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.
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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
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