De Moivre's Theorem — Formula, Proof & Examples

De Moivre’s Theorem

A formula useful for finding powers and roots of complex numbers.

De Moivre's Theorem: [r(cosθ+isinθ)]^n = r^n(cosnθ+isinnθ), with examples for (1+i)^2 = 8-8i and i^(3/8) roots.

Key Formula

\bigl[r(\cos\theta + i\sin\theta)\bigr]^n = r^n\bigl(\cos(n\theta) + i\sin(n\theta)\bigr)

Where:

$r$ = The modulus (absolute value) of the complex number, representing its distance from the origin in the complex plane.
$\theta$ = The argument (angle) of the complex number, measured in radians or degrees from the positive real axis.
$n$ = The exponent — can be any integer. When finding nth roots, n is replaced by 1/k where k is the root index.
$i$ = The imaginary unit, where i² = −1.

Worked Example

Problem: Compute (1 + i)⁶ using De Moivre's Theorem.

Step 1: Convert 1 + i to polar form. Find the modulus r.

r = \sqrt{1^2 + 1^2} = \sqrt{2}

Step 2: Find the argument θ. Since both the real and imaginary parts are positive, the angle is in the first quadrant.

\theta = \arctan\!\left(\frac{1}{1}\right) = \frac{\pi}{4}

Step 3: Write the polar form and apply De Moivre's Theorem with n = 6.

\bigl[\sqrt{2}\,(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4})\bigr]^6 = (\sqrt{2})^6\bigl(\cos\tfrac{6\pi}{4} + i\sin\tfrac{6\pi}{4}\bigr)

Step 4: Simplify the modulus and angle. (√2)⁶ = 2³ = 8, and 6π/4 simplifies to 3π/2.

= 8\bigl(\cos\tfrac{3\pi}{2} + i\sin\tfrac{3\pi}{2}\bigr)

Step 5: Evaluate the trigonometric values: cos(3π/2) = 0 and sin(3π/2) = −1.

= 8(0 + i(-1)) = -8i

Answer: (1 + i)⁶ = −8i

Another Example

This example uses De Moivre's Theorem to find roots (not powers) of a complex number, showing how the formula generates multiple equally-spaced solutions on a circle in the complex plane.

Problem: Find all three cube roots of 8 (that is, solve z³ = 8).

Step 1: Write 8 in polar form. Its modulus is 8 and its argument is 0.

8 = 8(\cos 0 + i\sin 0)

Step 2: Apply the root formula from De Moivre's Theorem. For the kth of n roots, use θ replaced by (θ + 2πk)/n and r replaced by r^(1/n), where k = 0, 1, …, n − 1.

z_k = 8^{1/3}\left(\cos\frac{0 + 2\pi k}{3} + i\sin\frac{0 + 2\pi k}{3}\right), \quad k = 0, 1, 2

Step 3: Compute each root. The modulus of every root is 8^(1/3) = 2.

z_0 = 2(\cos 0 + i\sin 0) = 2

Step 4: For k = 1, the angle is 2π/3.

z_1 = 2\!\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2\!\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}

Step 5: For k = 2, the angle is 4π/3.

z_2 = 2\!\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = 2\!\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}

Answer: The three cube roots of 8 are 2, −1 + i√3, and −1 − i√3.

Frequently Asked Questions

Does De Moivre's Theorem work for negative and fractional exponents?

Yes. The theorem holds for any integer n, including negative integers. For fractional exponents (roots), you apply the formula with n = 1/k and add multiples of 2π/k to the angle to obtain all k distinct roots. Each root lies on a circle of radius r^(1/k) and the roots are equally spaced around it.

Why do you need polar form for De Moivre's Theorem?

The theorem relies on the polar representation r(cos θ + i sin θ) because it separates the magnitude from the angle. Raising to a power then becomes two simpler operations: raise the modulus to that power and multiply the angle by that power. In rectangular form (a + bi), you would need to expand the product term by term, which is far more tedious for large exponents.

What is the connection between De Moivre's Theorem and Euler's formula?

Euler's formula states e^(iθ) = cos θ + i sin θ. Using it, De Moivre's Theorem becomes (re^(iθ))^n = r^n e^(inθ), which follows directly from the exponent rule for exponentials. Euler's formula therefore provides a concise proof and a shorthand notation for the theorem.

De Moivre's Theorem (polar form power) vs. Binomial expansion of (a + bi)ⁿ

	De Moivre's Theorem (polar form power)	Binomial expansion of (a + bi)ⁿ
Approach	Convert to polar form, then multiply the angle and raise the modulus	Expand (a + bi)ⁿ using the binomial theorem term by term
Efficiency for large n	Very efficient — only requires one angle multiplication and one power of r	Requires computing n + 1 binomial terms and simplifying powers of i
Finding roots	Naturally yields all n distinct nth roots	Does not directly produce all roots
Result form	Gives the answer in polar form (convert to rectangular if needed)	Gives the answer directly in rectangular form a + bi

Why It Matters

De Moivre's Theorem appears in precalculus and early college mathematics whenever you work with powers or roots of complex numbers. It is essential in electrical engineering for analyzing AC circuits and in signal processing where rotating phasors are described using polar complex numbers. The theorem also provides elegant proofs of trigonometric identities by equating real and imaginary parts after expanding both sides.

Common Mistakes

Mistake: Forgetting to multiply the angle by n. Students sometimes raise only the modulus to the nth power and leave the angle unchanged.

Correction: Both parts change: the modulus becomes rⁿ and the argument becomes nθ. Always multiply the angle by the exponent.

Mistake: When finding nth roots, students find only one root instead of all n distinct roots.

Correction: There are exactly n distinct nth roots. After computing the first root with k = 0, you must add 2π/n to the angle for each subsequent root (k = 1, 2, …, n − 1) to obtain them all.

Related Terms

Complex Numbers — The number system De Moivre's Theorem operates on
Polar Form of a Complex Number — Required representation before applying the theorem
Complex Number Formulas — Collection of formulas including De Moivre's
Root of a Number — Finding nth roots is a key application
Power — Computing integer powers is the primary use
Formula — De Moivre's Theorem is a named formula
Trigonometry — The theorem uses cosine and sine functions