# x^7=1 find xx^7-1=0 (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0 (x-1)(x^3){x^3+x^2+x+1+x^(-1)+x^(-2)+x^(-3)}=0 let x+x^(-1)=t x^3+x^2+x+1+x^(-1)+x^(-2)+x^(-3)=0 t^3+t^2+t-4=0 I can't go further. Can't we...

x^7=1

find x

x^7-1=0

(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0

(x-1)(x^3){x^3+x^2+x+1+x^(-1)+x^(-2)+x^(-3)}=0

let x+x^(-1)=t

x^3+x^2+x+1+x^(-1)+x^(-2)+x^(-3)=0

t^3+t^2+t-4=0

I can't go further. Can't we find x except De-Moivre's formula?

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

This is much easier if you turn to a way of representing complex numbers: A complex number can be represented using real and imaginary parts, or using modulus and angle:

`x=a+bi = re^(i theta)`

Think of it as the difference between rectangular coordinates and polar coordinates. We write `x=r e^(i theta)`, where r is a nonnegative real number.

Also, `r e^(i theta) = r e^(i theta + 2 pi i)`

It's like rotating around the circle an extra time, and still landing in the same spot.

`x^7 = (r e^(i theta))^7 = r^7 e^(7 i theta)`

If we want `x^7 = 1`, then we need `r^7 = 1` and `7 theta = 2 pi` (or 0, or `4 pi`, or `6 pi`, etc).

So, we need `r=1`, and `theta = 0, (2 pi)/7, (4 pi)/7, (6 pi)/7, (8 pi)/7, (10 pi)/7, (12 pi)/7`

So, the possibilities for x are:

`x=e^(0 pi i / 7)=1`

`x=e^(2 pi i / 7)`

`x=e^(4 pi i / 7)`

`x=e^(6 pi i / 7)`

`x=e^(8 pi i / 7)`

`x=e^(10 pi i / 7)`

`x=e^(12 pi i / 7)`

`x=e^(14 pi i / 7)=e^(2 pi i )=1`

An easier was to write this is, say `delta =e^(2 pi i / 7)`

Then the possibilites for x are:

`1, delta, delta^2, delta^3, delta^4, delta^5, delta^6`

If you want to factor, as you were doing before, you would get:

`(x-1)(x-delta)(x-delta^2)(x-delta^3)(x-delta^4)(x-delta^5)(x-delta^6)=0`