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I want to factorize the equation Z^n=1 over complex numbers
I reached (x-1)(X^4+X^3+X^2+x+1) andd then got stucked
First of all, I don't know where x and X have come from, since your equation is in terms of z...Originally Posted by Hyunqul
Anyway, let, then
Therefore
The solutions are all evenly spaced about a circle, so all have the same modulus and are separated by an angle of, so the solutions are
Each solutionimplies a factor of
, so that means the factorised form (finally) is
![\displaystyle \begin{align*} z^n &= 1 \\ z^n - 1 &= 0 \\ \left\{ z - \left[\cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)}\right]\right\} \cdot \cdot \cdot \left\{ z - \left[\cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)}\right] \right\} &= 0 \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} z^n &= 1 \\ z^n - 1 &= 0 \\ \left\{ z - \left[\cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)}\right]\right\} \cdot \cdot \cdot \left\{ z - \left[\cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)}\right] \right\} &= 0 \end{align*})
The nth roots of unity, z such thatare given by
such that
so we must have r= 1,
which is, of course, equivalent to
. That is, geometrically, the n nth roots of unity lie on the unit circle in the complex plane, equally space around the circle. The five fifth roots of unity, in particular, form a "pentagon" on the unit circle.
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