This question is killing me (1 Viewer)

[tiggerfamilytre]

This comes from Patel's book:

"Find the roots of z^6 + 1 = 0 and hence resolve z^6 + 1 into real quadratic factors; deduce that cos3x = 4(cosx - cos[pi/6])(cosx - cos[pi/2])(cosx - cos[{5pi}/6])"

Please help

 

[haboozin]

This comes from Patel's book:

"Find the roots of z^6 + 1 = 0 and hence resolve z^6 + 1 into real quadratic factors; deduce that cos3x = 4(cosx - cos[pi/6])(cosx - cos[pi/2])(cosx - cos[{5pi}/6])"

Please help

roots cis +-(pi/6) cis+-(pi/2) cis +-(5pi/6)

now equate to real quad factors:
(z^2 +1) (z^2 -2cos(pi/6)z +1) (z^2 + 2 cos(5pi/6)z +1)

i 'm not sure about the other part...

 

[LaCe]

I've done this question but i cant seem to find it
Firstly find all of the roots of z^6=-1
then go something like
z+ 1/z= ...
z^2 + 1/z^2 = ...

or u will find that the solutions are in conjugate pairs so z^6 +1 = (z-z1)(z-...)(...)(...)
Expand

 

[Slidey]

Double post. See my solution in this thread: http://www.boredofstudies.org/community/showthread.php?t=68498