1. ## Primitive roots

This problem appeared on my latest exam and I couldnt solve it.

Knowing that w is a 36-primitive unit root solve:

$w^{6}+w^{12}+3$

2. Originally Posted by kezman
This problem appeared on my latest exam and I couldnt solve it.

Knowing that w is a 36-primitive unit root solve:

$w^{6}+w^{12}+3$
First what is a 36 primitive root unit?
If you mean a primitive root of 36, that is wrong. 36 does not have a primitive root.

And solve for what?

3. Sorry maybe I wasnt clear.

36-unit roots (root of unity), are all the solutions of $z^{36} = 1$

Knowing that w is a 36-Primitive root of unity I have to solve

$w^{12} + w^6 + 3$

4. Originally Posted by kezman
$z^{36} = 1$
Knowing that w is a 36-Primitive root of unity I have to solve
$w^{12} + w^6 + 3$
But there is nothing to solve
$w^{12} + w^6 + 3 = ?.$
Is that what you mean?

5. Yes, sorry. The exam says Calculate $w^6 + w^{12} + 3$

6. In polar form we have $z = \cos \left( {\frac{\pi }{{18}}} \right) + i\sin \left( {\frac{\pi }{{18}}} \right).$
Now, $z^{12} = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right)\quad \& \quad z^6 = \cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right).$

Note that $\cos \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {\frac{\pi }{3}} \right) = 0\quad \& \quad \sin \left( {\frac{{2\pi }}{3}} \right) + \sin \left( {\frac{\pi }{3}} \right) = \sqrt 3.$
Now you can finish.