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2016 Paper 3 Q7
D: 1700.0 B: 1516.0
complex numbers roots of unity De Moivre regular polygon modulus factorization geometric proof product of distances

Let \(\omega = \e^{2\pi {\rm i}/n}\), where \(n\) is a positive integer. Show that, for any complex number \(z\), \[ (z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,. \] The points \(X_0, X_1, \ldots\, X_{n-1}\) lie on a circle with centre \(O\) and radius 1, and are the vertices of a regular polygon.

  1. The point \(P\) is equidistant from \(X_0\) and \(X_1\). Show that, if \(n\) is even, \[ |PX_0| \times |PX_1 |\times \,\cdots\, \times |PX_{n-1}| = |OP|^n +1\, ,\] where \(|PX_ k|\) denotes the distance from \(P\) to \(X_k\). Give the corresponding result when \(n\) is odd. (There are two cases to consider.)
  2. Show that \[ |X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,. \]

Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).

  1. \(P\) lies on the perpendicular bisect of \(1\) and \(\omega\), so \(p = re^{\pi i/n}\), where \(r\) can be positive or negative, but \(|r| = |OP|\). \begin{align*} && |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\ &&&= |p^n - 1| \\ &&&= |r^ne^{\pi i} - 1| \\ &&&= |-|OP|^n - 1| \tag{since \(n\) even} \\ &&&= |OP|^n+1 \end{align*} If \(n\) is odd, depending on the sign of \(r\) we get \(|OP|^n+1\) or \(||OP|^n-1|\).
  2. \(\,\) \begin{align*} && (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\ &&&= 1 + z +\cdots + z^{n-1} \\ && |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\ &&&= 1+1+1^2+\cdots + 1^{n-1} \\ &&&= n \end{align*}

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