Complex Numbers Question - Roots of Unity (1 Viewer)

[Yamiyo]

Can someone please show me how to do this question from the Cambridge book?

a) Use De Moivre's theorem to solve z^5 =-1.
[I can do this part; z= -1, cis(+)(-) pi/5, cis(+)(-) (3pi)/5; (+)(-)= plus or minus]
This is the part I don't get at all:
b) By grouping the roots in complex conjugate pairs, show that
z^5 + 1 = (z+1)(z^2 - 2z (cos pi/5) + 1)(z^2 - 2z (cos 3pi/5) +1)

 

[Sarsy]

I did this the other day, but unfortunately Iruka beat me to it. lol

 

[Yamiyo]

Since we know all the roots of z^5 +1 =0, we can factorize as

z^5 +1 = (z--1)(z - [cos pi/5 + i sin pi/5])(z - [cos -pi/5 + i sin -pi/5])(z - [cos 2*pi/5 + i sin 2*pi/5])(z - [cos -2*pi/5 + i sin -2*pi/5]).

Now just multiply the complex conjugate pairs, i.e., expand (z - [cos pi/5 + i sin pi/5])(z - [cos -pi/5 + i sin -pi/5]), and (z - [cos 2*pi/5 + i sin 2*pi/5])(z - [cos -2*pi/5 + i sin -2*pi/5]).

You should get some cancellation of factors if you make use of the fact that sin(-x) = -sin(x) but cos(-x) = cos(x).

Ooooohhhhhhh... that's actually suprisingly simple. Thanks a lot.