complex number Q (1 Viewer)

[bobo123]

say you got
factorise z^5-1 and then
prove : cos(2pi/5) + cos(4pi/5) = -1/2

instead of using the long method of equating the co-efficients on both sides to prove the second part
i found this other method in some other textbook


where cos(2kpi/n) =[ w^k+w^(n-k) ]/2

anyway whats the name of this method? cuz i havent been taught this way at school so i need something to back myself up on about this just incase i lose marks for doing something not taught at school

 

[chunder]

what textbook was that?

would you mind just saying what the w and k stand for?

 

[McLake]

I wouldn't recommend using that method straight away. I suggest doing "the long tedious way" and then using that to verify your answer ....

 

[bobo123]

well it wasnt really a textbook
just a booklet of paper

w is the root of the complex number

the whole method looks dodgy, no name, no proof.
but it works pretty good, which is good for lazy me
the long tedious way is alright too but just takes me a while

theres also a method for sin as well

anyway
i need a name or something
HELP aNYONE

 

[KeypadSDM]

Easy solution:
prove : cos(2pi/5) + cos(4pi/5) = -1/2
using z^5 -1 = 0
Sum of roots = 0
A legitimate assumtion is that:
Sum of real part of roots = 0 (admittedly i haven't proven this, but it shouldn't be too hard)
Real Parts of roots are:
Cos[2Pi/5], Cos[4Pi/5], Cos[6Pi/5], Cos[8Pi/5], Cos[10Pi/5] (Using deMoivre's theorem)

.: Cos[2Pi/5] + Cos[4Pi/5] + Cos[6Pi/5] + Cos[8Pi/5] + Cos[10Pi/5] = 0

But, Cos[10Pi/5] = 1
and; Cos[4Pi/5] = Cos[6Pi/5]
and; Cos[2Pi/5] = Cos[8Pi/5]

.: 2Cos[2Pi/5] + 2Cos[4Pi/5] + 1 = 0
.: Cos[2Pi/5] + Cos[4Pi/5] = -0.5

 

[spice girl]

yes the assumption is legitimate

Re(0) = Im(0) = 0

 

[KeypadSDM]

Thanks, it's good to have some strong backing