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2013 Paper 3 Q4
D: 1700.0 B: 1516.0
complex numbers roots of unity De Moivre product formula trigonometric identity factorization roots of polynomials

Show that \((z-\e^{i\theta})(z-\e^{-i\theta})=z^2 -2z\cos\theta +1\,\). Write down the \((2n)\)th roots of \(-1\) in the form \(\e^{i\theta}\), where \(-\pi <\theta \le \pi\), and deduce that \[ z^{2n} +1 = \prod_{k=1}^n \left(z^2-2z \cos\left( \tfrac{(2k-1)\pi}{2n}\right) +1\right) \,. \] Here, \(n\) is a positive integer, and the \(\prod\) notation denotes the product.

  1. By substituting \(z=i\) show that, when \(n\) is even, \[ \cos \left(\tfrac \pi {2n}\right) \cos \left(\tfrac {3\pi} {2n}\right) \cos \left(\tfrac {5\pi} {2n}\right) \cdots \cos \left(\tfrac{(2n-1) \pi} {2n}\right) = {(-1\vphantom{\dot A})}^{\frac12 n} 2^{1-n} \,. \]
  2. Show that, when \(n\) is odd, \[ \cos^2 \left(\tfrac \pi {2n}\right) \cos ^2 \left(\tfrac {3\pi} {2n}\right) \cos ^2 \left(\tfrac {5\pi} {2n}\right) \cdots \cos ^2 \left(\tfrac{(n-2) \pi} {2n}\right) = n 2^{1-n} \,. \] You may use without proof the fact that \(1+z^{2n}= (1+z^2)(1-z^2+z^4 - \cdots + z^{2n-2})\,\) when \(n\) is odd.

Solution: \begin{align*} && (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\ &&&= z^2-2\cos \theta z + 1 \end{align*} The \(2n\)th roots of \(-1\) are \(e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}\) or \(e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}\) \begin{align*} && z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\ &&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\ &&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \end{align*}

  1. \begin{align*} && i^{2n} + 1 &= \prod_{k=1}^n \left (i^2 - 2i \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \\ \Rightarrow && (-1)^n + 1 &= (-1)^n2^ni^n\prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) \\ \Rightarrow && \prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) &= 2^{1-n}(-1)^{n/2} \tag{if \(n\equiv 0\pmod{2}\)} \end{align*}
  2. When \(n\) is odd, we notice that two of the roots are \(i\) and \(-i\), if we exclude those, (ie by factoring out \(z^2+1\), we see that \begin{align*} && 1-z^2+z^4-\cdots + z^{2n-2} &= \prod_{k=1, 2k-1\neq n}^n \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=(n+1)/2}^{n} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=1}^{(n-1)/2} \left (z^2+2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ \Rightarrow && 1-i^2 + i^4 + \cdots + i^{2n-2} &= \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right) \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right)\\ \Rightarrow && n &= 2^{n-1} \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) \\ \Rightarrow && \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) &= n2^{1-n} \end{align*}

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