Complex numbers 2

Eulers formulae, de moivre, roots of unity

Showing 1-25 of 32 problems
2025 Paper 3 Q8
D: 1500.0 B: 1500.0

  1. Show that $$z^{m+1} - \frac{1}{z^{m+1}} = \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)$$ Hence prove by induction that, for \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Find similarly \(z^{2n} - \frac{1}{z^{2n}}\) as a product of \((z + \frac{1}{z})\) and a sum.
    1. By choosing \(z = e^{i\theta}\), show that $$\sin 2n\theta = 2\sin\theta \sum_{r=1}^n \cos(2r-1)\theta$$
    2. Use this result, with \(n = 2\), to show that \(\cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - \frac{1}{2}\).
    3. Use this result, with \(n = 7\), to show that \(\cos\frac{2\pi}{15} + \cos\frac{4\pi}{15} + \cos\frac{8\pi}{15} + \cos\frac{16\pi}{15} = \frac{1}{2}\).
  2. Show that \(\sin\frac{\pi}{14} - \sin\frac{3\pi}{14} + \sin\frac{5\pi}{14} = \frac{1}{2}\).

Show Solution
  1. \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right) \\ &= z^{m+1} + \frac{1}{z^{m-1}} - z^{m-1} - \frac{1}{z^{m+1}} + z^{m-1} - \frac{1}{z^{m-1}} \\ &= z^{m+1} - \frac{1}{z^{m+1}} \\ &= LHS \end{align*}. Claim: For \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Proof: (By Induction) Base Case: (\(n=1\)). \begin{align*} LHS &= z^2 - \frac{1}{z^2} \\ &= (z-\frac1z)(z + \frac{1}{z}) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z + \frac{1}{z} \right) \\ &= (z - \frac1z) \sum_{r=1}^1 \left ( z^{2r-1} + \frac{1}{z^{2r-1}} \right) \\ &= RHS \end{align*} as required. Inductive step: Suppose our result is true for some \(n=k\), then consider \(n = k+1\). \begin{align*} RHS &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k+1} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) \\ &= \left(z - \frac{1}{z}\right)\sum_{r=1}^{k} \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right) + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k} - \frac{1}{z^{2k}} + \left(z - \frac{1}{z}\right)\left(z^{2k+1} + \frac{1}{z^{2k+1}}\right) \\ &= z^{2k+2} - \frac{1}{z^{2k+2}} \\ &= LHS \end{align*}. Therefore if our result is true for \(n=k\) is true, it is true for \(n=k+1\). Since it is also true for \(n=1\) it is true for all \(n \geq 1\) but the principle of mathematical induction. Since \(\displaystyle z^{m+1} - \frac{1}{z^{m+1}} = \left(z + \frac{1}{z}\right)\left(z^m - \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)\), we must have \(\displaystyle z^{2n}-\frac{1}{z^{2n}} = \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right)\)
    1. Since $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ we have \begin{align*} && e^{2n\theta i} - e^{-2n\theta i} &= \left(e^{\theta i} - e^{-\theta i}\right)\sum_{r=1}^n \left(e^{(2r-1)\theta i} + e^{-(2r-1)\theta i}\right) \\ \Rightarrow && 2i \sin 2n \theta &= 2i \sin \theta \sum_{r=1}^n 2 \cos (2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2\sin \theta \sum_{r=1}^n \cos (2r-1) \theta \end{align*}
    2. When \(n = 2, \theta = \frac{\pi}{5}\) we have: \begin{align*} &&\sin \frac{4\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}) \\ &&\sin \frac{\pi}{5} &= 2 \sin \frac{\pi}{5} (\cos \frac{\pi}{5} - \cos \frac{2\pi}{5}) \\ &&\frac12 &= \cos \frac{\pi}{5} - \cos \frac{2 \pi}{5} \\ \Rightarrow && \cos \frac{2\pi}{5} &= \cos \frac{\pi}{5} - \frac12 \end{align*}
    3. When \(n = 7, \theta = \frac{\pi}{15}\) we have: \begin{align*} && \sin \frac{14 \pi}{15} &= 2 \sin \frac{\pi}{15} \sum_{r=1}^7 \cos (2r-1) \frac{\pi}{15} \\ \Rightarrow && \frac12 &= \cos \frac{\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}+ \cos \frac{7 \pi}{15}+ \cos \frac{9 \pi}{15}+ \cos \frac{11 \pi}{15}+ \cos \frac{13 \pi}{15} \\ &&&= -\cos \frac{16\pi}{15} + \cos \frac{3 \pi}{15} + \cos \frac{5 \pi}{15}- \cos \frac{8 \pi}{15}+ \cos \frac{9 \pi}{15}- \cos \frac{4 \pi}{15}- \cos \frac{2\pi}{15} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \cos \frac{\pi}{5} + \cos \frac{\pi}{3} + \cos \frac{3 \pi}{5} \\ &&&= - \left ( \cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15}\right) + \frac12 + \frac12 \\ \Rightarrow && \frac12 &= cos \frac{2\pi}{15}+\cos \frac{4\pi}{15}+\cos \frac{8\pi}{15}+\cos \frac{16\pi}{15} \end{align*}
  2. By using \(z = e^{i \theta}\) we have that: \begin{align*} && z^{2n}-\frac{1}{z^{2n}} &= \left ( z + \frac{1}{z} \right) \sum_{r=1}^n \left (z^{2r-1}-\frac{1}{z^{2r-1}} \right ) \\ \Rightarrow && e^{2n \theta i} - e^{-2n \theta i} &= (e^{\theta i} + e^{-\theta i}) \sum_{r=1}^n (e^{(2r-1)\theta i} - e^{(2r-1) \theta i}) \\ \Rightarrow && 2i \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n 2i \sin(2r-1) \theta \\ \Rightarrow && \sin 2n \theta &= 2 \cos \theta \sum_{r=1}^n \sin(2r-1) \theta \end{align*} When \(n = 3, \theta = \frac{\pi}{14}\) we must have: \begin{align*} &&\sin \frac{3 \pi}{7} &= 2 \cos \frac{\pi}{14}( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \left (\frac{\pi}{2} - \frac{\pi}{14} \right)( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ &&&= 2 \sin \frac{3\pi}{7} ( \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14}) \\ \Rightarrow && \frac12 &= \sin \frac{\pi}{14}+\sin \frac{3\pi}{14}+\sin \frac{5\pi}{14} \end{align*} as required.
2023 Paper 3 Q3
D: 1500.0 B: 1500.0

