1. ## complex numbers

if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729

i am not really sure how to do this

2. ## Re: complex numbers

$w$ is a cube root of unity so we know it's magnitude is $1$

Thus we can write $w = e^{i \theta}$

let $f(w) = 2 + 5w + 2w^2$

a little work shows that

$f(e^{i\theta}) = (5+4\cos(\theta))e^{i \theta}$

and thus

$\left(f(w)\right)^6 = (5+4\cos(\theta))^6 e^{6i \theta}= 729$

and as $(5+4\cos(\theta))^6 \in \mathbb{R}$ it must be that

$(5+4\cos(\theta))^6 = 729$

$5+4\cos(\theta) = 3$

$\cos(\theta) = -\dfrac 1 2$

$\theta = \dfrac{2\pi}{3},~\dfrac{4\pi}{3}$

$w = e^{i2\pi/3},~e^{i 4\pi/3}$

and these are exactly the two non-real cube roots of unity.

Actually you'll need to reverse these steps as you are given $w$ is a cube root of unity and must show that $f(w)^6 = 729$

It pretty much just amounts to plugging in the two non-real cube roots of unity and grinding through the algebra using the one shortcut I noted.

3. ## Re: complex numbers

I am confused by tour terminology. could you do it in terms of cos(a) and isin(a)

4. ## Re: complex numbers

Originally Posted by edwardkiely
if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729
Originally Posted by edwardkiely
I am confused by tour terminology. could you do it in terms of cos(a) and isin(a)
There are two non-real cube roots of unity: $\omega=-\frac{1}{2}+\bf{i}\frac{\sqrt3}{2}$ is one of them.
Now $\omega^2=-\frac{1}{2}-\bf{i}\frac{\sqrt3}{2}$ Can you show that?

You calculate $2+5\omega +2\omega^2$. If that is $3$ you are done because $3^6=729$.

5. ## Re: complex numbers

$\omega ^3-1=(\omega -1)\left(\omega ^2+\omega +1\right)$

and since $\omega \neq 1$ we get $\omega ^2+\omega +1=0$

so that $2\omega ^2+2\omega +2=0$ and $2+5\omega +2\omega ^2=3\omega$

$\left(2+5\omega +2\omega ^2\right)^6=3^6\omega ^6=729$

6. ## Re: complex numbers

how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.

7. ## Re: complex numbers

Originally Posted by edwardkiely
if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729
i am not really sure how to do this
Originally Posted by edwardkiely
how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.
That was given.

There are exactly three cube roots of unity, two of which are non-real:
$\large{\exp \left( {\dfrac{{2\pi\bf{i} }}{{\sqrt 3 }}} \right) = \cos \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) \pm\bf{i} \sin \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) = \dfrac{1}{2} \pm \bf{i}\dfrac{{\sqrt 3 }}{2}}$

Those are the only two non-real.

8. ## Re: complex numbers

Originally Posted by edwardkiely
how do you know w=-.5 +i(√3/2) there are many complex numbers with a magnitude of 1 . and i can do the rest.
Idea's answer is far more elegant and most likely what they wanted you to come up with.