4U Roots of Unity/Polynomials Question

nerrma

New Member

Dec 2, 2018

Part (a) and (b) are straightforward enough. Any help for part (c)?

Screenshot_2018-12-02 TxtBk Chapter Complex Nosv2 pdf.png

$$ \noindent Since $ |z|=1, $ we can use the identity: $z^n+z^{-n}=\cos n\theta $ (you can prove this by subbing in $z= $ cis$\theta $ and using DMT$. \\ \\$ Dividing the LHS by $z^3, $ we obtain $: \\ z^3+z^{-3} = 2\cos 3\theta \qquad $ Using the above identity$ \\ \\ $ On the RHS, we divide each factor by $z,$ this is equivalent to dividing the whole side by $ z^3: \\ \left ( \frac{z^2 +1}{z} \right )\left ( \frac{z^2-\sqrt{3}z+1}{z} \right )\left ( \frac{z^2+\sqrt{3}z+1}{z} \right )=\left ( z+z^{-1} \right )\left ( z+z^{-1}-\sqrt{3} \right )\left ( z+z^{-1}+\sqrt{3} \right ) \\ =(2\cos \theta)\left ( 2\cos \theta-2\cos \frac{\pi}{6} \right )\left ( 2\cos \theta-2\cos \frac{5\pi}{6} \right ) \\ =8 \cos \theta \left ( \cos \theta - \cos\frac{\pi}{6} \right )\left ( \cos \theta - \cos\frac{5\pi}{6} \right ) \\ \\ \therefore \cos 3\theta =4\cos \theta \left ( \cos \theta - \cos\frac{\pi}{6} \right )\left ( \cos \theta - \cos\frac{5\pi}{6} \right )$

4U Roots of Unity/Polynomials Question (1 Viewer)

nerrma

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jathu123

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