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 that computes high powers in a line and manufactures trig identities to order.
Builds on Modulus, argument and loci and The binomial expansion.
IN THIS TOPIC
- State De Moivre's theorem and use it to evaluate powers of complex numbers.
- Choose modulus-argument form for powers instead of repeated bracket expansion.
- Derive multiple-angle identities such as cos 3θ in terms of cos θ.
WHAT YOU PROBABLY THINK
To find (1 + i)⁸ there is no alternative to multiplying out the brackets eight times.
The theorem
Since multiplying complex numbers multiplies moduli and adds arguments, raising to a power does both n times over. That observation is De Moivre's theorem:
and for a general number, |zn| = |z|n with arg zn = n arg z. High powers collapse to two small calculations, which retires the opening claim.
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: (1 + i)² = 2i, squared gives −4, squared again gives 16.
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: the real parts give cos 3θ, the imaginary parts sin 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.
Real parts: cos 3θ = cos³θ − 3 cos θ sin²θ.
Replace sin²θ with 1 − cos²θ: 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.
YOUR TURN
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: (√3 + i)⁶ = −64.
THE EXAM BIT
- Convert to modulus-argument 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.
- Convert powers of sin back via sin²θ = 1 − cos²θ when the target is all in cos.
CHECK YOURSELF
Use De Moivre's theorem to evaluate (1 + i)¹⁰.
Show a hint
Modulus √2, argument π/4; reduce the final argument by full turns.
Show the answer
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Powers in modulus-argument form: raise the modulus, multiply the argument.
Expand (cos θ + i sin θ) to the n binomially, 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.
CHECK YOUR PROGRESS
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- State De Moivre's theorem and use it to evaluate powers of complex numbers.
- Choose modulus-argument form for powers instead of repeated bracket expansion.
- 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.
No animated video for this topic yet; these notes stand alone.