  1. Show that, if \(a\) and \(b\) are complex numbers, with \(b \neq 0\), and \(s\) is a positive real number, then the points in the Argand diagram representing the complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\) form an isosceles triangle. Given three points which form an isosceles triangle in the Argand diagram, explain with the aid of a diagram how to determine the values of \(a\), \(b\) and \(s\) so that the vertices of the triangle represent complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\).
  2. Show that, if the roots of the equation \(z^3 + pz + q = 0\), where \(p\) and \(q\) are complex numbers, are represented in the Argand diagram by the vertices of an isosceles triangle, then there is a non-zero real number \(s\) such that \[\frac{p^3}{q^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}\,.\]
  3. Sketch the graph \(y = \dfrac{(3x-1)^3}{(9x+1)^2}\), identifying any stationary points.
  4. Show that if the roots of the equation \(z^3 + pz + q = 0\) are represented in the Argand diagram by the vertices of an isosceles triangle then \(\dfrac{p^3}{q^2}\) is a real number and \(\dfrac{p^3}{q^2} > -\dfrac{27}{4}\).

2022 Paper 2 Q7
D: 1500.0 B: 1500.0

Let \(\mathrm{h}(z) = nz^6 + z^5 + z + n\), where \(z\) is a complex number and \(n \geqslant 2\) is an integer.

  1. Let \(w\) be a root of the equation \(\mathrm{h}(z) = 0\).
    1. Show that \(|w^5| = \sqrt{\dfrac{\mathrm{f}(w)}{\mathrm{g}(w)}}\), where \[\mathrm{f}(z) = n^2 + 2n\operatorname{Re}(z) + |z|^2 \quad \text{and} \quad \mathrm{g}(z) = n^2|z|^2 + 2n\operatorname{Re}(z) + 1.\]
    2. By considering \(\mathrm{f}(w) - \mathrm{g}(w)\), prove by contradiction that \(|w| \geqslant 1\).
    3. Show that \(|w| = 1\).
  2. It is given that the equation \(\mathrm{h}(z) = 0\) has six distinct roots, none of which is purely real.
    1. Show that \(\mathrm{h}(z)\) can be written in the form \[\mathrm{h}(z) = n(z^2 - a_1 z + 1)(z^2 - a_2 z + 1)(z^2 - a_3 z + 1),\] where \(a_1\), \(a_2\) and \(a_3\) are real constants.
    2. Find \(a_1 + a_2 + a_3\) in terms of \(n\).
    3. By considering the coefficient of \(z^3\) in \(\mathrm{h}(z)\), find \(a_1 a_2 a_3\) in terms of \(n\).
    4. How many of the six roots of the equation \(\mathrm{h}(z) = 0\) have a negative real part? Justify your answer.

2021 Paper 3 Q7
D: 1500.0 B: 1500.0

  1. Let \[ z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}}, \] where \(\theta\) and \(\phi\) are real, and \(\theta - \phi \neq 2n\pi\) for any integer \(n\). Show that \[ z = i\cot\!\bigl(\tfrac{1}{2}(\phi - \theta)\bigr) \] and give expressions for the modulus and argument of \(z\).
  2. The distinct points \(A\) and \(B\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\) and \(B\) are represented by the complex numbers \(a\) and \(b\), and \(O\) is at the origin. The point \(X\) is represented by the complex number \(x\), where \(x = a + b\) and \(a + b \neq 0\). Show that \(OX\) is perpendicular to \(AB\). If the distinct points \(A\), \(B\) and \(C\) in the complex plane, which are represented by the complex numbers \(a\), \(b\) and \(c\), lie on a circle with radius \(1\) and centre \(O\), and \(h = a + b + c\) represents the point \(H\), then \(H\) is said to be the orthocentre of the triangle \(ABC\).
  3. The distinct points \(A\), \(B\) and \(C\) lie on a circle with radius \(1\) and centre \(O\). In the complex plane, \(A\), \(B\) and \(C\) are represented by the complex numbers \(a\), \(b\) and \(c\), and \(O\) is at the origin. Show that, if the point \(H\), represented by the complex number \(h\), is the orthocentre of the triangle \(ABC\), then either \(h = a\) or \(AH\) is perpendicular to \(BC\).
  4. The distinct points \(A\), \(B\), \(C\) and \(D\) (in that order, anticlockwise) all lie on a circle with radius \(1\) and centre \(O\). The points \(P\), \(Q\), \(R\) and \(S\) are the orthocentres of the triangles \(ABC\), \(BCD\), \(CDA\) and \(DAB\), respectively. By considering the midpoint of \(AQ\), show that there is a single transformation which maps the quadrilateral \(ABCD\) onto the quadrilateral \(QRSP\) and describe this transformation fully.

