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

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1. Answer to Question 1

   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.

   1. The number is already in modulus-argument form with \(r=1\), so multiply the angle by 5. \(\cos\tfrac{5\pi}6+i\sin\tfrac{5\pi}6=\) \(-\dfrac{\sqrt3}2+\dfrac12i\) This point is on the unit circle in the second quadrant.
   2. \(|1+i|=\sqrt2\) and \(\arg(1+i)=\tfrac\pi4\); the angle becomes \(8\times\tfrac\pi4=2\pi\). \(1+i=\sqrt2\left(\cos\tfrac\pi4+i\sin\tfrac\pi4\right)\), so \((1+i)^8=(\sqrt2)^8(\cos2\pi+i\sin2\pi)=16\times1=\) 16 So \((1+i)^8=16\), a real number.
   3. \(|\sqrt3-i|=2\) and the point is in the fourth quadrant. \(\sqrt3-i=2\left(\cos(-\tfrac\pi6)+i\sin(-\tfrac\pi6)\right)\), so the sixth power is \(2^6(\cos(-\pi)+i\sin(-\pi))=64\times(-1)=\) \(-64\) So \((\sqrt3-i)^6=-64\).
   4. \(|-1+i|=\sqrt2\) and the point is in the second quadrant; \((\sqrt2)^5=4\sqrt2\). \(-1+i=\sqrt2\left(\cos\tfrac{3\pi}4+i\sin\tfrac{3\pi}4\right)\). The fifth power is \(4\sqrt2\left(\cos\tfrac{15\pi}4+i\sin\tfrac{15\pi}4\right)=4\sqrt2\left(\cos(-\tfrac\pi4)+i\sin(-\tfrac\pi4)\right)=4\sqrt2\cdot\tfrac{\sqrt2}2(1-i)=\) \(4-4i\) Since \(\tfrac{15\pi}4-4\pi=-\tfrac\pi4\), the result is \(4-4i\).

2. Answer to Question 2

   Method: in exponential form \(re^{i\theta}\), the modulus is \(r\) and the argument is \(\theta\). Multiplying multiplies the moduli and adds the exponents, exactly like the laws of indices.

   1. Use \(\cos\tfrac\pi3=\tfrac12\) and \(\sin\tfrac\pi3=\tfrac{\sqrt3}2\). \(3\left(\tfrac12+\tfrac{\sqrt3}2i\right)=\) \(\dfrac32+\dfrac{3\sqrt3}2i\)
   2. \(1-i\) is in the fourth quadrant with reference angle \(\tfrac\pi4\). \(|1-i|=\sqrt2\), \(\arg=-\tfrac\pi4\), so \(\sqrt2\,e^{-i\pi/4}\)
   3. Multiply moduli and add exponents: \(6e^{i(\pi/4+\pi/6)}=\) \(6e^{5i\pi/12}\). Divide: \(3e^{i(\pi/2-\pi/3)}=\) \(3e^{i\pi/6}\).
   4. Use the laws of indices, then Euler's formula. \(\left(e^{i\pi/6}\right)^{12}=e^{2\pi i}=\cos2\pi+i\sin2\pi=\) 1. Also \(e^{i\pi}=\cos\pi+i\sin\pi=-1\), so \(e^{i\pi}+1=\) 0. The second result links the five most important numbers \(0\), \(1\), \(e\), \(i\) and \(\pi\).

3. Answer to Question 3

   Method: by de Moivre, \((\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta\). Expanding the left-hand side with the binomial theorem and comparing real and imaginary parts gives multiple angle formulae; use \(s^2=1-c^2\) or \(c^2=1-s^2\) to keep one function.

