STEP complex numbers and polynomials: guide and practice problems
Complex numbers turn trigonometry into algebra. STEP 3 especially likes roots of unity, de Moivre's theorem and loci in the Argand diagram.
Key ideas
- The non-real nth roots of unity add to -1; pairing w^k with w^(n-k) gives 2cos(2πk/n).
- Substituting x = 2cos θ can turn a cubic into cos 3θ = constant.
- Vieta's formulas give sums, products and sums of reciprocals of roots without finding the roots.
- A locus |z - a| = λ|z - b| with λ ≠ 1 is a circle: put z = x + iy and complete the square.
Common traps
- Mixing up the sign in the sum of roots (minus the next coefficient over the leading one).
- Choosing the wrong root after solving a quadratic for a cosine (check the sign).
- Forgetting that arg is only defined up to multiples of 2π.
Practice problems
Original problems in the style of STEP. Try each one for at least 20 minutes before you open a hint; write a full solution, then compare.
Problem 1: A cubic with cosine roots
- Show that \(4\cos^3\theta-3\cos\theta=\cos3\theta\).
- By substituting \(x=2\cos\theta\), show that the roots of \(x^3-3x+1=0\) are \(2\cos\tfrac{2\pi}9\), \(2\cos\tfrac{4\pi}9\) and \(2\cos\tfrac{8\pi}9\).
- Hence find the values of \(\cos\tfrac{2\pi}9\cos\tfrac{4\pi}9\cos\tfrac{8\pi}9\) and of \(\sec\tfrac{2\pi}9+\sec\tfrac{4\pi}9+\sec\tfrac{8\pi}9\).
Hint 1
Expand \(\cos(2\theta+\theta)\), or use de Moivre.
Hint 2
With \(x=2\cos\theta\) the cubic becomes \(2\cos3\theta+1=0\). Then use the product of the roots and the sum of their reciprocals.
Full solution
(i) \(\cos3\theta=\cos2\theta\cos\theta-\sin2\theta\sin\theta=(2c^2-1)c-2(1-c^2)c=4c^3-3c\).
(ii) \(x^3-3x+1=8\cos^3\theta-6\cos\theta+1=2\cos3\theta+1\). So \(\cos3\theta=-\tfrac12\): \(3\theta=\tfrac{2\pi}3,\tfrac{4\pi}3,\tfrac{8\pi}3\) give \(\theta=\tfrac{2\pi}9,\tfrac{4\pi}9,\tfrac{8\pi}9\), three different values of \(2\cos\theta\), so these are the three roots.
(iii) Product of roots \(=-1\), so \(8\cos\tfrac{2\pi}9\cos\tfrac{4\pi}9\cos\tfrac{8\pi}9=-1\): the product is \(-\tfrac18\). Sum of reciprocals of the roots \(=\dfrac{\text{sum of pair products}}{\text{product}}=\dfrac{-3}{-1}=3\); since each root is \(2\cos\theta\), \(\sum\tfrac1{2\cos\theta}=3\), so \(\sum\sec\theta=6\).
Results: Product \(-\frac18\); sum of secants \(6\).
2 more complex numbers and polynomials problems
Fifth roots of unity and cos 72°; A circle in the Argand diagram. Each with two hints and a full solution.
Next steps
Related topics: Trigonometry · Sequences, series and induction. Stuck? Read the problem-solving strategies. Ready for the real thing? Official STEP past papers.
STEP is run by OCR. A Level Math Revision is independent: it is not affiliated with or endorsed by OCR, the University of Cambridge or any university. Every problem here is our own, written in the style of STEP; for real papers use OCR's official past papers.