Find three cube roots for the following complex number. Leave your answers in trigonometric form. (Enter your answers from smallest to largest angle measure.)125i

To find the cube roots of a complex number, we first need to convert the complex number into its trigonometric form. The trigonometric form of a complex number is r(cos θ + i sin θ), where r is the magnitude of the complex number and θ is the angle it makes with the positive x-axis.

Step 1: Convert the complex number into trigonometric form The given complex number is 125i, which can be written as 0 + 125i. This corresponds to a point on the complex plane with x-coordinate 0 and y-coordinate 125. The magnitude r of this complex number is √(0² + 125²) = 125. The angle θ it makes with the positive x-axis can be found using the formula θ = atan2(y, x) = atan2(125, 0) = π/2. Therefore, the trigonometric form of the complex number is 125(cos π/2 + i sin π/2).

Step 2: Find the cube roots The cube roots of a complex number r(cos θ + i sin θ) are given by the formula r^(1/3)[cos((θ + 2πk)/3) + i sin((θ + 2πk)/3)], where k = 0, 1, 2. Substituting r = 125 and θ = π/2 into this formula gives the three cube roots:

1. 125^(1/3)[cos((π/2 + 2π0)/3) + i sin((π/2 + 2π0)/3)] = 5(cos π/6 + i sin π/6)
2. 125^(1/3)[cos((π/2 + 2π1)/3) + i sin((π/2 + 2π1)/3)] = 5(cos 5π/6 + i sin 5π/6)
3. 125^(1/3)[cos((π/2 + 2π2)/3) + i sin((π/2 + 2π2)/3)] = 5(cos 3π/2 + i sin 3π/2)

So, the three cube roots of 125i are 5(cos π/6 + i sin π/6), 5(cos 5π/6 + i sin 5π/6), and 5(cos 3π/2 + i sin 3π/2).