 to 1 less than the n so k = 0, 1, 2 and n = 3.
Principle root: $\displaystyle \displaystyle w_0=8^{1/3}\left(cos\left(\frac{\pi+2\pi*0}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*0}{3}\right)\right)=\cdots$
$\displaystyle \displaystyle w_1=8^{1/3}\left(cos\left(\frac{\pi+2\pi*1}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*1}{3}\right)\right)=\cdots$
$\displaystyle \displaystyle w_2=8^{1/3}\left(cos\left(\frac{\pi+2\pi*2}{3}\right)+\math bf{i}sin\left(\frac{\pi +2\pi*2}{3}\right)\right)=\cdots$

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