Show Solution
  1. \(\,\) \begin{align*} && z &= \frac{e^{i \theta} + e^{i \phi}}{e^{i \theta} - e^{i \phi}} \\ &&&= \frac{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})}{e^{i\frac12(\theta +\phi)}(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})} \\ &&&= \frac{(e^{i \frac12(\theta-\phi)} + e^{-i\frac12(\theta- \phi)})/2}{i(e^{i \frac12(\theta-\phi)} - e^{-i\frac12(\theta- \phi)})/2i} \\ &&&= \frac{\cos \frac12(\theta-\phi)}{i \sin \frac12(\theta-\phi)} \\ &&&= -i \cot \tfrac12(\theta-\phi) \end{align*} Therefore \(|z| = \cot \tfrac12(\theta-\phi)\) and \(\arg z = \frac{\pi}{2}\)
  2. Since \(a,b\) lie on the unit circle, wlog \(a = e^{i \theta}, b = e^{i\phi}\). Not that the line \(OX\) has vector \(a+b\) and \(AB\) has vector \(b-a\) and not their ratio has argument \(\frac{\pi}{2}\) and hence they are perpendicular.
  3. \(AH\) has vector \(h - a = (a+b+c) - a = b+c\) which we've already established is perpendicular to \(c-b\) which is the vector for \(BC\) (unless \(b+c = 0\) in which case \(h = a\)).
  4. \(p = a +b+c, q = b+c+d\) etc. The midpoint of \(AQ = \frac12(a+b+c+d)\) which is the same as the midpoint of \(BR\), \(CS\) and \(DP\). Therefore we could say the transformation is reflection in the point \(\frac12(a+b+c+d)\)
2020 Paper 2 Q7
D: 1500.0 B: 1500.0

In this question, \(w = \dfrac{2}{z-2}\).

  1. Let \(z\) be the complex number \(3 + t\mathrm{i}\), where \(t \in \mathbb{R}\). Show that \(|w - 1|\) is independent of \(t\). Hence show that, if \(z\) is a complex number on the line \(\operatorname{Re}(z) = 3\) in the Argand diagram, then \(w\) lies on a circle in the Argand diagram with centre \(1\). Let \(V\) be the line \(\operatorname{Re}(z) = p\), where \(p\) is a real constant not equal to \(2\). Show that, if \(z\) lies on \(V\), then \(w\) lies on a circle whose centre and radius you should give in terms of \(p\). For which \(z\) on \(V\) is \(\operatorname{Im}(w) > 0\)?
  2. Let \(H\) be the line \(\operatorname{Im}(z) = q\), where \(q\) is a non-zero real constant. Show that, if \(z\) lies on \(H\), then \(w\) lies on a circle whose centre and radius you should give in terms of \(q\). For which \(z\) on \(H\) is \(\operatorname{Re}(w) > 0\)?

2020 Paper 3 Q3
D: 1500.0 B: 1500.0

Given distinct points \(A\) and \(B\) in the complex plane, the point \(G_{AB}\) is defined to be the centroid of the triangle \(ABK\), where the point \(K\) is the image of \(B\) under rotation about \(A\) through a clockwise angle of \(\frac{1}{3}\pi\). Note: if the points \(P\), \(Q\) and \(R\) are represented in the complex plane by \(p\), \(q\) and \(r\), the centroid of triangle \(PQR\) is defined to be the point represented by \(\frac{1}{3}(p+q+r)\).

  1. If \(A\), \(B\) and \(G_{AB}\) are represented in the complex plane by \(a\), \(b\) and \(g_{ab}\), show that \[ g_{ab} = \frac{1}{\sqrt{3}}(\omega a + \omega^* b), \] where \(\omega = \mathrm{e}^{\frac{\mathrm{i}\pi}{6}}\).
  2. The quadrilateral \(Q_1\) has vertices \(A\), \(B\), \(C\) and \(D\), in that order, and the quadrilateral \(Q_2\) has vertices \(G_{AB}\), \(G_{BC}\), \(G_{CD}\) and \(G_{DA}\), in that order. Using the result in part (i), show that \(Q_1\) is a parallelogram if and only if \(Q_2\) is a parallelogram.
  3. The triangle \(T_1\) has vertices \(A\), \(B\) and \(C\) and the triangle \(T_2\) has vertices \(G_{AB}\), \(G_{BC}\) and \(G_{CA}\). Using the result in part (i), show that \(T_2\) is always an equilateral triangle.

