Question

The complex cube roots of 6sqrt3 6i

Answer 1

The complex cube roots of 6√3 + 6i are 2√3 + 2i, -√3 - 3i, and -√3 + i.

Step-by-step explanation:

To find the complex cube roots, we can use the polar form of a complex number. Given a complex number a + bi, its polar form is r(cos θ + i sin θ), where r is the magnitude and θ is the argument.

For 6√3 + 6i, the magnitude r is 2√3 and the argument θ is 60° (or π/3 radians). The cube root of a complex number in polar form is found by taking the cube root of the magnitude and dividing the argument by 3.

So, the cube roots are:

2√3 cis(π/9) or 2√3 + 2i

2√3 cis(7π/9) or -√3 - 3i

2√3 cis(13π/9) or -√3 + i

These are the three complex cube roots of 6√3 + 6i. Each root represents a distinct solution, and together they form a set of complex numbers that, when cubed, result in the original complex number.