Find all the cube roots of -125i in polar form. convert all results to a+bi form.

The cube roots can be found easily by converting the complex number into polar form.
z = -125i
z = 125<270, where angle is in degrees.
z^(1/3) = 125<((270 + k(360))/3)
        = 125<(90 + 120k)

For k = 0:
z^(1/3) = 125<90 = 125i
For k = 1:
z^(1/3) = 125<210 = 125(-sqrt(3)/2 - i/2) = -125sqrt(3)/2 - 125/2*i
For k = 2:
z^(1/3) = 125<330 = 125(sqrt(3)/2 - i/2) = 125sqrt(3)/2 - 125/2*i

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

I'm just guessing, but my answer would c. 0 to 300.

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.