Complex Number Question (1 Viewer)

[ifireballx]

Let w=cis 2pi/9

i. Show that w^k is a solution of z^9-1=0 where k is an integer.

ii. Prove that w + w^2 + w^3 + ... +w^8 = -1

iii. Hence show that cos(pi/9) cos(2pi/9) cost (4pi/9) = 1/8

 

[Axio]

Let w=cis 2pi/9

i. Show that w^k is a solution of z^9-1=0 where k is an integer.

ii. Prove that w + w^2 + w^3 + ... +w^8 = -1

iii. Hence show that cos(pi/9) cos(2pi/9) cost (4pi/9) = 1/8

i. w^k=cis2pik/9
solutions of z^9=1:
1 is a solution, and other solutions are spaced equally around unit circle.
Therefore, z=cis2pik/9 are the solutions where k = 0,1,2...8.
and therefore z=w^k. (Not sure if this is a proof...)

ii. R.T.P w^8 +w^7... +1=0

0=z^9 -1 =(z^3-1)(z^6+z^3+1)=(z-1)(z^2 +z +1)(z^6 +z^3 +1)=(z-1)(z^8+z^5 +z^2 +z^7 +z^4 +z +z^6 +z^3 +1)

sub in w.
therefore w^8 +w^7...+1 = 0

 

[Sy123]