Question

Can someone help with a complex number problem in euler's form and using de moivre's theorem? Medals given

Answer 1

Hey, are you there? What's your question @chau88

Answer 2

Ooh interesting, I know that \[e^{i \theta} = \cos \theta + i \sin \theta\] so you will have to apply this

Answer 3

This question, for (a), (b) and (c) I have queries.
(a) I got, w^2=e^4ipi/5, w^3=e^6ipi/5; w^4=e^8ipi/5

Answer 4

That looks decent, let me see

Answer 5

but my answer is different from the given answer, which is w^4=e^4ipi/5, w^3=e^-4pi/5i; w^4=e^-2pi/5 i

Answer 6

\[\huge w^2 = \left( e^{2i \pi/5} \right)^2 = e^{4i \pi/5}\] right?

Answer 7

yes

Answer 8

\[\huge w^3 = \left( e^{2i \pi/5} \right)^3 = e^{-4i \pi/5}\] this is for cubed is it

Answer 9

wait. why?

Answer 10

That's what you have given for the answer in the book

Answer 11

did you use the anticlockwise?

Answer 12

because if you multiply it, you should get 6ipi/5

Answer 13

Ah, I see. Thanks. @Astrophysics , what about (b), wo far I have e^2ipi/5 + e^4i pi/5+ e^-4ipi/5+e^-2ipi/5. can I group them into 2cos2theta+2 cos 4theta?

Answer 14

How do I get -1?

Answer 15

@ganeshie8

Answer 16

I think you can use Euler's formula for b, but I'm not sure

Answer 17

identity*

Answer 18

is it z^n+1/z^n=2cos ntheta?
I used that, but I got 12 cos^2 theta -8 and not -1 :(

Answer 19

Note that\[1, \omega, \omega^2, \omega^3, \omega^4 \]are the roots of the equation\[x^5 - 1= 0\]The sum of roots of this equation is \(0\).

Answer 20

Nice!

Answer 21

@Beauregard , how can you tell that this is the fifth root of unity?