Show Solution
  1. Note that the vector \(\overrightarrow{AB}\) is \(b-a\), and if we rotate this by \(\frac13\pi\) we get \(e^{-i\pi/3}(b-a)\) after rotating it. Therefore the point \(K\) is represented by \(a + e^{-i\pi/3}(b-a)\) and so \(G_{AB}\) is \begin{align*} && g_{ab} &= \tfrac13(a + b + a + e^{-i\pi/3}(b-a)) \\ &&&= \tfrac13((1+ e^{-i\pi/3})b+(2-e^{-i\pi/3})a)\\ &&&= \tfrac13((1+\tfrac12 - \tfrac{\sqrt3}{2}i)b + ((2-\tfrac12+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac13((\tfrac32 - \tfrac{\sqrt3}{2}i)b + ((\tfrac32+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac1{\sqrt3}((\tfrac{\sqrt3}2 - \tfrac{1}{2}i)b + ((\tfrac{\sqrt3}2+\tfrac{1}{2}i)a) \\ &&&= \frac{1}{\sqrt3}(\omega^* b + \omega a) \end{align*}
  2. First note that \(Q_1\) is a parallelogram iff \(c - a = (b-a) + (d-a)\) ie \(a + c = b+d\) (indeed this is true for all quadrilaterals), so. \begin{align*} && Q_1 &\text{ is a parallelogram} \\ \Longleftrightarrow && a + b &= c + d \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega - \omega^*)(a + c) &= \frac{1}{\sqrt{3}}(\omega -\omega^*)(b + d) \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega a + \omega^*b)+\frac{1}{\sqrt{3}}(\omega c + \omega^*d) &=\frac{1}{\sqrt{3}}(\omega b + \omega^*c)+\frac{1}{\sqrt{3}}(\omega d + \omega^*a) \\ \Longleftrightarrow && g_{ab}+g_{cd} &=g_{bc}+g_{da} \\ \Longleftrightarrow && Q_2 &\text{ is a parallelogram} \\ \end{align*}
  3. We consider \(\frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}}\) so \begin{align*} && \frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}} &= \frac{(\omega a + \omega^*b)-(\omega b + \omega^* c)}{(\omega c + \omega^*a)-(\omega b + \omega^* c)} \\ &&&= \frac{\omega a- \omega^* c -(\omega- \omega^*)b }{\omega^*a-\omega b -(\omega^* -\omega )c} \\ &&&= \frac{\omega^2 a- c -(\omega^2- 1)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- \omega^4 c -(1- \omega^4)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- (1-\omega^2) c -\omega^2b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&= \omega^2 \end{align*} Therefore the triangle is equilateral.
2018 Paper 3 Q7
D: 1700.0 B: 1516.0

  1. Use De Moivre's theorem to show that, if \(\sin\theta\ne0\)\,, then \[ \frac{ \left(\cot \theta + \rm{i}\right)^{2n+1} -\left(\cot \theta - \rm{i}\right)^{2n+1}}{2\rm{i}} = \frac{\sin \left(2n+1\right)\theta} {\sin^{2n+1}\theta} \,, \] for any positive integer \(n\). Deduce that the solutions of the equation \[ \binom{2n+1}{1}x^{n}-\binom{2n+1}{3}x^{n-1} +\cdots + \left(-1\right)^{n}=0 \] are \[x=\cot^{2}\left(\frac{m\pi}{2n+1}\right) \] where \( m=1\), \(2\), \(\ldots\) , \(n\,\).
  2. Hence show that \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
  3. Given that \(0<\sin \theta <\theta <\tan \theta\) for \(0 < \theta < \frac{1}{2}\pi\), show that \[ \cot^{2}\theta<\frac{1}{\theta^{2}}<1+\cot^{2}\theta. \] Hence show that \[ \sum^\infty_{m=1} \frac{1}{m^2}= \frac{\pi^2}{6}\,.\]

Show Solution
  1. \begin{align*} \frac{\left(\cot \theta + i\right)^{2n+1} -\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1} -\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\ &= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\ &=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} \end{align*} Notice that: \begin{align*} (\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\ &= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\ \end{align*} Therefore if \(\theta\) satisfies \(\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0\) then \(z = \cot^2 \theta\) satisfies the equation. But \(\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n\) are \(n\) distinct all the roots must be \(\cot^2 \frac{m \pi}{2n+1}\).
  2. Notice that the sum of the roots will be \(\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}\) and so \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
  3. For \(0 < \theta < \frac{1}{2}\pi\), \begin{align*} && 0 < \sin \theta < \theta < \tan \theta \\ \Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\ \Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\ \end{align*} Therefore \begin{align*} && \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\ \Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\ \Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} \end{align*}
2017 Paper 3 Q2
D: 1700.0 B: 1500.1

The transformation \(R\) in the complex plane is a rotation (anticlockwise) by an angle \(\theta\) about the point represented by the complex number \(a\). The transformation \(S\) in the complex plane is a rotation (anticlockwise) by an angle \(\phi\) about the point represented by the complex number \(b\).

  1. The point \(P\) is represented by the complex number~\(z\). Show that the image of \(P\) under \(R\) is represented by \[ \e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,. \]
  2. Show that the transformation \(SR\) (equivalent to \(R\) followed by \(S\)) is a rotation about the point represented by \(c\), where \[ %\textstyle c\,\sin \tfrac12 (\theta+\phi) = a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi \,, \] provided \(\theta+\phi \ne 2n\pi\) for any integer \(n\). What is the transformation \(SR\) if \(\theta +\phi = 2\pi\)?
  3. Under what circumstances is \(RS =SR\)?

