# Complex Numbers and Cubic Polynomial Roots

• Sep 17th 2010, 12:12 PM
Complex Numbers and Cubic Polynomial Roots
Hello everyone,

Let w be one of the three complex numbers having the property that
w^3 = -4 + 4*sqrt(3)i.

What is the |w|?

Also if a = 2Re(w) how can one show that a is a root of the polynomial:
p(z) = z^3 - 12z + 8.

• Sep 17th 2010, 01:15 PM
Plato
First note that $\displaystyle \left| {z^3 } \right| = \sqrt {16 + 48} = 8$. So $\displaystyle |z|=?$

I do not follow the second part of the question.
Is it related to first part?
• Sep 17th 2010, 01:45 PM
Quote:

Hello everyone,

Let w be one of the three complex numbers having the property that
w^3 = -4 + 4*sqrt(3)i.

What is the |w|?

Also if a = 2Re(w) how can one show that a is a root of the polynomial:
p(z) = z^3 - 12z + 8.

Can you double-check your polynomial ?

Should it be $\displaystyle p(z)=z^3-12z+8\sqrt{3}$ ?

EDIT: no, it isn't, I used a 60 degree angle instead of a 30 degree one!
• Sep 17th 2010, 02:07 PM
Thanks!

Part b is related to the first part, so

a= 2*Re(w) where w is the be one of the three complex numbers having the property that w^3 = -4 + 4*sqrt(3)i.
Then I must show that a is a root of the polynomial z^3 - 12*z + 8.

And the polynomial is p(z) = z^3 - 12z + 8

• Sep 17th 2010, 02:21 PM
mr fantastic
Quote:

Thanks!

Part b is related to the first part, so

a= 2*Re(w) where w is the be one of the three complex numbers having the property that w^3 = -4 + 4*sqrt(3)i.
Then I must show that a is a root of the polynomial z^3 - 12*z + 8.

And the polynomial is p(z) = z^3 - 12z + 8

1. Using polar representation, $\displaystyle a = 4 \cos \left( \frac{2 \pi}{9} \right)$.
2. Using standard trig identities, $\displaystyle \cos^3 (\theta) = \frac{3}{4} \cos (\theta) + \frac{1}{4} \cos (3 \theta)$.