Key takeaway: De Moivre’s theorem says [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ), so a power of a complex number raises the modulus and multiplies the angle.
Grade 12 – PrecalculusN-CN.B.5Numbers and Operations
Definition of De Moivre's Theorem
De Moivre’s theorem turns a heavy job, raising a complex number to a high power, into a light one. Instead of multiplying out a binomial dozens of times , convert to polar form, handle the modulus and angle separately, and convert back if needed.
The rule works because multiplying complex numbers multiplies their moduli and adds their angles. Using the same factor \(n\) times therefore raises the modulus to the \(n\)th power and adds the angle to itself \(n\) times , which is the same as multiplying it by \(n\). Geometrically, each extra power rotates the point by another \(\theta\) around the origin.
Formula and notation
For a real \(r\ge 0\), an angle \(\theta\) and an integer \(n\):
\[\bigl[r(\cos\theta+i\sin\theta)\bigr]^{n}=r^{n}\bigl(\cos n\theta+i\sin n\theta\bigr).\]
Replacing \(n\) by \(\frac{1}{n}\) and adding multiples of a full turn leads to the \(n\)th roots.
Key facts
- De Moivre’s theorem accepts degrees or radians, but the unit must stay consistent throughout.
- Reduce the final angle by full turns to see the answer in simplest form.
- The theorem also works for negative integer powers, which give reciprocals.
- With r = 1 it describes points that march around the unit circle in equal steps.
De Moivre's Theorem example
Compute \((\sqrt{3}-i)^6\).
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Modulus: \(r=\sqrt{3+1}=2\). Angle: the point is in quadrant IV with reference angle \(30^{\circ}\), so \(\theta=-30^{\circ}\).
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Apply the theorem: \(2^6\bigl(\cos(-180^{\circ})+i\sin(-180^{\circ})\bigr)\).
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Evaluate: \(\cos(-180^{\circ})=-1\) and \(\sin(-180^{\circ})=0\), so the result is \(64\cdot(-1)\).
\((\sqrt{3}-i)^6=\)\(-64\).
Common mistake
Watch out
Gwen computes \((\sqrt{3}-i)^6\) as \(2(\cos(-180^{\circ})+i\sin(-180^{\circ}))=-2\), raising only the angle.
Why it is wrong
De Moivre’s theorem changes both parts. The modulus must be raised to the power as well: \(2^6=64\), not 2. Both parts change: the modulus is powered and the angle is multiplied, giving \(-64\).
Related terms and lessons
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