Show Solution
  1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
  2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
  3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.
2016 Paper 3 Q7
D: 1700.0 B: 1516.0

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 \[ |X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,. \]

Show 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*}
2015 Paper 3 Q4
D: 1700.0 B: 1516.0

  1. If \(a\), \(b\) and \(c\) are all real, show that the equation \[ z^3+az^2+bz+c=0 \tag{\(*\)} \] has at least one real root.
  2. Let \[ S_1= z_1+z_2+z_3, \ \ \ \ S_2= z_1^2 + z_2^2 + z_3^2, \ \ \ \ S_3= z_1^3 + z_2^3 + z_3^3\,, \] where \(z_1\), \(z_2\) and \(z_3\) are the roots of the equation \((*)\). Express \(a\) and \(b\) in terms of \(S_1\) and \(S_2\), and show that \[ 6c =- S_1^3 + 3 S_1S_2 - 2S_3\,. \]
  3. The six real numbers \(r_k\) and \(\theta_k\) (\(k=1, \ 2, \ 3\)), where \(r_k>0\) and \(-\pi < \theta_k <\pi\), satisfy \[ \textstyle \sum\limits _{k=1}^3 r_k \sin (\theta_k) = 0\,, \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^2 \sin (2\theta_k) = 0\,, \ \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^3 \sin (3\theta_k) = 0\, . \] Show that \(\theta_k=0\) for at least one value of \(k\). Show further that if \(\theta_1=0\) then \(\theta_2 = - \theta_3\,\).

Show Solution
  1. Let \(z \in \mathbb{R}\) and let \(z \to \pm \infty\) then \(z^3 + az^2 + bz + c\) changes sign, therefore somewhere it must have a real root.
  2. \begin{align*} &&z^3 + az^2 + bz + c &= (z-z_1)(z-z_2)(z-z_3) \\ && &= z^3 - (z_1+z_2+z_3)z^2 + (z_1z_2 + z_2z_3+z_3z_1)z - (z_1z_2z_3) \\ \\ \Rightarrow && S_1 &= z_1+z_2+z_3 \\ &&&= -a \\ \\ \Rightarrow && S_2 &= z_1^2+z_2^2+z_3^2 \\ &&&= (z_1+z_2+z_3)^2 - 2(z_1z_2 + z_2z_3+z_3z_1) \\ &&&= a^2 - 2b \\ \Rightarrow && a &= -S_1 \\ && b &= \frac12 \l S_1^2 - S_2\r \\ \\ && 0 &= z_i^3 + az_i^2+bz_i+c \\ \Rightarrow && 0 &= S_3 + aS_2+bS_1+3c \\ &&&= S_3 -S_1S_2 + \frac12 \l S_1^2 - S_2\r S_1 + 3c \\ \Rightarrow && 0 &= 2S_3 - 3S_1S_2 + S_1^3 + 6c \end{align*}
  3. Let \(z_k= r_ke^{i \theta_k}\), then we have \(\textrm{Im}(S_k) = 0\) and so the polynomial with roots \(z_k\) has real coefficients, and therefore at least one root is real. This root will have \(\theta_k = 0\). Moreover, since if \(w\) is a root of a real polynomial \(\overbar{w}\) is also a root, and therefore if \(\theta_1 = 0\), we must have that \(z_2\) and \(z_3\) are complex conjugate, ie \(\theta_2 = - \theta_3\)
2013 Paper 3 Q4
D: 1700.0 B: 1516.0

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.

Show 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*}
2013 Paper 3 Q8
D: 1700.0 B: 1484.0

Evaluate \(\displaystyle \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)}\) where \(\alpha\) is a fixed angle and \(n\ge2\). The fixed point \(O\) is a distance \(d\) from a fixed line \(D\). For any point \(P\), let \(s\) be the distance from \(P\) to \(D\) and let \(r\) be the distance from \(P\) to \(O\). Write down an expression for \(s\) in terms of \(d\), \(r\) and the angle \(\theta\), where \(\theta\) is as shown in the diagram below.

TikZ diagram
The curve \(E\) shown in the diagram is such that, for any point \(P\) on \(E\), the relation \(r = k s\) holds, where \(k\) is a fixed number with \(0< k <1\). Each of the \(n\) lines \(L_1\), \(\ldots\,\), \(L_n\) passes through \(O\) and the angle between adjacent lines is \(\frac \pi n\). The line \(L_j\) (\(j=1\), \(\ldots\,\), \(n\)) intersects \(E\) in two points forming a chord of length \(l_j\). Show that, for \(n\ge2\), \[ \sum_{j=1}^n \frac 1 {l_j} = \frac {(2-k^2)n} {4kd}\,. \]

