What are the fourth roots of −3+33√i ? Enter your answer by filling in the boxes. Enter the roots in order of increasing angle measure in simplest form.

In order of increasing angle measure, the fourth roots of -3 + 3√3·i are presented as follows;

Step-by-step explanation:

The root of a complex number a + b·i is given as follows;

r = √(a² + b²)

θ = arctan(b/a)

The roots are;

·[cos((θ + 2·k·π)/n) + i·sin((θ + 2·k·π)/n)]

Where;

k = 0, 1, 2,..., n -2, n - 1

For z = -3 + 3√3·i, we have;

r = √((-3)² + (3·√3)²) = 6

θ = arctan((3·√3)/(-3)) = -π/3 (-60°)

Therefore, we have;

When k = 0, the fourth root is presented as follows;

When k = 1 the fourth root is presented as follows;

When k = 2, the fourth root is presented as follows;

When k = 3, the fourth root is presented as follows;

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says