Find the cube roots of 8(cos pi/3 + i sin pi/3)?

The cube roots are:

2(cos pi/9 + i sin pi/9)

2(cos 7pi/9 + i sin 7pi/9)

2(cos 13pi/9 + i sin 13pi/9)

Explanation: think of the complex number as a 2-dimensional vector. Then 8(cos pi/3 + i sin pi/3) is the vector with length 8 which forms an angle of pi/3 with the positive x axis. When you multiply complex numbers, the length of the product is the product of the lengths of the factors, so the cube root of a vector with length 8 would have a length of 2. The angle of the product is the sum of the angles of the factors, so the cube root of a vector with angle pi/3 would have to be an angle which multiplied by 3 gives pi/3 *or* differs by a multiple of 2pi. The obvious answer is pi/9, but 7pi/9 times 3 gives you 21pi/9 which = 2pi + pi/3, and 13pi/9 times 3 gives you 39pi/9 which = 4pi + pi/3.

Cube Root Of Pi

Let r(cos x + i sin x ) be a cube root of 8(cos pi/3 + i sin pi/3) then

r^3 (cos 3x + i sin 3x ) = 8(cos pi/3 + i sin pi/3)

Equating modulus and argument we have r^3 = 8 so r = 2 and 3x = pi/3 + 2kpi Where k is any integer.

Hence x = pi/9 + 2kpi/3, and k = 0,1,2 gives distinct values.

So the 3 cube roots of 8(cos pi/3 + i sin pi/3) are 2( cos x +isinx) where x is given above.