The solutions of the equation x^n=1, where n is a positive integer, are called the nth roots of 1 or "nth roots of unity."

Question

Explain the relationship between the solutions of the equation x^n=1 and the zeros of the complex polynomial P(x)=x^n-1. How many nth roots of 1 are there?

Answer 1

1^1/n =

1(cos 0 + i sin 0)

1(cos 2π/n + i sin 2π/n)

1(cos 4π/n + i sin 4π/n)

....

until 

1(cos n(2π/n) + i sin n(2π)/n)

Answer 2

n = 1   indicates 1 real solution.

 

n = 2 indicates 2 real solutions.

 

x³ - 1 = 0 = (x - 1)(x² + x + 1)

 

n = 3 indicates 1 real solution and 2 complex solutions.

 

x⁴ - 1 = 0

 

(x² + 1)(x² - 1)

 

n = 4 indicates 2 real solutions and 2 complex solutions.

 

x⁶ - 1 = 0

 

(x³ - 1)(x³ + 1) = (x - 1)(x² + x + 1)(x + 1)(x² - x + 1)

 

n = 6 indicates 2 real solutions and 4 complex solutions.

 

Does this help you some?

 

You can also use Descartes rule of signs to do this.

Answer 3

Hi Gabriele,

 

This explanation may be a little more sophisticated than you need so I welcome others to contribute.

 

This requires using the polar form of complex numbers as follows:

 

A complex number may be represented in the polar form by using Euler's equation:

 

re^iθ = r cos θ * i r sin θ

 

so 1 would simply be 

 

1 = e^i2π

 

taking the nth root

 

1^1/n = e^i2π/n

 

so wherever 

 

θ = 2π/n

 

is a root. So there are n roots. 

 

If this explanation is too advanced for your current level you may try to simply try to find the algebraic factors, although this may be difficult for larger values of n.