Question

Please, help
Let \(\xi = e^{2\pi i/5}\)
a) find the minimal polynomials for \(cos (2\pi/5)~~and~~sin(2\pi/5)\) over \(\mathbb Q\)
b) find the minimal polynomial for \(\xi\) over \(\mathbb Q\)
I do not know how to start.

Answer 1

@wio @Zarkon

Answer 2

ask a QH lol :)

Answer 3

My attempt:
\(\xi = e^{2i\pi/5} = cos(2\pi/5)+isin(2\pi/5)\). Take the real part cos (2pi/5)

Answer 4

Let \(y= 2\pi/5\)
we have the polynomial will be
1+cos y + cos (2y) + cos (3y)+cos(4y)+cos(5y) =0

Answer 5

\(cos (5y) = cos (2\pi)=1\)
\(cos(4y) = cos (8\pi/5) = cos(2\pi-2\pi/5) = cos y\)
\(cos(3y) = cos (6\pi/5) = cos(2\pi -4\pi/5) = cos(2y)\)

Answer 6

hence , 1+cos y + cos (2y) + cos (2y) + cos(y) +1 =0
or 2+ 2 cosy + 2 cos (2y) =0
but \(cos(2y) = 2 cos^2(y) -1\)

Answer 7

hence the minimum polynomial for cos (2pi/5) is
2 + cos(y) +4cos^2(y) -2 =0
or \(4cos^2(2\pi/5)+cos(2\pi/5)=0\)

Answer 8

But not sure about it, and not know how to do for sin. Appreciate any comments.

Answer 9

I'm not entirely sure what a minimal polynomial means, but I do know of one important polynomial that xi satisfies since these are all vectors in the 2D plane that are equally spaced 72 degrees apart so they're in equilibrium in a sense: \[\Large \xi^0+\xi^1+\xi^2+\xi^3+\xi^4=0\]

Answer 10

Very similar question w/ solution here: https://math.dartmouth.edu/~m111w15/111-W15-HW2-solns.pdf

The ans will be different though since \[\alpha =2\cos (\frac{ 2\pi }{ 5 })\] in the example.

Good luck!

Answer 11

Thank you very much