   1. \((c+is)^3=c^3+3c^2(is)+3c(is)^2+(is)^3=c^3+3ic^2s-3cs^2-is^3\)
   2. Real part: \(c^3-3cs^2=c^3-3c(1-c^2)=\) \(4c^3-3c\). By de Moivre the real part is also \(\cos3\theta\).
   3. Imaginary part: \(3c^2s-s^3=3(1-s^2)s-s^3=\) \(3s-4s^3\), which equals \(\sin3\theta\).
   4. \(\theta=\tfrac\pi6\): \(\cos\tfrac\pi2=0\) and \(4\left(\tfrac{\sqrt3}2\right)^3-3\cdot\tfrac{\sqrt3}2=\tfrac{3\sqrt3}2-\tfrac{3\sqrt3}2=0\). \(\sin\tfrac\pi2=1\) and \(3\cdot\tfrac12-4\cdot\tfrac18=1\). Both correct.

4. Answer to Question 4

   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 \(2i\sin\theta\).

   1. Use the binomial coefficients 1, 3, 3, 1 and pair \(z^3\) with \(\tfrac1{z^3}\) and \(3z\) with \(\tfrac3z\). \(\left(z+\tfrac1z\right)^3=z^3+3z+\tfrac3z+\tfrac1{z^3}=\left(z^3+\tfrac1{z^3}\right)+3\left(z+\tfrac1z\right)=2\cos3\theta+6\cos\theta\). The left side is \(8\cos^3\theta\). So \(\cos^3\theta=\tfrac14(\cos3\theta+3\cos\theta)\). Divide both sides by 8 to obtain the result.
   2. Use the coefficients 1, 4, 6, 4, 1 with alternating signs; \((2i)^4=16\). \(\left(z-\tfrac1z\right)^4=z^4-4z^2+6-\tfrac4{z^2}+\tfrac1{z^4}=2\cos4\theta-8\cos2\theta+6\). The left side is \((2i\sin\theta)^4=16\sin^4\theta\). So \(\sin^4\theta=\tfrac18(\cos4\theta-4\cos2\theta+3)\). This formula lets us integrate \(\sin^4\theta\), which cannot be done directly.
   3. Integrate each cosine term; at \(\tfrac\pi2\) the sine terms are \(\sin2\pi=0\) and \(\sin\pi=0\). \(\tfrac18\left[\tfrac{\sin4\theta}4-2\sin2\theta+3\theta\right]_0^{\pi/2}=\tfrac18\left(0-0+\tfrac{3\pi}2\right)=\) \(\dfrac{3\pi}{16}\) So the integral equals \(\tfrac{3\pi}{16}\approx0.589\).

5. Answer to Question 5

   1. \(8=8(\cos0+i\sin0)\), so \(z=2\left(\cos\tfrac{2\pi k}3+i\sin\tfrac{2\pi k}3\right)\) for \(k=0,1,2\): \(2\), \(-1+\sqrt3\,i\), \(-1-\sqrt3\,i\)
   2. \(-8i=8\left(\cos(-\tfrac\pi2)+i\sin(-\tfrac\pi2)\right)\), so \(z=2\left(\cos\left(-\tfrac\pi6+\tfrac{2\pi k}3\right)+i\sin(\dots)\right)\): arguments \(-\tfrac\pi6,\tfrac\pi2,\tfrac{7\pi}6\). The roots are \(\sqrt3-i\), \(2i\), \(-\sqrt3-i\).
   3. Each set of roots forms an equilateral triangle centred on the origin, so the sum is 0 in both cases (there is no \(z^2\) term, so the sum of the roots is 0).

