Hello,
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...
Anyway, let $\displaystyle \displaystyle z = r\,e^{i\theta}$, then
$\displaystyle \displaystyle \begin{align*} z^n &= 1 \\ \left(r\,e^{i\theta}\right)^n &= e^{2\pi i} \\ r^ne^{i\theta n} &= e^{2 \pi i} \\ r^n = 1\textrm{ and }e^{i\theta n} &= e^{2 \pi i} \\ r = 1 \textrm{ and } i\theta n &= 2 \pi i \\ r = 1 \textrm{ and } \theta &= \frac{2 \pi}{n} \end{align*}$
Therefore $\displaystyle \displaystyle z = e^{\frac{2 \pi i}{n}} = \cos{\left(\frac{2 \pi}{n}\right)} + i\sin{\left(\frac{2 \pi }{n}\right)}$
The solutions are all evenly spaced about a circle, so all have the same modulus and are separated by an angle of $\displaystyle \displaystyle \frac{2\pi}{n}$, so the solutions are
$\displaystyle \displaystyle \begin{align*} z_1 &= \cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)} \\ z_2 &= \cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)} \\ z_3 &= \cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)} \\ &\vdots \\ z_n &= \cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)} \\ &= \cos{(2\pi)} + i\sin{(2\pi)} \\ &= 1 + 0i \\ &= 1 \end{align*}$
Each solution $\displaystyle \displaystyle z_i$ implies a factor of $\displaystyle \displaystyle \left(z - z_i\right)$, so that means the factorised form (finally) is
$\displaystyle \displaystyle \begin{align*} z^n &= 1 \\ z^n - 1 &= 0 \\ \left\{ z - \left[\cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)}\right]\right\} \cdot \cdot \cdot \left\{ z - \left[\cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)}\right] \right\} &= 0 \end{align*}$
The nth roots of unity, z such that $\displaystyle z^n= 1$ are given by $\displaystyle z= re^{i\theta}$ such that $\displaystyle z^n= r^ne^{ni\theta}= 1= (1)e^{i0}$ so we must have r= 1, $\displaystyle n\theta= 0$ which is, of course, equivalent to $\displaystyle n\theta= 2\pi$. 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.