Complex Numbers with multiple powers

• Oct 15th 2010, 05:52 AM
jordanrs
Complex Numbers with multiple powers
Hello,

Im stuck on the following question

Consider the equation $\displaystyle z^n = 1$, for the four cases $\displaystyle N = 2, 3, 4 and 5$. What are the solutions to this equation in each case? Guess (and if possible prove) how many solutions there are for arbitrary N.

Im just unsure of what its asking, for $\displaystyle Z^2 = -1$ surely Z will just be +/- i but for the other powers i dont really know what the solutions would be

Any help most appreciated.
• Oct 15th 2010, 07:21 AM
Plato
This gives too much away but…
Let $\displaystyle \zeta = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi }}{n}} \right) + i\,\sin \left( {\frac{{2\pi }}{n}} \right)$.

Now, each power $\displaystyle \zeta^k$ for $\displaystyle k=0,1,\cdots,n-1$ is a different complex number.

But $\displaystyle (\zeta^k)^n=1$, so how many are there?
• Oct 16th 2010, 04:18 AM
HallsofIvy
Quote:

Originally Posted by Plato
This gives too much away but…
Let $\displaystyle \zeta = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi i}}{n}} \right) + i\,\sin \left( {\frac{{2\pi i}}{n}} \right)$.

Typo: there should be no i inside the trig functions.
$\displaystyle \zeta = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi}}{n}} \right) + i\,\sin \left( {\frac{{2\pi}}{n}} \right)$.

Quote:

Now, each power $\displaystyle \zeta^k$ for $\displaystyle k=0,1,\cdots,n-1$ is a different complex number.

But $\displaystyle (\zeta^k)^n=1$, so how many are there?