Question

What can be concluded about M, the number of non-real roots of the equation `x^11=1`?

Answer 1

Real root: 1 only. The other 10 complex roots are
`cis((2k)/11pi), k =1, 2, 3, ..., 9, 10`.

Explanation:

The equation is `x^11-1=`. The number of changes in signs of the

coefficients is 1. So, the number of positive real roots cannot e

exceed 1.

Changing x to -x, the equation becomes `-x^11-1=0` and the

number of sign changes is now 0. So, there is no negative root.

Also, complex roots occur in conjugate pairs, and so, the number of

complex roots is even.

Thus, there is only one real root and this is 1, observing that the

sum of the coefficients is 0.

Overall, the 11 11th roots of unity are

`cis ((2k/11)pi), k = 0, 1, 2, 3, ...10,` .

and, here, k = 0, gives a root as `cis 0 = cos 0 + i sin 0 =1`