Complex Number Question (1 Viewer)

ifireballx · Dec 1, 2014

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 · Dec 1, 2014

ifireballx said:
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 · Dec 1, 2014

$\\ \omega = \cos \left(\frac{2\pi}{9}\right) + i\sin \left(\frac{2\pi}{9}\right) \\ \\ $i) Due to De Moivere's theorem: $ \\ \omega^k = \left(\cos \left(\frac{2\pi}{9}\right) + i\sin \left(\frac{2\pi}{9}\right) \right)^k = \cos \left(\frac{2\pi k}{9} \right) + i\sin \left(\frac{2\pi k}{9} \right) \\ \\ $Now, if$ \ k \ $is an integer$ \ (\omega ^k)^9 = \cos (2\pi k) + i\sin(2\pi k) = 1 \\ \\ \therefore \ (w^k)^9 - 1 = 0 \ \Rightarrow \ \omega^k \ $is a root of$ \ z^9 - 1 = 0$

$\\ $ii) Since$ \ z^9 - 1 = 0 \ $is a polynomial with a degree of$ \ 9 \ $it has$ \ 9 \ $roots, and$ \\ \omega^k \ $is a root of the polynomial for all integers$ \ k \ $so, since$ \ \omega^k \ $takes 9 distinct values which are$ \\ (1, \omega, \omega^2, \omega^3, \omega^4, \omega^5, \omega^6, \omega^7, \omega^8) \\ \\ $So, these are all the roots of$ \ z^9 - 1 = 0 \\ $So, the sum of roots is$ \ 0 \ $which can be written as$ \\ 1 + \omega + \omega^2 + \dots + \omega^8 = 0 \\ $So,$ \ \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + \omega^7 + \omega^8 = -1$

$\\ $iii) Now,$ \\\\ \left(2\cos \frac{2\pi}{9}\right) \left(2\cos \frac{4\pi}{9} \right) \left(2\cos \frac{6\pi}{9} \right) \left(2\cos \frac{8\pi}{9} \right) \\\\ = (\omega + \omega^8)(\omega^2 + \omega^7)(\omega^3 + \omega^6)(\omega^4 + \omega^5) = \omega^{10}(1 + \omega)(1 + \omega^3)(1 + \omega^5)(1 + \omega^7)\\ = \omega(1 + \omega)(1 + \omega^3)(1 + \omega^5)(1 + \omega^7)$

$\\ $Then just expand or factorise and simplify, to get that product equal to$ \ - 1 \ $and then noticing that$ \ 2\cos \frac{6\pi}{9} = -1 \ $and$ \ \cos \frac{8\pi}{9} = \cos \frac{\pi}{9}$

Complex Number Question (1 Viewer)

ifireballx

New Member

Axio

=o

Sy123

This too shall pass

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