**1 Answer**

written 3.5 years ago by |

We have $x^5=1=\cos 0+i \sin 0$

$\therefore x^5 = \cos (2k\pi) + i\sin (2k\pi)$

$\therefore x = [\cos (2k\pi) + i\sin (2k\pi)]^{1/5} = \cos \dfrac {2k\pi}{5}+i\sin \dfrac {2k\pi}{5}$

Putting k=0,1,2,3,4 we get the five roots,

$x_0 = \cos 0 + i\sin 0=1$

$x_1 = \cos \dfrac {2\pi}{5}+ i\sin \dfrac {2\pi}{5}$

$x_2=\cos \dfrac {4\pi}{5}+i\sin \dfrac {4\pi}{5}$

$x_3 = \cos \dfrac {6\pi}{5}+isin\dfrac {6\pi}{5}$

$x_4=\cos \dfrac {8\pi}{5}+ i\sin \dfrac {8\pi}{5}$

Putting $x_1 =a$ we see that $x_2 = \alpha^2 , \ x_3 = \alpha^3, \ x^4=\alpha^4$

The roots are $1, \alpha, \alpha^2, \alpha^3, \alpha^4$ and hence

$x^5-1 = (x-1) (x-\alpha) (x-\alpha^2)(x-\alpha^3)(x-\alpha^4)$

$(x^5-1)/(x-1)= (x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)$

Hence, $(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4) = 1+x+x^2+x^3+x^5$

Putting, x=1 we get,

$(1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4)=5$