Question

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

Answer 1

Use de Moivre's theorem.
-i = e^((3/2)i \[\pi\])
The first cube root is 5e^(((3/2)i [\pi\])/3)
The next you would add 2pi before dividing by 3 to get 5e^(((7/2)i [\pi\])/3)
and the third one, similarly 5e^(((11/2)i [\pi\])/3).

In rectangular form, they would be 5i, -(5i+5sqrt(3))/2, and -(5i-5sqrt(3))/2.

Answer 2

Sorry, not used to the special characters. Here's a cleaned-up version.

Use de Moivre's theorem. -i = e^((3/2)iπ)

The first cube root is 5e^(((3/2)i π)/3) The next you would add 2pi before dividing by 3 to get 5e^(((7/2)i π)/3) and the third one, similarly 5e^(((11/2)i π)/3). In rectangular form, they would be 5i, -(5i+5sqrt(3))/2, and -(5i-5sqrt(3))/2.