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De Moivre and complex roots

Use De Moivre’s theorem to find powers and all the roots of a complex number.

For integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ). Keep the angular unit consistent. Convert to rectangular form only after applying the power rule.

The nth roots of a nonzero number have modulus r^(1/n) and arguments (θ + 2kπ)/n for k = 0,…,n − 1. They are equally spaced on a circle. Taking only θ/n misses the remaining roots.

Worked example

Find the cube roots of 8.

  1. Write 8 with modulus 8 and argument 0.
  2. Each root has modulus 2.
  3. The arguments are 0, 2π/3 and 4π/3.

Answer: 2, −1 + √3 i, −1 − √3 i

Revise first: Modulus and argument.

Finding every root, without counting one twice

Write z = r(cos θ + i sin θ) before taking powers or roots. Powers multiply the argument; taking an nth root divides the argument but also introduces n choices. An argument describes a direction, so adding a full turn leaves the same complex number.

For zⁿ = R(cos α + i sin α), the roots have modulus R^(1/n) and arguments (α + 2kπ)/n, with k = 0, …, n − 1. Taking just the principal argument loses roots. Taking k = n repeats the first root.

The cube roots of −1

  1. The modulus is 1. Take α = π, so the arguments are (π + 2kπ)/3.
  2. For k = 0, 1, 2 this gives π/3, π and 5π/3.
  3. The roots are 1/2 + (√3/2)i, −1 and 1/2 − (√3/2)i. Each cubes to −1.
  4. Their sum is zero, agreeing with the missing z² coefficient in z³ + 1.
Why does k = 3 add no new root?

Its argument is 7π/3, the same direction as π/3. There are exactly three distinct roots.

Take the idea further

Rotate all the roots by π/6. What equation do the rotated roots satisfy? Since w = ze^(iπ/6), w³ = −i.

Explore the diagram and test a prediction

More advanced methods

Practise de moivre and complex roots

Course mapping

These specification references show where the topic occurs. The questions cover only some parts of each topic.

Next practice: Complex loci.