6. Answer to Question 6

   1. \(\omega^k=\cos\tfrac{k\pi}3+i\sin\tfrac{k\pi}3\): \(1\), \(\tfrac12+\tfrac{\sqrt3}2i\), \(-\tfrac12+\tfrac{\sqrt3}2i\), \(-1\), \(-\tfrac12-\tfrac{\sqrt3}2i\), \(\tfrac12-\tfrac{\sqrt3}2i\)
   2. \(\omega^2=-\tfrac12+\tfrac{\sqrt3}2i\) and \(\omega-1=-\tfrac12+\tfrac{\sqrt3}2i\). They are equal. \(\omega^3=e^{i\pi}=-1\).
   3. Pair each root with its negative: \(1+(-1)=0\), \(\omega+\omega^4=0\), \(\omega^2+\omega^5=0\). The total is 0. (The coefficient of \(z^5\) in \(z^6-1\) is zero.)
   4. The product of the roots of \(z^6-1=0\) is \((-1)^6\times(-1)=\) \(-1\). Check: the arguments sum to \(\tfrac\pi3(0+1+2+3+4+5)=5\pi\), giving \(\cos5\pi=-1\).

7. Answer to Question 7

   1. \(z=5(\cos6+i\sin6)\). \(\cos6\approx0.9602\) and \(\sin6\approx-0.2794\). So \(z\approx\) \(4.80-1.40i\) cm.
   2. One turn needs \(3T=2\pi\), so \(T=\dfrac{2\pi}3\approx\) 2.09 s.
   3. \(\dfrac{dz}{dt}=15ie^{3it}\), and \(|ie^{3it}|=1\) so the speed is \(|15|=\) 15 cm s\(^{-1}\), which does not depend on \(t\).
   4. Multiplying by \(i\) rotates a number through \(90^\circ\). Since \(\dfrac{dz}{dt}=3i\,z\) and \(3\) is a positive real number, the velocity points \(90^\circ\) anticlockwise from the position, which is the direction of the tangent.

8. Answer to Question 8

   1. \(|z|^5=1024\), so \(|z|=4\). The positions are \(4\left(\cos\tfrac{2\pi k}5+i\sin\tfrac{2\pi k}5\right)\) for \(k=0,1,2,3,4\).
   2. The chord subtending an angle \(\tfrac{2\pi}5\) at the centre has length \(2\times4\sin\tfrac\pi5=8\sin36^\circ\approx\) 4.70 m.
   3. \(5\times4.70=\) 23.5 m (23.51 m to 2 decimal places).
   4. Each triangle has area \(\tfrac12\times4\times4\times\sin72^\circ\). Five triangles: \(40\sin72^\circ\approx\) 38.04 m\(^2\).

9. Answer to Question 9

   1. \(-64=64(\cos\pi+i\sin\pi)\), so \(|z|=2\) and \(\arg z=\dfrac{\pi+2\pi k}6\): the arguments are \(\tfrac\pi6,\tfrac\pi2,\tfrac{5\pi}6,\tfrac{7\pi}6,\tfrac{3\pi}2,\tfrac{11\pi}6\).
   2. \(\sqrt3+i\), \(2i\), \(-\sqrt3+i\), \(-\sqrt3-i\), \(-2i\), \(\sqrt3-i\)
   3. In a regular hexagon the side equals the distance from the centre to a corner, so the side is 2 dm. Check: \(|2i-(\sqrt3+i)|=|{-\sqrt3}+i|=2\).
   4. Six equilateral triangles of side 2: \(6\times\tfrac{\sqrt3}4\times4=6\sqrt3\approx\) 10.4 dm\(^2\).

10. Answer to Question 10

    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 \(2i\sin\theta\).

    1. \(\left(z+\tfrac1z\right)^4=z^4+4z^2+6+\tfrac4{z^2}+\tfrac1{z^4}=2\cos4\theta+8\cos2\theta+6\). The left side is \(16\cos^4\theta\). So \(\cos^4\theta=\tfrac18(\cos4\theta+4\cos2\theta+3)\).
    2. \(I=100\cos^4\omega t=\) \(12.5\cos4\omega t+50\cos2\omega t+37.5\)
    3. Over a full period the mean of \(\cos4\omega t\) and \(\cos2\omega t\) is zero, so the mean is the constant term, 37.5 W m\(^{-2}\).

