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itsrainingmacey
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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!
 one year ago
 one year ago
itsrainingmacey Group Title
how would i find the 5th roots of 2i? (leave the answers in polar form and the angle in degrees) thank you!
 one year ago
 one year ago

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KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
First, you need to change 2i into polar form. Can you show me what you get when it's in polar form?
 one year ago

itsrainingmacey Group TitleBest ResponseYou've already chosen the best response.0
Is it 2cis0? I'm not entirely sure.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
On a graph, dw:1353369315067:dw2i 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\[\Large 2e^{\frac{3i\pi}{2}}\]
 one year ago

itsrainingmacey Group TitleBest ResponseYou've already chosen the best response.0
Oh! Okay, I see! I'm going to work it out with the information you have just given me. Thank you!
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
As a hint, it's a lot easier to take roots of \(2e^{\frac{3i\pi}{2}}\)
 one year ago

itsrainingmacey Group TitleBest ResponseYou've already chosen the best response.0
So far, I have 2cis3pi/10 and 2cis7pi/10. Am I on the right track?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Remember that you need to have \[\Large \sqrt[5]{2}\]out in front of those, but otherwise, I think those are correct.
 one year ago
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