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how would i find the 5th roots of -2i? (leave the answers in polar form and the angle in degrees) thank you!

Mathematics
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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.
On a graph, |dw:1353369315067:dw|-2i is located there. The angle from the positive real axis, to the negative imaginary axis, is \(3\pi/2\). So the polar form should be \[2\text{ cis} (3\pi/2)\]or\[\Large 2e^{\frac{3i\pi}{2}}\]

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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?
Remember that you need to have \[\Large \sqrt[5]{2}\]out in front of those, but otherwise, I think those are correct.

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