Complex cube roots of unity (1 Viewer)

[scardizzle]

Prove that w, a complex cube root of unity, is a repeated root of P(x) = 3x^5 + 2x^4 + x^3 - 6x^2 - 5x - 4.
EDIT(left out the other half of the q):Hence find the zeroes to the equation over the complex number field

 

[life92]

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

Therefore, P'(x) = 15x^4 + 8x^3 + 3x^2 - 12x - 5
P'(w) = 15w^4 + 8w^3 + 3w^2 - 12w - 5
= 15w + 8 + 3w^2 - 12w - 5 since w^4 = w^3 * w, where w^3 = 1
= 3 + 3w^2 +3w
= 3 (1+w+w^2)
= 0 , since 1+w+w^2 =0
Therefore, w is a repeated root of multiplicity 2.

This answers the question, but if you want to find out if w is root of multiplicity >2, then differentiate again and sub it in.

 

[inedible]

If w is a cube root of unity then, by definition, w^3=1 and w^2+w+1=0
If w is a root of P(x) then P(w)=0
L.H.S
=3(w^2)(w^3) + 2w(w^3) + w^3 - 6w^2 - 5w - 4
=3w^2 + 2w + 1 - 6w^2 - 5w - 4 (since w^3=1)
=-3w^2 - 3w - 3
=-3(w^2 + w +1)
=-3(o)
=0
=R.H.S

Q.E.D

 

[life92]

If you differentiate again, then you will find that w is not a root of multiplicity 3.

Also, w cannot be a root of multiplicity >2 as complex roots are always in conjugate pairs.

So therefore one of the factors is (x-w)^2
Since w is a double root, then the conjugate of w is also a double root.
I will use w_ to denote the conjugate of w

So lets let alpha = w
and beta = w_

alpha + beta = -1
alpha * beta = 1

So we have the quadratic equation x^2 + x + 1 as a factor
And since w and w_ are double roots, then
(x^2+x+1)^2 is a factor.

Now by inspection, the last factor must be (3x-4)
Therefore, the roots are w, w_ and 4/3, where w = -1/2 + i root3 / 2, w_ = -1/2 - i root3 / 2

Hope that helps