11. Answer to Question 11

    1. \(-16=16(\cos\pi+i\sin\pi)\). \(|z|=2\) and \(\arg z=\tfrac{\pi+2\pi k}4\): \(\tfrac\pi4,\tfrac{3\pi}4,\tfrac{5\pi}4,\tfrac{7\pi}4\). Positions \(\sqrt2+\sqrt2\,i\), \(-\sqrt2+\sqrt2\,i\), \(-\sqrt2-\sqrt2\,i\), \(\sqrt2-\sqrt2\,i\).
    2. Neighbouring points differ by \(2\sqrt2\) in one coordinate only, so every side is \(2\sqrt2\approx2.83\) nm and the angles are right angles. It is a square (the roots are equally spaced round a circle with four of them).
    3. Each lies at distance equal to the modulus: 2 nm.

12. Answer to Question 12

    1. \(|z|=14\), vertices \(14e^{2\pi ik/7}\) for \(k=0,\dots,6\).
    2. The vertices are \(14\omega^k\) with \(\omega=e^{2\pi i/7}\). Their sum is the geometric series \(14\cdot\dfrac{1-\omega^7}{1-\omega}=14\cdot\dfrac{1-1}{1-\omega}=\) 0.
    3. Side \(=2\times14\sin\tfrac\pi7=28\sin\tfrac\pi7\approx\) 12.15 mm.
    4. The longest diagonal skips two vertices and subtends \(3\times\tfrac{2\pi}7\) at the centre: \(28\sin\tfrac{3\pi}7\approx\) 27.30 mm. (The coin has constant width because each vertex is opposite a side.)

13. Answer to Question 13

    Method: powers of \(\sin\theta\) can be written as sums of sines of multiples of \(\theta\) by rearranging a multiple angle formula. Each term then represents a separate frequency in the signal.

    1. Make \(\sin^3\theta\) the subject of the triple angle formula. Rearrange: \(4\sin^3\theta=3\sin\theta-\sin3\theta\), so \(\sin^3\theta=\tfrac14(3\sin\theta-\sin3\theta)\).
    2. Cube the input: \(x^3=8\sin^3\theta\), then use part a). \(y=8\sin^3\theta=2(3\sin\theta-\sin3\theta)=\) \(6\sin\theta-2\sin3\theta\) So the output contains the original frequency and a new frequency three times higher.
    3. Compare the two amplitudes. The amplitudes are 6 V and 2 V, so the third harmonic is \(\dfrac26=\dfrac13\approx\) 33.3 % of the fundamental. So the distortion is large: one third of the size of the main signal.

14. Answer to Question 14

    1. \(\phi_0=-\tfrac\pi2\) so \(z=60e^{-i\pi/2}=-60i\), directly below the centre. Height \(=67.5-60=\) 7.5 m.
    2. The top has argument \(\tfrac\pi2\), so \(-\tfrac\pi2+\tfrac{2\pi k}{32}=\tfrac\pi2\) gives \(k=16\). Height \(=67.5+60=\) 127.5 m.
    3. \(\phi_8=-\tfrac\pi2+\tfrac\pi2=0\), so the capsule is level with the centre: 67.5 m.
    4. The positions are \(-60i\times\omega^k\) with \(\omega=e^{2\pi i/32}\), a geometric series with ratio \(\omega\ne1\) and \(\omega^{32}=1\), so the sum is 0. By symmetry the capsules balance about the centre.

