De Moivre’s theorem and roots of unity – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: write the number in modulus-argument form \(r(\cos\theta+i\sin\theta)\), then use de Moivre: \(\left[r(\cos\theta+i\sin\theta)\right]^n=r^n(\cos n\theta+i\sin n\theta)\). Reduce the angle \(n\theta\) by multiples of \(2\pi\) before evaluating. Mark scheme as shown in brackets.

   1. \(1-i\) is in the fourth quadrant, at angle \(\tfrac\pi4\) below the real axis. \(|1-i|=\sqrt2\), \(\arg=-\tfrac\pi4\) [1]: \(\sqrt2\left(\cos(-\tfrac\pi4)+i\sin(-\tfrac\pi4)\right)\) [1]
   2. Raise the modulus to the power 10 and multiply the argument by 10. \((\sqrt2)^{10}=32\) [1] and angle \(-\tfrac{10\pi}4=-\tfrac{5\pi}2\equiv-\tfrac\pi2\), so \(32(\cos(-\tfrac\pi2)+i\sin(-\tfrac\pi2))=\) \(-32i\) [1] Since \(-\tfrac{5\pi}2+2\pi=-\tfrac\pi2\), the answer is \(-32i\).

2. Answer to Question 2 (5 marks)

   Method: with \(z=\cos\theta+i\sin\theta\), use \(z^n+\dfrac1{z^n}=2\cos n\theta\) and \(z^n-\dfrac1{z^n}=2i\sin n\theta\). Expand with the binomial theorem, pair the terms \(z^k\) and \(z^{-k}\), and compare with the power of \(2\cos\theta\) or \(2i\sin\theta\). Mark scheme as shown in brackets.

   1. \(\left(z-\tfrac1z\right)^4=z^4-4z^2+6-4z^{-2}+z^{-4}\) [1]. This is \(2\cos4\theta-8\cos2\theta+6\) [1]. The left side is \(16\sin^4\theta\), giving the result [1].
   2. \(\tfrac18\left[\tfrac{\sin4\theta}4-2\sin2\theta+3\theta\right]_0^{\pi/4}\) [1] \(=\tfrac18\left(0-2+\tfrac{3\pi}4\right)=\) \(\dfrac{3\pi-8}{32}\) [1]

   Answer: \(\displaystyle\int_0^{\pi/4}\sin^4\theta\,d\theta=\dfrac{3\pi-8}{32}\approx0.0445\).

   In (b), the term \(\tfrac{\sin4\theta}4\) vanishes at both limits because \(\sin\pi=0\).

3. Answer to Question 3 (5 marks)

   Method: write the right-hand side in modulus-argument form, take the cube root of the modulus, and add multiples of \(\tfrac{2\pi}3\) to the argument to get all three roots. Mark scheme as shown in brackets.

   1. \(-27i=27\left(\cos(-\tfrac\pi2)+i\sin(-\tfrac\pi2)\right)\), so \(r=3\) [1] and \(\theta=-\tfrac\pi6+\tfrac{2\pi k}3\) [1]: \(3\left(\cos(-\tfrac\pi6)+i\sin(-\tfrac\pi6)\right)\), \(3\left(\cos\tfrac\pi2+i\sin\tfrac\pi2\right)\), \(3\left(\cos(-\tfrac{5\pi}6)+i\sin(-\tfrac{5\pi}6)\right)\) [1]
   2. Equilateral triangle on a circle of radius 3 [1]. Area \(=\tfrac12\cdot3\cdot3\cdot\sin120^\circ\times3=\dfrac{27\sqrt3}4\approx\) 11.7 [1]

   The three roots are equally spaced at \(120^\circ\) on the circle of radius 3, so they form an equilateral triangle.

4. Answer to Question 4 (3 marks)

   Method: replace \(\cos3\theta\) by \(4c^3-3c\) where \(c=\cos\theta\), collect on one side and factorise completely. Then solve each factor in the interval \(0\le\theta\le\pi\). Mark scheme as shown in brackets.

   1. \(4c^3-3c=c\) gives \(4c^3-4c=0\) [1], \(4c(c-1)(c+1)=0\) [1]. So \(\cos\theta=0,1,-1\) and \(\theta=\) \(0,\ \tfrac\pi2,\ \pi\) [1]

   Check: \(\cos\tfrac{3\pi}2=0=\cos\tfrac\pi2\), \(\cos0=1\) and \(\cos3\pi=-1=\cos\pi\).

   The common error is dividing both sides by \(\cos\theta\), which loses \(\theta=\tfrac\pi2\). There are no other solutions, because each of \(\cos\theta=0\), \(1\) and \(-1\) has exactly one solution in \([0,\pi]\).

5. Answer to Question 5 (3 marks)

   Method: the \(n\)th roots of unity are \(e^{2k\pi i/n}\) for \(k=0,1,\dots,n-1\); scaling by 5 multiplies every modulus by 5. Mark scheme as shown in brackets.

   1. For sixth roots, the angles are multiples of \(\tfrac{2\pi}6=\tfrac\pi3\). \(5e^{ik\pi/3}\), \(k=0,\dots,5\) [1]
   2. Adjacent corners and the centre form an equilateral triangle, since the angle at the centre is \(\tfrac\pi3\). The hexagon is made of six such triangles. The side equals the radius: 5 cm [1]. Area \(=6\times\tfrac12\times5\times5\times\sin60^\circ=\tfrac{75\sqrt3}2\approx\) 65.0 cm\(^2\) [1]

Total : 20 marks