# Roots of Unity

• Aug 28th 2010, 09:48 PM
Lukybear
Roots of Unity
Suppose w^3 = 1 and k is a positive integer

i) Find two possible values of 1 + w^k + w^2k

I have no idea where to begin? Please give me some hints.
• Aug 28th 2010, 09:52 PM
undefined
Quote:

Originally Posted by Lukybear
Suppose w^3 = 1 and k is a positive integer

i) Find two possible values of 1 + w^k + w^2k

I have no idea where to begin? Please give me some hints.

Suppose k is a multiple of 3. Then 2k is also a multiple of 3. So 1 + w^k + w^(2k) = 3.

Suppose k gives remainder 1 when divided by 3. Then 2k gives remainder 2 when divided by 3.

Suppose k gives remainder 2 when divided by 3. Then 2k gives remainder 1 when divided by 3.

In either of the two latter cases you get 1 + w + w^2.
• Aug 28th 2010, 09:53 PM
Prove It
$w^3 = 1$

$w^3 - 1 = 0$

$w^3 - 1^3 = 0$

$(w - 1)(w^2 + w + 1) = 0$

So $w - 1 = 0$ or $w^2 + w + 1 = 0$

$w = 1$ or

$w^2 + w + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 = 0$

$\left(w + \frac{1}{2}\right)^2 + \frac{3}{4} = 0$

$\left(w + \frac{1}{2}\right)^2 = -\frac{3}{4}$

$w + \frac{1}{2} = \pm\frac{\sqrt{3}i}{2}$

$w = \frac{-1 \pm \sqrt{3}i}{2}$.

So the three solutions are

$w = 1, w = \frac{-1 - \sqrt{3}i}{2}, w = \frac{-1 + \sqrt{3}i}{2}$.
• Aug 28th 2010, 11:38 PM
Lukybear
Quote:

Originally Posted by undefined
Suppose k is a multiple of 3. Then 2k is also a multiple of 3. So 1 + w^k + w^(2k) = 3.

Suppose k gives remainder 1 when divided by 3. Then 2k gives remainder 2 when divided by 3.

Suppose k gives remainder 2 when divided by 3. Then 2k gives remainder 1 when divided by 3.

In either of the two latter cases you get 1 + w + w^2.

Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right? Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

Thanks really much.
• Aug 29th 2010, 12:22 AM
undefined
Quote:

Originally Posted by Lukybear
Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right? Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

Thanks really much.

Dealing with positive integers, it is not possible to have a remainder that is greater than (or equal to) the divisor! But 4 and 1 are congruent mod 3. I used modular arithmetic to get that 1 * 2 congruent to 2 mod 3, and 2 * 2 congruent to 1 mod 3.
• Aug 29th 2010, 03:32 PM
HallsofIvy
Quote:

Originally Posted by Lukybear
Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right?

k can't "give a remainder of 4" when divided by 3. The "remainder" when dividing by the integer n must be less than n.

Quote:

If you get a "remainder of 4" then you have divided incorrectly- increase your divisor by 1 and you will get a remainder of 1. Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

Thanks really much.
If "k, divided by 3, gives a remainder of 1" we can write k= 3n+ 1 for some integer n. Then 2k= 2(3n+1)= 6n+ 2= 3(2n)+ 2.

If "2, divided by 3, gives a remaider of 2" we can write k= 3m+ 2 for some integer m. Then 2k= 2(3m+2)= 6m+ 4= 6m+ 3+ 1= 3(6m+1)+ 1.