15. Answer to Question 15

    1. The real part of \((c+is)^5\) is \(c^5-10c^3s^2+5cs^4\). Put \(s^2=1-c^2\): \(c^5-10c^3+10c^5+5c(1-2c^2+c^4)=c^5-10c^3+10c^5+5c-10c^3+5c^5=\) \(16c^5-20c^3+5c\).
    2. \(5\theta=\tfrac\pi2\) and \(\tfrac{3\pi}2\) respectively, and \(\cos\tfrac\pi2=\cos\tfrac{3\pi}2=0\).
    3. \(c(16c^4-20c^2+5)=0\) with \(c\ne0\) for these angles, so \(16c^4-20c^2+5=0\); put \(x=c^2\) to get \(16x^2-20x+5=0\).
    4. \(x=\dfrac{20\pm\sqrt{400-320}}{32}=\dfrac{5\pm\sqrt5}8\). Since \(\tfrac\pi{10}\) is smaller than \(\tfrac{3\pi}{10}\), its cosine is larger, so we take the larger root: \(\cos^2\tfrac\pi{10}=\dfrac{5+\sqrt5}8\approx0.9045\), \(\cos\tfrac\pi{10}=\sqrt{\dfrac{5+\sqrt5}8}\approx0.9511\).

16. Answer to Question 16

    1. \(|w|=\sqrt{64+192}=16\) and \(\arg w=\pi-\arctan\sqrt3=\tfrac{2\pi}3\): \(16\left(\cos\tfrac{2\pi}3+i\sin\tfrac{2\pi}3\right)\).
    2. \(z=2\left(\cos\left(\tfrac\pi6+\tfrac{k\pi}2\right)+i\sin(\dots)\right)\): arguments \(\tfrac\pi6,\tfrac{2\pi}3,\tfrac{7\pi}6,\tfrac{5\pi}3\). In Cartesian form: \(\sqrt3+i\), \(-1+\sqrt3\,i\), \(-\sqrt3-i\), \(1-\sqrt3\,i\).
    3. The roots have equal modulus 2 and consecutive arguments differ by \(\tfrac\pi2\). A quadrilateral inscribed in a circle with equal angles of \(90^\circ\) at the centre is a square (diagonals are equal diameters that cross at right angles and bisect each other).
    4. The diagonal is a diameter of length 4, so the area is \(\tfrac12\times4\times4=\) 8. (Side \(2\sqrt2\), area 8.)

17. Answer to Question 17

    1. By de Moivre, \(z^k=\cos k\theta+i\sin k\theta\). Adding for \(k=0,1,\dots,n-1\) gives the real part \(C\) and imaginary part \(S\).
    2. The series is geometric with first term 1 and common ratio \(z\), so the sum is \(\dfrac{1-z^n}{1-z}\) (valid for \(z\ne1\)).
    3. \(1-e^{in\theta}=e^{in\theta/2}\left(e^{-in\theta/2}-e^{in\theta/2}\right)=-2i\sin\tfrac{n\theta}2\,e^{in\theta/2}\). Likewise \(1-z=-2i\sin\tfrac\theta2\,e^{i\theta/2}\). The quotient is \(\dfrac{\sin(n\theta/2)}{\sin(\theta/2)}e^{i(n-1)\theta/2}\). The real part is \(C=\dfrac{\sin\tfrac{n\theta}2\cos\tfrac{(n-1)\theta}2}{\sin\tfrac\theta2}\).
    4. \(C=\dfrac{\sin\tfrac{7\pi}8\cos\tfrac{3\pi}4}{\sin\tfrac\pi8}=\dfrac{\sin\tfrac\pi8\cdot\left(-\tfrac{\sqrt2}2\right)}{\sin\tfrac\pi8}=\) \(-\dfrac{\sqrt2}2\approx-0.707\). Check: the terms are \(1,\tfrac{\sqrt2}2,0,-\tfrac{\sqrt2}2,-1,-\tfrac{\sqrt2}2,0\), with sum \(-\tfrac{\sqrt2}2\).

