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WASpecialist MathematicsSyllabus dot point

How can we raise a complex number to a high integer power without expanding the binomial?

State and apply de Moivre's theorem to evaluate integer powers of complex numbers and derive trigonometric identities

WACE Specialist Unit 3 de Moivre's theorem: raising a complex number in polar form to an integer power, its proof by induction, evaluating powers quickly, and deriving multiple-angle trigonometric identities, with a worked example.

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  1. What this dot point is asking
  2. The theorem
  3. Evaluating powers
  4. Deriving trigonometric identities

What this dot point is asking

SCSA wants you to apply de Moivre's theorem to integer powers and to use it to derive trigonometric identities by equating real and imaginary parts.

The theorem

For n≥1n \ge 1 this follows by induction from the polar multiplication rule: each extra factor multiplies the modulus by rr and adds θ\theta to the argument. The base case n=1n = 1 is trivial, and if the result holds for n=kn = k then

[r cis θ]k+1=[r cis θ]k⋅r cis θ=rk cis(kθ)⋅r cis θ=rk+1 cis((k+1)θ).[r\,\text{cis}\,\theta]^{k+1} = [r\,\text{cis}\,\theta]^k \cdot r\,\text{cis}\,\theta = r^k\,\text{cis}(k\theta)\cdot r\,\text{cis}\,\theta = r^{k+1}\,\text{cis}((k+1)\theta).

It extends to negative integers using 1z=1r cis(−θ)\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta) and to n=0n = 0 trivially.

Evaluating powers

To compute znz^n, convert zz to polar form, apply the theorem, then convert back if a Cartesian answer is required. This is far faster than expanding (x+iy)n(x + iy)^n for large nn.

Deriving trigonometric identities

Take r=1r = 1 and write (cos⁡θ+isin⁡θ)n(\cos\theta + i\sin\theta)^n two ways: once by de Moivre as cos⁡nθ+isin⁡nθ\cos n\theta + i\sin n\theta, and once by the binomial theorem. Equating real parts gives a formula for cos⁡nθ\cos n\theta and equating imaginary parts gives sin⁡nθ\sin n\theta, both in terms of powers of cos⁡θ\cos\theta and sin⁡θ\sin\theta.