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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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2006 Paper 1 Q4
D: 1500.0 B: 1514.2
curve sketching inequality trigonometric differentiation regular polygon optimisation small angle approximation geometric proof

By sketching on the same axes the graphs of \(y=\sin x\) and \(y=x\), show that, for \(x>0\):

  1. \(x>\sin x\,\);
  2. \(\dfrac {\sin x} {x} \approx 1\) for small \(x\).
A regular polygon has \(n\) sides, and perimeter \(P\). Show that the area of the polygon is \[ \displaystyle \frac{P^2} { {4n \tan \l\dfrac{ \pi} { n} \r}} \;. \] Show by differentiation (treating \(n\) as a continuous variable) that the area of the polygon increases as \(n\) increases with \(P\) fixed. Show also that, for large \(n\), the ratio of the area of the polygon to the area of the smallest circle which can be drawn around the polygon is approximately \(1\).

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2001 Paper 2 Q7
D: 1600.0 B: 1470.8
complex numbers Argand diagram regular polygon rotation circumcentre geometric construction multiplication by cis

In an Argand diagram, \(O\) is the origin and \(P\) is the point \(2+0\mathrm{i}\). The points \(Q\), \(R\) and \(S\) are such that the lengths \(OP\), \(PQ\), \(QR\) and \(RS\) are all equal, and the angles \(OPQ\), \(PQR\) and \(QRS\) are all equal to \({5{\pi}}/6\), so that the points \(O\), \(P\), \(Q\), \(R\) and \(S\) are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that \(Q\) is the point \(2 + \sqrt 3 + \mathrm{i}\) and find \(S\). The point \(C\) is the centre of the circle that passes through the points \(O\), \(P\) and \(Q\). Show that, if the polygon is rotated anticlockwise about \(O\) until \(C\) first lies on the real axis, the new position of \(S\) is $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$

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1997 Paper 3 Q3
D: 1700.0 B: 1504.6
complex numbers roots of unity De Moivre geometric proof regular polygon factorization vectors summation

By considering the solutions of the equation \(z^n-1=0\), or otherwise, show that \[(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},\] where \(z\) is any complex number and \(\omega={\rm e}^{2\pi i/n}\). Let \(A_1,A_2,A_3,\dots,A_n\) be points equally spaced around a circle of radius \(r\) centred at \(O\) (so that they are the vertices of a regular \(n\)-sided polygon). Show that \[\overrightarrow{OA_1}+\overrightarrow{OA_2}+\overrightarrow{OA_3} +\dots+\overrightarrow{OA_n}=\mathbf0.\] Deduce, or prove otherwise, that \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]

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1988 Paper 3 Q10
D: 1700.0 B: 1500.0
differential equation polar coordinates pursuit curve symmetry exponential function regular polygon first order ODE

Four greyhounds \(A,B,C\) and \(D\) are held at positions such that \(ABCD\) is a large square. At a given instant, the dogs are released and \(A\) runs directly towards \(B\) at constant speed \(v\), \(B\) runs directly towards \(C\) at constant speed \(v\), and so on. Show that \(A\)'s path is given in polar coordinates (referred to an origin at the centre of the field and a suitable initial line) by \(r=\lambda\mathrm{e}^{-\theta},\) where \(\lambda\) is a constant. Generalise this result to the case of \(n\) dogs held at the vertices of a regular \(n\)-gon (\(n\geqslant3\)).

Solution:

TikZ diagram
It's straightforward to see that \(\dot{r} = -\frac{v}{\sqrt{2}}\) and \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v\delta t/\sqrt{2}}{r - v \delta t/\sqrt{2}} = \frac{v}{r\sqrt{2}} \delta t + o(\delta t^2) \Rightarrow \dot{\theta} = \frac{v}{r \sqrt{2}}\) Solving this system, we can see that \(r(t) = \frac{l - vt}{\sqrt{2}}\) and \(\frac{\d r}{\d \theta} = -r \Rightarrow r = \lambda e^{-\theta}\) where \(\lambda = \frac{l}{\sqrt{2}}\) where \(l\) is the initial square side length.
TikZ diagram
By the cosine rule, we can see that \(r(t + \delta t)^2 = r^2+(v\delta t)^2 - 2rv\delta t \cos (\frac12 (\pi - \frac{2\pi}{n}))\), ie: \(\frac{r(t + \delta t)^2 - r^2}{\delta t} = - 2r v \sin(\frac{\pi}{n}) \Rightarrow \dot{r} = - v \sin (\frac{\pi}{n})\). We can also observe that \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v \delta t \sin (\frac{\pi}{2} - \frac{\pi}{n})}{r - v \delta t \cos (\frac{\pi}{2} - \frac{\pi}{n})} \Rightarrow \dot{\theta} = \frac{v}{r} \cos (\frac{\pi}{n})\). Combining these, we can see that \(\frac{\d r}{\d \theta} = - r \tan (\frac{\pi}{n}) \Rightarrow r = \lambda e^{-\tan(\frac{\pi}{n})t}\) where \(\lambda\) is the initial radius of the circumscribed circle.

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