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2015-01-12 Roots of Unity.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage{graphicx} \usepackage{wrapfig} \usepackage{txfonts} \usepackage[hmargin=2cm,vmargin=2cm]{geometry} \parindent0em \setlength{\parskip}{0.3cm} \begin{document} \hspace*{-0.5cm}{\large Roots of Unity Exercise} \hspace*{-0.5cm}{\bf Q24} Solve $x^8+1=0$, express each root in the form $r$~cis~$\theta$ and then $A+iB$. Decompose $x^8+1$ into real quadratic factors and deduce that\\ $\cos 4\theta=8\left[\cos\theta-\cos\frac{\pi}{8}\right]\left[\cos\theta-\cos\frac{3\pi}{8}\right]\left[\cos\theta-\cos\frac{5\pi}{8}\right]\left[\cos\theta-\cos\frac{7\pi}{8}\right].$ \bigskip \hspace*{-0.5cm}{\bf Solution:} By $\omega=$~cis~$\left(\frac{\pi+2k\pi}{8}\right),$ we found 8 roots: $\omega_1, \omega_2, \ldots, \omega_8$. $ \omega_1=$~cis~$\frac{\pi}{8}=\cos\frac{\pi}{8}+i\sin\frac{\pi}{8},\\ \omega_2=$~cis~$\frac{3\pi}{8}=\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8},\\ \omega_3=$~cis~$\frac{5\pi}{8}=\cos\frac{5\pi}{8}+i\sin\frac{5\pi}{8},\\ \omega_4=$~cis~$\frac{7\pi}{8}=\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8},\\ \omega_5=$~cis~$\frac{9\pi}{8}=\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8}=\cos\frac{7\pi}{8}-i\sin\frac{7\pi}{8}=\overline{\omega_4},\\ \omega_6=$~cis~$\frac{11\pi}{8}=\cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8}=\cos\frac{5\pi}{8}-i\sin\frac{5\pi}{8}=\overline{\omega_3},\\ \omega_7=$~cis~$\frac{13\pi}{8}=\cos\frac{13\pi}{8}+i\sin\frac{13\pi}{8}=\cos\frac{3\pi}{8}-i\sin\frac{3\pi}{8}=\overline{\omega_2},\\ \omega_8=$~cis~$\frac{15\pi}{8}=\cos\frac{15\pi}{8}+i\sin\frac{15\pi}{8}=\cos\frac{\pi}{8}-i\sin\frac{\pi}{8}=\overline{\omega_1}. $ \bigskip For each of the four conjugation pairs:\\ $ \omega_1+\omega_8=\omega_1+\overline{\omega_1}=2\cos\frac{\pi}{8},\quad \omega_1\cdot\omega_8=\omega_1\cdot\overline{\omega_1}=1,\\ \omega_2+\omega_7=\omega_2+\overline{\omega_2}=2\cos\frac{3\pi}{8},\quad \omega_2\cdot\omega_7=\omega_2\cdot\overline{\omega_2}=1,\\ \omega_3+\omega_6=\omega_3+\overline{\omega_3}=2\cos\frac{5\pi}{8},\quad \omega_3\cdot\omega_6=\omega_3\cdot\overline{\omega_3}=1,\\ \omega_4+\omega_5=\omega_4+\overline{\omega_4}=2\cos\frac{7\pi}{8},\quad \omega_4\cdot\omega_5=\omega_4\cdot\overline{\omega_4}=1. $ \bigskip Because $\omega_n$ is a root of $x^8+1=0$, we can decompose $x^8+1$ into real factors, then group them in conjugate pairs: $ x^8+1=(x-\omega_1)(x-\omega_8)\cdot(x-\omega_2)(x-\omega_7)\cdot(x-\omega_3)(x-\omega_6)\cdot(x-\omega_4)(x-\omega_5)\\ \\ =\left[x^2-(\omega_1+\omega_8)x+\omega_1\omega_8\right]\cdot \left[x^2-(\omega_2+\omega_7)x+\omega_2\omega_7\right]\cdot \left[x^2-(\omega_3+\omega_6)x+\omega_3\omega_6\right]\cdot \left[x^2-(\omega_4+\omega_5)x+\omega_4\omega_5\right]\\ \\ =\left[x^2-2\cos\frac{\pi}{8}x+1\right]\cdot \left[x^2-2\cos\frac{3\pi}{8}x+1\right]\cdot \left[x^2-2\cos\frac{5\pi}{8}x+1\right]\cdot \left[x^2-2\cos\frac{7\pi}{8}x+1\right] $ Note: $x^8+1$ above does not need to be zero; we just make use of the roots of $x^8+1=0$ to factorise $x^8+1$. \bigskip Now divide both sides by $x^4$: $ x^4+\frac{1}{x^4} =\left[x-2\cos\frac{\pi}{8}+\frac{1}{x}\right]\cdot \left[x-2\cos\frac{3\pi}{8}+\frac{1}{x}\right]\cdot \left[x-2\cos\frac{5\pi}{8}+\frac{1}{x}\right]\cdot \left[x-2\cos\frac{7\pi}{8}+\frac{1}{x}\right] $ \bigskip Since $x$ is arbitrary, we let$~~x=\cos\theta+i\sin\theta,\quad \frac{1}{x}=x^{-1}=\cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta=\overline{x}.\quad x+\frac{1}{x}=x+\overline{x}=2\cos\theta. $ Likewise,$~~x^4=\cos 4\theta~~$and$\quad x^4+\frac{1}{x^4}=2\cos 4\theta. $ We now have:\quad $ 2\cos 4\theta =\left[2\cos\theta-2\cos\frac{\pi}{8}\right]\cdot \left[2\cos\theta-2\cos\frac{3\pi}{8}\right]\cdot \left[2\cos\theta-2\cos\frac{5\pi}{8}\right]\cdot \left[2\cos\theta-2\cos\frac{7\pi}{8}\right] $ \bigskip Divide both sides by 2: $ \cos 4\theta =8\left[\cos\theta-\cos\frac{\pi}{8}\right] \left[\cos\theta-\cos\frac{3\pi}{8}\right] \left[\cos\theta-\cos\frac{5\pi}{8}\right] \left[\cos\theta-\cos\frac{7\pi}{8}\right] $ \quad \ldots Q.E.D. \end{document}