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First, you need to change -2i into polar form. Can you show me what you get when it's in polar form?

Is it 2cis0? I'm not entirely sure.

Oh! Okay, I see! I'm going to work it out with the information you have just given me. Thank you!

As a hint, it's a lot easier to take roots of \(2e^{\frac{3i\pi}{2}}\)

So far, I have 2cis3pi/10 and 2cis7pi/10.
Am I on the right track?