Show Solution
\begin{align*} \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)} &= e^{2i\alpha} \sum_{r=0}^{n-1} \left (\e^{2i\pi/n} \right)^r \\ &= e^{2i\alpha} \frac{1-\left (\e^{2i\pi/n} \right)^n}{1-\e^{2i\pi/n} } \\ &= 0 \end{align*} \(d = s + r \cos \theta\) ie \(s = d - r \cos \theta\) Therefore \(d = \frac{r}{k} + r \cos \theta \Rightarrow r = \frac{kd}{1+k \cos \theta}\). The \(l_j\) will come from \(r(\alpha + \frac{j \pi}{n} )+r(\alpha + \pi + \frac{j \pi}{n} )\) \begin{align*} && l_j &= r(\alpha + \frac{(j-1) \pi}{n} )+r(\alpha + \pi + \frac{(j-1) \pi}{n} ) \\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1+k \cos \left ( \alpha+\pi+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{kd}{1+k \cos \left ( \alpha + \frac{(j-1) \pi}{n}\right)}+\frac{kd}{1-k \cos \left ( \alpha+ \frac{(j-1) \pi}{n}\right)}\\ &&&= \frac{2kd}{1-k^2 \cos^2 \left ( \alpha + \frac{(j-1) \pi}{n}\right)}\\ \Rightarrow && \sum_{j=1}^n \frac 1 {l_j} &= \sum_{j=0}^{n-1} \frac{1-k^2 \cos^2 \left ( \alpha + \frac{j \pi}{n}\right)}{2kd} \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \cos^2 \left ( \alpha + \frac{j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{k^2}{2kd} \sum_{j=0}^{n-1} \frac{1+ \cos \left ( 2\alpha + \frac{2j \pi}{n}\right)}{2} \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \sum_{j=0}^{n-1}\cos \left ( 2\alpha + \frac{2j \pi}{n}\right) \\ &&&= \frac{n}{2kd}-\frac{nk^2}{2kd}-\frac{k^2}{4kd} \underbrace{\textrm{Re} \left ( \sum_{j=0}^{n-1}e^{ 2i(\alpha + \frac{j \pi}{n})} \right)}_{=0} \\ &&&= \frac{n}{2kd} - \frac{nk^2}{4kd} \\ &&&= \frac{n(2-k^2)}{4kd} \end{align*}
2011 Paper 3 Q3
D: 1700.0 B: 1484.0

Show that, provided \(q^2\ne 4p^3\), the polynomial \[ \hphantom{(p\ne0, \ q\ne0)\hspace{2cm}} x^3-3px +q \hspace {2cm} (p\ne0, \ q\ne0) \] can be written in the form \[ a(x-\alpha)^3 + b(x-\beta)^3\,, \] where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(pt^2 -qt +p^2=0\), and \(a\) and \(b\) are constants which you should express in terms of \(\alpha\) and \(\beta\). Hence show that one solution of the equation \(x^3-24x+48=0\,\) is \[ x= \frac{2 (2-2^{\frac13})}{1-2^{\frac13}} \] and obtain similar expressions for the other two solutions in terms of \(\omega\), where \(\omega = \mathrm{e}^{2\pi\mathrm{i}/3}\,\). Find also the roots of \(x^3-3px +q=0\) when \(p=r^2\) and \(q= 2r^3\) for some non-zero constant \(r\).

2010 Paper 3 Q3
D: 1700.0 B: 1545.2

For any given positive integer \(n\), a number \(a\) (which may be complex) is said to be a primitive \(n\)th root of unity if \(a^n=1\) and there is no integer \(m\) such that \(0 < m < n\) and \(a^m = 1\). Write down the two primitive 4th roots of unity. Let \({\rm C}_n(x)\) be the polynomial such that the roots of the equation \({\rm C}_n(x)=0\) are the primitive \(n\)th roots of unity, the coefficient of the highest power of \(x\) is one and the equation has no repeated roots. Show that \({\rm C}_4(x) = x^2+1\,\).

  1. Find \({\rm C}_1(x)\), \({\rm C}_2(x)\), \({\rm C}_3(x)\), \({\rm C}_5(x)\) and \({\rm C}_6(x)\), giving your answers as unfactorised polynomials.
  2. Find the value of \(n\) for which \({\rm C}_n(x) = x^4 + 1\).
  3. Given that \(p\) is prime, find an expression for \({\rm C}_p(x)\), giving your answer as an unfactorised polynomial.
  4. Prove that there are no positive integers \(q\), \(r\) and \(s\) such that \({\rm C}_q(x) \equiv {\rm C}_r(x) {\rm C}_s(x)\,\).

Show Solution
The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.
  1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
  2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
  3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
  4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.
2009 Paper 3 Q6
D: 1700.0 B: 1473.1

Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.

2005 Paper 3 Q6
D: 1700.0 B: 1516.0

In this question, you may use without proof the results \[ 4 \cosh^3 y - 3 \cosh y = \cosh (3y) \ \ \ \ \text{and} \ \ \ \ \mathrm{arcosh} \, y = \ln ( y+\sqrt{y^2-1}). \] \noindent[ {\bf Note: } \(\mathrm{arcosh}y\) is another notation for \(\cosh^{-1}y\,\)] Show that the equation \(x^3 - 3a^2x = 2a^3 \cosh T\) is satisfied by \( 2a \cosh \l \frac13 T \r\) and hence that, if \(c^2\ge b^3>0\), one of the roots of the equation \(x^3-3bx=2c\) is \(\ds u+\frac{b}{u}\), where \(u = (c+\sqrt{c^2-b^3})^{\frac13}\;\). Show that the other two roots of the equation \(x^3-3bx=2c\) are the roots of the quadratic equation \[\ds x^2 + \Big( u+\frac{b}{u}\Big) x + u^2+\frac{b^2}{u^2}-b=0\, ,\] and find these roots in terms of \(u\), \(b\) and \(\omega\), where \(\omega = \frac{1}{2}(-1 + \mathrm{i}\sqrt{3})\). Solve completely the equation \(x^3-6x=6\,\).