18. Answer to Question 18

    1. The numbers \(1,\omega,\dots,\omega^4\) are the roots of \(z^5-1=0\), which has no \(z^4\) term, so their sum is 0. (Or sum the geometric series: \(\dfrac{1-\omega^5}{1-\omega}=0\).)
    2. \(\omega^4=\overline\omega\) because \(\omega^5=1\), so \(\omega+\omega^4=2\operatorname{Re}\omega=2\cos72^\circ\). Likewise \(\omega^3=\overline{\omega^2}\) and \(b=2\cos144^\circ\). From part (a), \(a+b=-1\).
    3. \(ab=\omega^3+\omega^4+\omega^6+\omega^7=\omega^3+\omega^4+\omega+\omega^2=\) \(-1\).
    4. \(a\) and \(b\) are the roots of \(t^2+t-1=0\), so \(t=\dfrac{-1\pm\sqrt5}2\). Since \(a=2\cos72^\circ>0\), \(a=\dfrac{\sqrt5-1}2\), so \(\cos72^\circ=\dfrac{\sqrt5-1}4\approx0.309\).

19. Answer to Question 19

    1. \(z=\cos\tfrac{k\pi}3+i\sin\tfrac{k\pi}3\) for \(k=0,1,2,3,4,5\).
    2. The roots are \(1,-1,e^{\pm i\pi/3},e^{\pm2i\pi/3}\). A conjugate pair \(e^{\pm i\alpha}\) gives the real quadratic \(z^2-2\cos\alpha\,z+1\). For \(\alpha=\tfrac\pi3\): \(z^2-z+1\) (using \(2\cos\tfrac\pi3=1\)). For \(\alpha=\tfrac{2\pi}3\): \(z^2+z+1\) (using \(2\cos\tfrac{2\pi}3=-1\)). So \(f(z)=(z-1)(z+1)(z^2-z+1)(z^2+z+1)\).
    3. \(z=\dfrac{-1\pm\sqrt{-3}}2=-\tfrac12\pm\tfrac{\sqrt3}2i=e^{\pm2\pi i/3}\). Expansion: \((z^2-z+1)(z^2+z+1)=z^4+z^2+1\) and \((z-1)(z+1)=z^2-1\). Then \((z^2-1)(z^4+z^2+1)=z^6-1\). Correct.
    4. The roots of \(z^2-z+1=0\) are \(\tfrac12\pm\tfrac{\sqrt3}2i\). Their sum is \(1=2\cos\tfrac\pi3\), so \(\cos\tfrac\pi3=\tfrac12\).

20. Answer to Question 20

    1. \(z=0\) gives \(1=0\), false. Expanding, the \(z^5\) terms cancel, leaving \(5z^4+10z^3+10z^2+5z+1=0\), degree 4.
    2. Divide by \(z^5\) (valid as \(z\ne0\)): \(\left(\dfrac{z+1}z\right)^5=1\). Also \(w=1+\tfrac1z\ne1\).
    3. \(w-1=\dfrac1z\), so \(z=\dfrac1{w-1}\). The fifth roots of unity other than 1 are \(w=e^{i\theta}\), \(\theta=\tfrac{2\pi k}5\), \(k=1,2,3,4\).
    4. \(e^{i\theta}-1=e^{i\theta/2}\cdot2i\sin\tfrac\theta2\), so \(\dfrac1{e^{i\theta}-1}=\dfrac{e^{-i\theta/2}}{2i\sin(\theta/2)}=\dfrac{\cos\tfrac\theta2-i\sin\tfrac\theta2}{2i\sin\tfrac\theta2}=-\dfrac12-\dfrac i2\cot\tfrac\theta2\). With \(\tfrac\theta2=\tfrac{k\pi}5\) the roots are \(z=-\tfrac12-\tfrac i2\cot\tfrac{k\pi}5\), \(k=1,2,3,4\). (Equivalently \(-\tfrac12+\tfrac i2\cot\tfrac{k\pi}5\) because \(\cot\tfrac{(5-k)\pi}5=-\cot\tfrac{k\pi}5\).) Check \(k=1\): \(\cot36^\circ\approx1.3764\), \(z\approx-0.5-0.688i\), and \((z+1)^5=(0.5-0.688i)^5\) and \(z^5\) both have modulus \(0.8507^5\approx0.4473\) with matching argument: verified numerically.