A real number raised to the power 1/n will have n roots; but some of them may be complex numbers. Up to now, we have represented complex numbers in the form:
Try to find all three roots of 1 using x1/3 = y. To do this, solve y3 = x, that is, solve:x = a + ci
That is, we need to solve,y3 = (b + di)(b + di)(b + di) = a + ci = x (b + di)(b2 + 2bdi + d2i2) = (b + di)(b2 - d2 + 2bdi) = (b3 - bd2 + 2b2di + b2di - d3i + 2bd2i2) = (b3 - bd2 - 2bd2) + (2b2d + b2d - d3)i = (b3 - 3bd2) + (3b2d - d3)i = a + ci = x
By factoring the equation for c:a = (b3 - 3bd2) = 1 c = (3b2d - d3) = 0
we see that there are two solutions for d,0 = d(3b2 - d2)
By substitution into the equation for a first for d = 0:d = 0 d2 = 3b2
Then for d2 = 3b2:(b3 - 3b(0)) = 1 b = 1
The factor d can take two values:b3 - 3b(3b2) = -8b3 = 1 b = (-1/8)1/3 = -1/2
then there are three possibilities:d2 = 3/4 d = ±0.5√3
Giving us the three roots of y:if b = 1 then d = 0 if b = -1/2 then d = +0.5√3 or d = -0.5√3
{1, -0.5(1 + √3i), -0.5(1 - √3i)}
Finding all the roots can be done more easily if we represent complex numbers in polar form.