2000 Paper 2 Q4
D: 1600.0 B: 1500.0

Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]

Show Solution
\begin{align*} && LHS &= (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) \\ &&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\ &&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\ &&&= RHS \end{align*} Therefore we can see \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta\). \begin{align*} && (5-i)^2(1+i) &= (24-10i)(1+i) \\ &&&= (24+10) + i(24-10) \\ &&&= 34+14i \\ \Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\ \Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\ \Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\ \end{align*} Consider \((1+i)(4-i)^3(20-i)\) \begin{align*} && (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\ &&&= 1985+i \\ \Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right) \end{align*}
2000 Paper 3 Q3
D: 1700.0 B: 1484.0

Given that \(\alpha = \e^{\mathrm{i} \pi/3}\) , prove that \(1 + \alpha^2 = \alpha\). A triangle in the Argand plane has vertices \(A\), \(B\), and \(C\) represented by the complex numbers \(p\), \(q\alpha^2\) and \(- r\alpha\) respectively, where \(p\), \(q\) and \(r\) are positive real numbers. Sketch the triangle~\(ABC\). Three equilateral triangles \(ABL\), \(BCM\) and \(CAN\) (each lettered clockwise) are erected on sides \(AB\), \(BC\) and \(CA\) respectively. Show that the complex number representing \(N\) is \mbox{\(( 1 - \alpha) p- \alpha^2 r\)} and find similar expressions for the complex numbers representing \(L\) and \(M\). Show that lines \(LC\), \(MA\) and \(NB\) all meet at the origin, and that these three line segments have the common length \(p+q+r\).

1997 Paper 3 Q3
D: 1700.0 B: 1504.6

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 \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]

1996 Paper 3 Q5
D: 1700.0 B: 1516.0

Show, using de Moivre's theorem, or otherwise, that \[ \tan7\theta=\frac{t(t^{6}-21t^{4}+35t^{2}-7)}{7t^{6}-35t^{4}+21t^{2}-1}\,, \] where \(t=\tan\theta.\)

  1. By considering the equation \(\tan7\theta=0,\) or otherwise, obtain a cubic equation with integer coefficients whose roots are \[ \tan^{2}\left(\frac{\pi}{7}\right),\ \tan^{2}\left(\frac{2\pi}{7}\right)\ \mbox{ and }\tan^{2}\left(\frac{3\pi}{7}\right) \] and deduce the value of \[ \tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right)\,. \]
  2. Find, without using a calculator, the value of \[ \tan^{2}\left(\frac{\pi}{14}\right)+\tan^{2}\left(\frac{3\pi}{14}\right)+\tan^{2}\left(\frac{5\pi}{14}\right)\,. \]

1995 Paper 1 Q4
D: 1484.0 B: 1500.0

By applying de Moivre's theorem to \(\cos5\theta+\mathrm{i}\sin5\theta,\) expanding the result using the binomial theorem, and then equating imaginary parts, show that \[ \sin5\theta=\sin\theta\left(16\cos^{4}\theta-12\cos^{2}\theta+1\right). \] Use this identity to evaluate \(\cos^{2}\frac{1}{5}\pi\), and deduce that \(\cos\frac{1}{5}\pi=\frac{1}{4}(1+\sqrt{5}).\)

Show Solution
\begin{align*} && (\cos \theta + i \sin \theta)^n &= \cos n \theta + i \sin n \theta \\ n = 5: && \cos 5 \theta + i \sin 5 \theta &= (\cos \theta + i \sin \theta)^5 \\ \textrm{Im}: && \sin 5 \theta &= \binom{5}{1}\cos^4 \theta \sin \theta + \binom{5}{3} \cos^2 \theta (- \sin^3 \theta) + \binom{5}{5} \sin^5 \theta \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta \sin^2 \theta+\sin^4 \theta) \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta (1-\cos^2 \theta)+(1-\cos^2 \theta)^2) \\ &&&= \sin \theta((5+10+1)\cos^4 \theta +(-10-2)\cos^2 \theta + 1) \\ &&&= \sin \theta(16\cos^4 \theta -12\cos^2 \theta + 1) \\ \end{align*} Suppose \(\theta= \frac{\pi}{5}\), then \(\sin 5 \theta = 0, \sin \theta \neq 0\), therefore if \(c = \cos \theta\) we must have \begin{align*} && 0 &= 16c^4-12c^2+1 \\ \Rightarrow && c^2 &= \frac{3 \pm \sqrt{5}}{8} \\ &&&= \frac{6\pm 2\sqrt{5}}{16} \\ &&&= \frac{(1 \pm \sqrt{5})^2}{16} \\ \Rightarrow && c &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*} Since \(c > 0\) we either have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\) or \(\cos \frac15 \pi = \frac{\sqrt{5}-1}4\), however \(\sqrt{5}-1 < 1.5\) and so \(\frac{\sqrt{5}-1}{4} < \frac12 = \cos \frac13 \pi\) we must have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\)
1995 Paper 2 Q6
D: 1600.0 B: 1516.0

If \(u\) and \(v\) are the two roots of \(z^{2}+az+b=0,\) show that \(a=-u-v\) and \(b=uv.\) Let \(\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).\) Show that \(\alpha\) is a root of \(z^{6}-1=0\) and express the roots in terms of \(\alpha.\) The number \(\alpha+\alpha^{2}+\alpha^{4}\) is a root of a quadratic equation \[ z^{2}+Az+B=0 \] where \(A\) and \(B\) are real. By guessing the other root, or otherwise, find the numerical values of \(A\) and \(B\). Show that \[ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2}, \] and evaluate \[ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}, \] making it clear how you determine the sign of your answer.

