roots of unity

Find the nth roots of unity, n > 0 (the solutions to the equation z^n = 1). Show that their sum is zero. Hence show

cos(2pi/n) + cos(4pi/n) + cos(6pi/n) + ....... + cos(2(n-1)pi/n) = -1

sin(2pi/n) + sin(4pi/n) + sin(6pi/n) + ...... + sin(2(n-1)pi/n) = 0

Reply 1

Jonny W

nth roots of unity: e^(2i pi m/n) for m = 0, 1, 2, ..., (n - 1).

(Sum of roots of z^n - 1 = 0)
= -(Coefficient of z^(n - 1) in z^n - 1)
= 0

So e^(0) + e^(2i pi/n) + e^(4i pi/n) + ... + + e^(2(n - 1)i pi/n) = 0.

Take the real and imaginary parts.

Reply 2

JamesF

0 = z^n - 1

Can be factorised
(z-1)(z-a)(z-b)......

Equating coefficients of the z^(n-1) term in the first polynomial, we see that
0 = 1 + a + b.....

Reply 3

JamesF

Awww, too slow :frown:

Reply 4

madhapper

cos(2pi/n) + cos(4pi/n) + cos(6pi/n) + ....... + cos(2(n-1)pi/n) = -1

sin(2pi/n) + sin(4pi/n) + sin(6pi/n) + ...... + sin(2(n-1)pi/n) = 0

how does one show that then?

Reply 5

popat

Jonny W

i dont understand this could you explain again or is there another way

Reply 6

evariste_3086

popat

i dont understand this could you explain again or is there another way

e^(0) + e^(2i pi/n) + e^(4i pi/n) + ... + + e^(2(n - 1)i pi/n) = 0.(1)
welll e^0=1
e^(2pi/n)=cos(2pi/n)+isin(2pi/n)
e^(4pi/n)=cos(4pi/n)+isin(4pi/n)
.
.
.
e^(2(n-1)pi/n)=cos(2(n-1)pi/n)+isin(2(n-1)pi/n)

so (1) tells us that
1+cos(2pi/n)+isin(2pi/n)+.....+cos(2(n-1)pi/n)+isin(2(n-1)pi/n)=0
for this to be true need the real part of both sides equal that is
1+cos(2pi/n)+cos(4pi/n)......+cos(2(n-1)pi/n)=0
and the imaginary parts of both sides to be equal
sin(2pi/n)+sin(4pi/n)+.......sin(2(n-1)pi/n)=0