What are the three different cube roots of the number "1"?

Question

I heard that there is one real root and two imaginary?

Answer 1

Actually, there is one real root and two complex roots.

If you know complex analysis, finding the three roots is straightforward: your answer is in the form exp(iA) where A is an angle in the x-y plane, and i^2 = -1. Then cubing the expression gives you exp(i3A)=1. The argument is 1 for any angle A between 0 and 2pi such that 3A=n2pi. So A = 0, 2pi/3, and 4pi/3.

Then, exp(iA) = exp(0), exp(i2pi/3), exp(i4pi/3)

exp(0) = 1 (you already knew that one);
exp(i2pi/3) = cos(2pi/3)+isin(2pi/3) = -1/2 +i*sqrt(3)/2
exp(i4pi/3) = cos(4pi/3)+isin(4pi/3) = -1/2 - i*sqrt(3)/2

Okay, now suppose you don't know complex analysis. You can still solve for the roots by assuming an expression in the form a+ib, such that (a+ib)^3 = 1, and solve for all a, b real:

(a+ib)(a+ib)(a+ib) = (a^2-b^2+i2ab)(a+ib)
= a^3+ia^2b-ab^2-ib^3+i2a^2b-2ab^2
= a^3-3ab^2 +i (3a^2b-b^3) = 1

Separating real and imaginary terms,
a^3-3ab^2 = 1
3a^2b-b^3 = 0

Factoring;
a(a^2-3b^2) = 1;
b(3a^2-b^2) = 0

First trivial solution: a=1, b=0.

To get the other two solutions, divide eq 2 by b, then bring b^2 to the other side such that
3a^2=b^2; b=+-a*sqrt(3). Then,
a(a^2-3*3a^2) = 1;
-8a^3 = 1; a = -1/2

b=+-sqrt(3)*a = +-sqrt(3)/2

So all solutions in the form a+bi are:
1
-1/2+i*sqrt(3)/2
-1/2-i*sqrt(3)/2

Good luck, work hard, and stay away from drugs.

Answer 2

The three cube roots of 1 are

1
(-1 + i√3)/2
(-1 - i√3)/2

Answer 3

The cube root of 1 is 1 (because 1x1x1 = 1). There aren't 3 different cube roots.