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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!
 2 years ago
 2 years 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!
 2 years ago
 2 years 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?
 2 years ago

itsrainingmacey Group TitleBest ResponseYou've already chosen the best response.0
Is it 2cis0? I'm not entirely sure.
 2 years 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\[\Large 2e^{\frac{3i\pi}{2}}\]
 2 years 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!
 2 years ago

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