Let w, a 20th units primitive root.
w + w^3 + w^7 + w^9
is a pure imaginary complex.
Im not sure how to "attack" this problem.
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.
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.
Consider taking the complex conjugate of the expression. For any root of unity , the conjugate satsifies , so . Furthermore, . Conjugating gives which is the negative of the original expression. But a quantity which is negated by complex conjugation is a pure imaginary.