Show Solution
\begin{align*} 0 &= z^2+az+b \\ &= (z-u)(z-v) \\ &= z^2-(u+v)z+uv \end{align*} Therefore by comparing coefficients, \(a = -u-v\) and \(b = uv\). Suppose \(\alpha = \cos(2\pi/7) + i \sin (2\pi/7)\), then by De Moivre, \(\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1\), ie \(\alpha^7-1 = 0\). Notice that \((\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1\) and \begin{align*} P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6) \\ &= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\ &= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\ &= 2 \end{align*} Therefore it is a root of \(x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}\) Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$ And \(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2\) since it is positive it is \(\frac{\sqrt{7}}{2}\)
1992 Paper 3 Q8
D: 1700.0 B: 1515.1

Show that \[ \sin(2n+1)\theta=\sin^{2n+1}\theta\sum_{r=0}^{n}(-1)^{n-r}\binom{2n+1}{2r}\cot^{2r}\theta, \] where \(n\) is a positive integer. Deduce that the equation \[ \sum_{r=0}^{n}(-1)^{r}\binom{2n+1}{2r}x^{r}=0 \] has roots \(\cot^{2}(k\pi/(2n+1))\) for \(k=1,2,\ldots,n\). Show that

  • sep}{3mm}
  • \(\bf (i)\) \({\displaystyle \sum_{k=1}^{n}\cot^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3}},\)
  • \(\bf (ii)\) \({\displaystyle \sum_{k=1}^{n}\tan^{2}\left(\frac{k\pi}{2n+1}\right)=n(2n+1)},\)
  • \(\bf (iii)\) \({\displaystyle \sum_{k=1}^{n}\mathrm{cosec}^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{2n(n+1)}{3}}.\)

1991 Paper 1 Q3
D: 1516.0 B: 1500.0

A path is made up in the Argand diagram of a series of straight line segments \(P_{1}P_{2},\) \(P_{2}P_{3},\) \(P_{3}P_{4},\ldots\) such that each segment is \(d\) times as long as the previous one, \((d\neq1)\), and the angle between one segment and the next is always \(\theta\) (where the segments are directed from \(P_{j}\) towards \(P_{j+1}\), and all angles are measured in the anticlockwise direction). If \(P_{j}\) represents the complex number \(z_{j},\) express \[ \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \] as a complex number (for each \(n\geqslant2\)), briefly justifying your answer. If \(z_{1}=0\) and \(z_{2}=1\), obtain an expression for \(z_{n+1}\) when \(n\geqslant2\). By considering its imaginary part, or otherwise, show that if \(\theta=\frac{1}{3}\pi\) and \(d=2\), then the path crosses the real axis infinitely often.

Show Solution
\begin{align*} && | \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} | &= d \\ && \arg \left ( \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \right) &= \arg (z_{n+1}-z_{n}) - \arg(z_{n}-z_{n-1}) \\ &&&= \theta \\ \Rightarrow && \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} &= d e^{i \theta} \end{align*} \begin{align*} && z_1 &= 0 \\ && z_2 &= 1 \\ && \frac{z_3-z_2}{z_2-z_1} &= de^{i \theta} \\ \Rightarrow && z_3 &= de^{i \theta} + 1 \\ && \frac{z_4-z_3}{z_3-z_2} &= de^{i \theta} \\ \Rightarrow && z_4 &= (d e^{i \theta})^2 + d e^{i \theta} + 1\\ \Rightarrow && z_{n+1} &= \frac{(de^{i \theta})^{n}-1}{de^{i \theta}-1} \end{align*} If \(d = 2, \theta = \tfrac13 \pi\), then, \(2e^{i \tfrac13 \pi} = 1 + \sqrt{3}i\) \begin{align*} \textrm{Im}(z_{n+1})) &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{2e^{i \tfrac13 \pi}-1}\right) \\ &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{\sqrt{3}i}\right) \\ &= -\frac{1}{\sqrt{3}}\textrm{Re} \left (2^n e^{i \frac{n}{3} \pi} \right) + \frac{1}{\sqrt{3}} \end{align*} Which clearly changes sign infinitely many times, ie crosses the origin infinitely many times.
1990 Paper 1 Q2
D: 1500.0 B: 1516.0

Let \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}.\) Show that \(1+\omega+\omega^{2}=0\) and calculate the modulus and argument of \(1+\omega^{2}.\) Let \(n\) be a positive integer. By evaluating \((1+\omega^{r})^{n}\) in two ways, taking \(r=1,2\) and \(3\), or otherwise, prove that \[ \binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots+\binom{n}{k}=\frac{1}{3}\left(2^{n}+2\cos\left(\frac{n\pi}{3}\right)\right), \] where \(k\) is the largest multiple of \(3\) less than or equal to \(n\). Without using a calculator, evaluate \[ \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} \] and \[ \binom{24}{2}+\binom{24}{5}+\cdots+\binom{24}{23}\,. \] {[}\(2^{25}=33554432.\){]}

Show Solution
Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}