Prove that the sum of the n nth roots of 1 is always 0 and the product of the n nth roots of 1 is +/- 1.
I have to use De Moivre's Theorem, and explain and justify all stages of my proofs. Can you please help me get started.
De Moivre's theorem states that for every real number `theta` and positive integer n,
(cos `theta` + i sin `theta` )^n = cos n`theta` ` ` + i sin n` `
Sum of n nth roots
Using the De Moivre's theorem,
in polar form unity can be written as,
1= cos 2`pi`m + i sin 2`pi` ` `m for a positive integer m.
Now nth roots of unity will be the roots of an equation of the form, x^n = 1
Using De Moivre's theorem:
for a complex number X = cos 2`pi` m/n + i sin 2`pi` m/n , X^n = cos 2`pi` m + i sin 2`pi` m = 1
i.e., X is a root of unity.
For summation of n nth roots of unity, we can take summation of all the values of X for n ranging between 0 and n.
Geometrically, all these n nth roots are equally spaced vectors around a unit circle and their sum is nothing but the center of the circle, i.e. 0 + i0
This can also be obtained by integrating X with limit 0 to n.
(hint: remember Integral (sinx )=-cos x and integral (cos x) = sinx; and that sin x = 0 for x =0 or 2 pi. and cos x will be 1 and -1 for 0 and 2 pi and thus sin terms would be 0 and cos terms will cancel out)
Product of n nth roots:
If X^n is the nth root of unity, other roots can be calculated by substituting n = 0 , 1,2...., n.
The product of such a series will be
X^n . X^(n-1)..............X^0 = X^(n +n-1+n-2+......1)
The exponent is an arithmetic series with numbers from 1 to n and summation of such series is given n.n/2
Thus product of n nth roots of unity = X^(n.n/2) = X^n . X^(n/2)
X^(n/2) can also be written as X^n. X^(1/2)
Thus the product = X^n. X^n. X^(1/2)
now X^n =1 and X =1 (for =0)..........thus the product reduces to X^(1/2) = (1)^(1/2) = `+-` 1