Hi Melody.
You need to read up on De Moivre's theorem.
The basic theorem says that one value of \(\displaystyle (\cos \theta+i\sin\theta)^{n} \text{ is }\cos n\theta+i\sin n\theta. \)
That's easy to prove when n is an integer, just use induction.
The 'one value of ', (because the angle is multivalued), becomes important when n is a fraction.
The basic result above could be expressed as
\(\displaystyle \{\cos(\theta+2k\pi)+i\sin(\theta + 2k\pi)\}^{n}=\cos(n\theta+2kn\pi)+i\sin(n\theta+2kn\pi)\).
k is any integer, and if n is also an integer, the 2kpi bit isn't needed, but if n is a fraction, (the theorem still holds if this the case), you now have fractional parts of 2pi, and therefore multiple values for the roots.
For square roots the numbers are 180 degees apart so there are just two possibles, cube roots are 120 degrees apart so there are three possibles, and so on.
Tiggsy
