Further Mathematics · AQA 7367 · Topic B

Complex numbers

Numbers of the form x + iy, where i² = −1. You add, multiply and divide them, find complex roots of polynomials, draw them on an Argand diagram, use modulus–argument and exponential form, de Moivre's theorem, nth roots and loci.

  • 11 sub-topics
  • 7 in Year 12, 4 in Year 13
  • 188 practice questions
  • 11 animated lessons
Practice questions (188) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. B1Solving polynomial equations with complex rootsY12
  2. B2Arithmetic with complex numbers in the form x + iyY12
  3. B3Complex conjugates and conjugate pairs of rootsY12
  4. B4Use and interpret Argand diagramsY12
  5. B5Convert between Cartesian and modulus–argument formY12
  6. B6Multiply and divide complex numbers in modulus–argument formY12
  7. B7Loci in the Argand diagram, such as |z − a| > r and arg(z − a) = θY12
  8. B8De Moivre's theorem: multiple angle formulae and sums of seriesY13
  9. B9Exponential form: e^(iθ) = cos θ + i sin θ and z = re^(iθ)Y13
  10. B10The nth roots of a complex number and regular polygonsY13
  11. B11Complex roots of unity to solve geometric problemsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 188 questions in this topic, free, no account needed.

Complex numbers

Easy

Complex arithmetic · Year 12 · 5 marks

z=3+4iz = 3 + 4\mathrm{i} and w=1−2iw = 1 - 2\mathrm{i}.
  1. (a)
    Find zwzw in the form a+bia + b\mathrm{i}.
    (2)
  2. (b)
    Find ∣z∣|z|.
    (1)
  3. (c)
    Find zw\dfrac{z}{w} in the form a+bia + b\mathrm{i}.
    (2)
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Part (a)2 marks

  1. Expand the brackets exactly as you would with xx: every term times every term.zwzw=3{}= 3−6i{}- 6\mathrm{i}+4i{}+ 4\mathrm{i}−8i2{}- 8\mathrm{i}^2
  2. Use i2=−1\mathrm{i}^2 = -1, so −8i2=+8-8\mathrm{i}^2 = +8, then collect real and imaginary parts.=11−2i= 11 - 2\mathrm{i}

Part (b)1 mark

  1. The modulus is the distance from the origin on an Argand diagram, so use Pythagoras.∣z∣=32+42=5|z| = \sqrt{3^2 + 4^2} = 5

Part (c)2 marks

  1. Multiply top and bottom by the conjugate of the denominator, which makes the denominator real.(3+4i)(1+2i)(1−2i)(1+2i)\dfrac{(3 + 4\mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})}=3+10i+8i25{}= \dfrac{3 + 10\mathrm{i} + 8\mathrm{i}^2}{5}
  2. Replace i2\mathrm{i}^2 with −1-1 and divide each part by 5.=−1+2i= -1 + 2\mathrm{i}

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Complex numbers

Medium

Roots of a cubic · Year 12 · 5 marks

f(z)=z3−3z2+z+5\mathrm{f}(z) = z^3 - 3z^2 + z + 5
  1. Given that 2−i2 - \mathrm{i} is a root of the equation f(z)=0\mathrm{f}(z) = 0, solve f(z)=0\mathrm{f}(z) = 0 completely.
    (5)
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  1. A polynomial with real coefficients has complex roots in conjugate pairs.2+i2 + \mathrm{i} is also a root.
  2. Two roots give two factors; multiplying them gives a real quadratic factor.(z−(2−i))(z−(2+i))\bigl(z - (2 - \mathrm{i})\bigr)\bigl(z - (2 + \mathrm{i})\bigr)=(z−2)2−i2{}= (z - 2)^2 - \mathrm{i}^2
  3. Expand (z−2)2(z-2)^2 and use i2=−1\mathrm{i}^2 = -1.=z2−4z+5= z^2 - 4z + 5
  4. Divide, or compare coefficients: the constant term 5 = 5 × 1 gives the last factor.z3−3z2+z+5z^3 - 3z^2 + z + 5=(z2−4z+5)(z+1){}= (z^2 - 4z + 5)(z + 1)
  5. Set each factor to zero to list all three roots.zz=−1, 2{}= -1,\ 2+i, 2{}+ \mathrm{i},\ 2−i{}- \mathrm{i}

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Complex numbers

Not needed for Year 12Hard

Multiple angles · Year 13 · 5 marks

  1. Use de Moivre's theorem to show that cos⁡3θ\cos 3\theta=4cos3⁡θ−3cos⁡θ{}= 4\cos^3\theta - 3\cos\theta.
    (5)
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  1. De Moivre's theorem with n=3n = 3.cos⁡3θ\cos 3\theta+isin⁡3θ{}+ \mathrm{i}\sin 3\theta=(cos⁡θ+isin⁡θ)3{}= (\cos\theta + \mathrm{i}\sin\theta)^3
  2. Expand with the binomial theorem, writing c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta.=c3= c^3+3c2(is){}+ 3c^2(\mathrm{i}s)+3c(is)2{}+ 3c(\mathrm{i}s)^2+(is)3{}+ (\mathrm{i}s)^3
  3. i2=−1\mathrm{i}^2 = -1, so the real terms are c3c^3 and −3cs2-3cs^2.Real parts: cos⁡3θ=c3−3cs2\cos 3\theta = c^3 - 3cs^2
  4. Use s2=1−c2s^2 = 1 - c^2 to write everything in cosines.=c3−3c(1−c2)= c^3 - 3c(1 - c^2)
  5. Collect the terms.=4cos3⁡θ−3cos⁡θ= 4\cos^3\theta - 3\cos\theta

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