complex number help (1 Viewer)

mathsbrain · Nov 19, 2012

Given z^6+z^3+1=[z^2-2zcos(2pi/9)+1][z^2-2zcos(4pi/9)+1][z^2-2zcos(8pi/9)+1], prove that
2cos(3x)+1=8[cos(x)-cos(2pi/9)][cos(x)-cos(4pi/9)][cos(x)-cos(8pi/9)]

mathsbrain · Nov 19, 2012

actually dont worry, got it, just let z=cis(x)

rolpsy · Nov 19, 2012

for future reference...

note that

$z^{n} + z^{-n} = 2\cos(n\theta)$

then

$\begin{align*}z^6 + z^3 + 1 &= \left (z^2 -2z\cos\left ( \frac{2\pi}{9} \right ) + 1\right )\left (z^2 -2z\cos\left ( \frac{4\pi}{9} \right ) + 1\right )\left (z^2 -2z\cos\left ( \frac{8\pi}{9} \right ) + 1\right )\\&= z^3\left (z -2\cos\left ( \frac{2\pi}{9} \right ) + z^{-1}\right )\left (z -2\cos\left ( \frac{4\pi}{9} \right ) + z^{-1}\right )\left (z -2\cos\left ( \frac{8\pi}{9} \right ) + z^{-1}\right )\\z^3 + 1 + z^{-3}&= \left (z -2\cos\left ( \frac{2\pi}{9} \right ) + z^{-1}\right )\left (z -2\cos\left ( \frac{4\pi}{9} \right ) + z^{-1}\right )\left (z -2\cos\left ( \frac{8\pi}{9} \right ) + z^{-1}\right )\\\left (z^3 + z^{-3} \right ) + 1&= \left (\left (z +z^{-1} \right ) -2\cos\left ( \frac{2\pi}{9} \right )\right )\left (\left (z +z^{-1} \right ) -2\cos\left ( \frac{4\pi}{9} \right )\right )\left (\left (z +z^{-1} \right ) -2\cos\left ( \frac{8\pi}{9} \right )\right )\\\end{align*}$

from which you can conclude

$\begin{align*}2\cos(3\theta) + 1 = 8\left (\cos(\theta) - \cos\left ( \frac{2\pi}{9} \right )\right )\left (\cos(\theta) - \cos\left ( \frac{4\pi}{9} \right )\right )\left (\cos(\theta) -\cos\left ( \frac{8\pi}{9} \right )\right )\end{align*}$

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complex number help (1 Viewer)

mathsbrain

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mathsbrain

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rolpsy

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