Maths › Complex numbers › De Moivre's theorem and trigonometric identities

De Moivre's theorem and trigonometric identities

Raise a complex number to a power by raising its length and multiplying its angle. One theorem computes high powers in a line and manufactures trig identities to order.

Year FMEDEXCEL 9FM0 CP2 2.6, 2.8, 2.9

Builds on Modulus, argument and loci and The binomial expansion.

IN THIS TOPIC

  • Use De Moivre's theorem to evaluate powers of complex numbers.
  • Write numbers in exponential form and use it for products and powers.
  • Derive multiple-angle identities such as cos 3θ in terms of cos θ.

COMMON MISCONCEPTION

To find (1 + i)⁸ there is no alternative to multiplying out the brackets eight times.

The theorem

Multiplying complex numbers multiplies moduli and adds arguments, so raising to a power does both n times over. That observation is De Moivre's theorem:

(cosθ+isinθ)n=cosnθ+isinnθ(\cos θ + \text{i} \sin θ)^{n} = \cos nθ + \text{i} \sin nθIN THE FORMULAE BOOKLET

The booklet prints it under Complex numbers, in the general form with the modulus attached, so find it there rather than reciting it. For a general number, |zn| = |z|n while arg zn = n arg z, up to whole turns. High powers collapse to two small calculations, with no repeated multiplication anywhere.

Powers of 1 + i marching round the Argand diagram: each multiplication stretches by root 2 and turns by 45 degrees1 + i2i−2 + 2i−4×√2 and +45° each step
FIG. 1The powers of 1 + i: each step stretches the arrow by √2 and turns it 45°, spiralling from 1 + i round to −4.

WORKED EXAMPLE

An eighth power in two lines

Evaluate (1 + i)⁸.

1 + i has modulus √2 and argument π/4.

So (1 + i)⁸ has modulus (√2)⁸ = 16 and argument 8 × π/4 = 2π, which is the direction of the positive real axis.

(1 + i)⁸ = 16. The bracket-expansion route agrees, since (1 + i)² = 2i, squared gives −4, and squared again gives 16.

Exponential form

Compare the Maclaurin series for ex, cos θ and sin θ and one identity drops out, cos θ + i sin θ = eiθ. Every complex number therefore has the compact exponential form z = r eiθ, with r the modulus and θ the argument. In that notation De Moivre stops being a theorem and becomes an index law, because (r eiθ)n = rn einθ. Products and quotients are just as painless. Setting r = 1 and θ = π gives eiπ = −1.

Use whichever form the question uses. Edexcel asks for answers in the form r eiθ often enough that the conversion needs to be automatic, and it is the natural home for roots of unity in the next lesson.

Identities to order

Run the theorem backwards and it manufactures trigonometry. Expand (cos θ + i sin θ)³ by the binomial theorem, and De Moivre says the answer is cos 3θ + i sin 3θ. Two expressions for one number must agree part by part, so the real parts give cos 3θ and the imaginary parts give sin 3θ.

Cubing a number on the unit circle: the length stays 1 and the angle triples, from 25 degrees round to 75z at 25°z³ at 75°same length, three times the turn
FIG. 2Cubing a number on the unit circle: the length stays at 1 while the angle triples, so cubes of cos θ + i sin θ know about 3θ.

WORKED EXAMPLE

cos 3θ from a cube

Express cos 3θ in terms of cos θ.

(cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ.

Equate real parts: cos 3θ = cos³θ − 3 cos θ sin²θ.

Replace sin²θ with 1 − cos²θ to get cos 3θ = 4 cos³θ − 3 cos θ.

The imaginary parts give sin 3θ = 3 sin θ − 4 sin³θ from the same expansion. One cube, two identities.

GUIDED PRACTICE

A power with a turn

Use De Moivre's theorem to evaluate (√3 + i)⁶.

Show the working

√3 + i has modulus 2 and argument π/6.

So the sixth power has modulus 2⁶ = 64 and argument 6 × π/6 = π.

An argument of π points along the negative real axis, so (√3 + i)⁶ = −64.

ASSESSMENT FOCUS

  • Convert to modulus-argument or exponential form before any power. The theorem does not apply to x + yi directly.
  • Reduce final arguments back into (−π, π] before interpreting the answer.
  • For identities, expand with the binomial theorem, then equate real or imaginary parts and say which you took.
  • Convert powers of sin back via sin²θ = 1 − cos²θ when the target is written all in cos.
  • Answers demanded in the form r eiθ need θ in radians and exact where possible.

CHECK YOURSELF

Use De Moivre's theorem to evaluate (1 + i)¹⁰, and give the answer in the form r eiθ.

Show a hint

Modulus √2, argument π/4; reduce the final argument by full turns.

Show the answer

Modulus (√2)¹⁰ = 32, argument 10π/4 = 5π/2, which reduces by a full turn to π/2, the positive imaginary axis. So (1 + i)¹⁰ = 32i, which is 32 eiπ/2.

Powers in modulus-argument form: raise the modulus, multiply the argument.

z = r eiθ turns De Moivre into an index law.

For identities, expand (cos θ + i sin θ)ⁿ binomially and equate parts with cos nθ + i sin nθ.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 52 KB · print edition · white background

Question workbook

PDF · 52 KB · screen edition · navy

Answer book

PDF · 58 KB · fully worked · navy
6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the de moivre's theorem and trigonometric identities questions page.

CHECK YOUR PROGRESS

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  • Use De Moivre's theorem to evaluate powers of complex numbers.
  • Write numbers in exponential form and use it for products and powers.
  • Derive multiple-angle identities such as cos 3θ in terms of cos θ.

Open the full revision checklist to see every objective in the course in one place.