# find the primitive root

• Jul 21st 2006, 02:10 PM
kezman
find the primitive root
Let w, a 20th units primitive root.

Prove that
w + w^3 + w^7 + w^9

is a pure imaginary complex.

Im not sure how to "attack" this problem.
• Jul 22nd 2006, 07:38 PM
ThePerfectHacker
Quote:

Originally Posted by kezman
Let w, a 20th units primitive root.

Prove that
w + w^3 + w^7 + w^9

is a pure imaginary complex.

Im not sure how to "attack" this problem.

That is not true, 1 is a 20th root of unity yet,
$\displaystyle 1+1^3+1^7+1^9$ is not a pure imaginary.
• Jul 23rd 2006, 11:12 AM
BubbleBrain_103
I think by primitive root, he means a generator for the multiplicative group of 20th roots of unity, i.e. a complex number of the form:

exp(I*k*Pi/10) where 0 < k < 20 and k is relatively prime to 20

expand your expression out using Euler's formula, and you get that the real part looks like

cos(k Pi/10) + cos(k 9 Pi/10) + cos(k 3 Pi/10) + cos(k 7 Pi/10)

And you should be able to show that, so long as k is relatively prime to 20,

cos(k Pi/10) + cos(k 9Pi/10) = 0 and cos(k 3Pi/10) + cos(k 7Pi/10) = 0

showing that the real part of your expression is zero.

I'm almost sure there's a slicker way to do it, explicitly using the relative primeness of k and 20, but my trig is a bit rusty. So at the least you could just boil it down to cases. That is, prove it for the cases
k = 1, 3, 7, 9, 11, 13, 17, 19, which are all the possible primitive roots. k = 1 for instance, you would write

cos(Pi/10) + cos(9Pi/10) = cos(Pi - 9Pi/10) + cos(9Pi/10) = -cos(9Pi/10) + cos(9Pi/10) = 0

using the identity cos(Pi-x) = -cos(x).

hope that helps. Try to find a slicker way before you resort to the cases method. And by the way, I verified the identities for all these k's on maple, so the theorem is true.
• Jul 23rd 2006, 12:37 PM
kezman
yes I want to avoid the cases method. Its to complicated and long.
• Jul 23rd 2006, 01:34 PM
BubbleBrain_103
Alright then...

If a and b are two real numbers, then the following condition is sufficient to guarantee that cos(a) = -cos(b):

a = Pi - b + n2Pi

where n2Pi is any integer multiple of 2Pi. Or, rewritten,

a + b = Pi + n2Pi

So in other words, if a + b is an ODD multiple of 2Pi, then cos(a) = -cos(b).

So in our case, let a = k*Pi/10, b = k*9*Pi/10, and we have

a + b = k(Pi/10+9*Pi/10) = k*10*Pi/10 = k*Pi

Note that k must be odd since it is assumed that k is relatively prime to 20. This shows then that

cos(k*Pi/10) + cos(k*9*Pi/10) = 0

whenever k is odd (but more importantly for us, when k is relatively prime to 20). Similarly you can show that

cos(k*3*Pi/10) + cos(k*7*Pi/10) = 0

which proves the theorem.
• Jul 30th 2006, 04:09 AM
rgep
Consider taking the complex conjugate of the expression. For any root of unity $\displaystyle \zeta$, the conjugate $\displaystyle \bar\zeta$ satsifies $\displaystyle \zeta\bar\zeta = 1$, so $\displaystyle \bar{w^i} = w^{20-i}$. Furthermore, $\displaystyle w^{10} = -1$. Conjugating $\displaystyle w + w^3 + w^7 + w^9$ gives $\displaystyle w^{19} + w^{17} + w^{13} + w^{11} = -w - w^3 - w^7 - w^9$ which is the negative of the original expression. But a quantity which is negated by complex conjugation is a pure imaginary.
• Jul 30th 2006, 04:56 PM
kezman
yes thats a good idea thansk.