In short

De Moivre's theorem states that for any integer n,

(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta).

In other words, raising a point on the unit circle to the nth power multiplies the angle by n. Combined with the binomial expansion, this gives formulas for \cos(n\theta) and \sin(n\theta) in terms of \cos\theta and \sin\theta. The theorem is the reason complex numbers are so powerful for trigonometry: a single algebraic identity replaces pages of angle-addition algebra.

Try computing (\cos 10° + i\sin 10°)^{36} by hand. Multiplying 36 complex numbers together sounds brutal. But think about what each multiplication does on the unit circle: it adds 10° to the angle. After 36 multiplications the angle reaches 360°, which is a full revolution back to 1. So the answer is just 1. No calculation needed -- just one clean rule that turns exponentiation into angle multiplication.

That rule is De Moivre's theorem. It is short enough to fit on an index card, yet it single-handedly unlocks formulas for \cos(n\theta) and \sin(n\theta) that pages of trigonometric identities cannot reach easily.

The statement

Start with a complex number on the unit circle:

z = \cos\theta + i\sin\theta

This has modulus 1 and argument \theta. What is z^2?

z^2 = z \cdot z

The moduli multiply: 1 \times 1 = 1. The arguments add: \theta + \theta = 2\theta. So:

z^2 = \cos(2\theta) + i\sin(2\theta)

What is z^3? Multiply z^2 by z again. Modulus: still 1. Argument: 2\theta + \theta = 3\theta. So:

z^3 = \cos(3\theta) + i\sin(3\theta)

The pattern is clear. Each multiplication by z adds another \theta to the argument, while the modulus stays at 1. After n multiplications:

De Moivre's Theorem

For any integer n (positive, negative, or zero) and any angle \theta,

(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta).

In Euler's notation, the theorem is transparent: (e^{i\theta})^n = e^{in\theta}, which is just the law of exponents. The full statement in trigonometric form carries more weight because it connects the exponent n with the multiple angle n\theta — something that is not obvious from \cos and \sin alone.

De Moivre's theorem illustrated on the unit circle for n equals 1, 2, 3A unit circle on the complex plane with three points marked. The first point z is at angle theta from the positive real axis. The second point z squared is at angle 2 theta. The third point z cubed is at angle 3 theta. All three lie on the unit circle. Lines from the origin to each point show the progression of the angle being multiplied. ReIm θ 2θ 3θ z z² z³ All three points lie on the unit circle — only the angle changes.
De Moivre's theorem on the unit circle. The point $z = \cos\theta + i\sin\theta$ sits at angle $\theta$. Squaring moves to $2\theta$. Cubing moves to $3\theta$. Each power multiplies the angle while the modulus stays at $1$. The point walks around the circle, $\theta$ radians at a time.

Proof by mathematical induction (positive integers)

The argument-addition reasoning above is convincing, but a clean inductive proof makes the result airtight.

Base case. When n = 1:

(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta = \cos(1 \cdot \theta) + i\sin(1 \cdot \theta). \;\checkmark

Inductive step. Assume the theorem holds for some positive integer k:

(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta). \quad \text{(inductive hypothesis)}

Now compute the (k+1)th power:

(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta)

Apply the inductive hypothesis to the first factor:

= [\cos(k\theta) + i\sin(k\theta)] \cdot [\cos\theta + i\sin\theta]

Expand the product:

= [\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta] + i[\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta]

The real part is \cos(k\theta + \theta) = \cos((k+1)\theta) by the angle-addition formula. The imaginary part is \sin(k\theta + \theta) = \sin((k+1)\theta).

= \cos((k+1)\theta) + i\sin((k+1)\theta). \;\checkmark

By mathematical induction, the theorem holds for all positive integers n. \square

Extension to zero and negative integers

Case n = 0: (\cos\theta + i\sin\theta)^0 = 1 = \cos(0) + i\sin(0). The theorem holds.

Case n < 0: Write n = -m where m is a positive integer. Then:

(\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)}

Multiply numerator and denominator by the conjugate \cos(m\theta) - i\sin(m\theta):

= \frac{\cos(m\theta) - i\sin(m\theta)}{\cos^2(m\theta) + \sin^2(m\theta)} = \cos(m\theta) - i\sin(m\theta)

since \cos^2 + \sin^2 = 1. Now use the identities \cos(-m\theta) = \cos(m\theta) and \sin(-m\theta) = -\sin(m\theta):

= \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta). \;\checkmark

So De Moivre's theorem holds for all integers — positive, negative, and zero.

Application 1: Computing powers directly

The most immediate use of De Moivre's theorem is computing high powers of complex numbers. Without it, raising \cos\theta + i\sin\theta to the 10th power would require nine successive multiplications. With it, the answer is instant: \cos(10\theta) + i\sin(10\theta).

For a general complex number z = r(\cos\theta + i\sin\theta), the nth power is:

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

The modulus gets raised to the nth power; the argument gets multiplied by n.

Application 2: Expressing \cos(n\theta) and \sin(n\theta)

This is the deeper application. De Moivre's theorem, combined with the binomial theorem, lets you express \cos(n\theta) and \sin(n\theta) as polynomials in \cos\theta and \sin\theta.

The idea: expand (\cos\theta + i\sin\theta)^n using the binomial theorem, then separate real and imaginary parts. The real part equals \cos(n\theta) and the imaginary part equals \sin(n\theta).

Take n = 3 as an illustration. By De Moivre:

(\cos\theta + i\sin\theta)^3 = \cos(3\theta) + i\sin(3\theta)

By the binomial theorem (with c = \cos\theta and s = \sin\theta for brevity):

(c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3
= c^3 + 3c^2 s\, i + 3c s^2\, i^2 + s^3\, i^3
= c^3 + 3c^2 s\, i - 3c s^2 - s^3 i
= (c^3 - 3cs^2) + i(3c^2 s - s^3)

Equate real and imaginary parts with \cos(3\theta) + i\sin(3\theta):

\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta
\sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta

These are the triple-angle formulas. You can derive them by applying the angle-addition formula twice, but the De Moivre route gives both formulas simultaneously and generalises to any n with no extra work.

Using \sin^2\theta = 1 - \cos^2\theta, the cosine formula becomes:

\cos(3\theta) = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta

This is the formula used in solving cubic equations and in Chebyshev polynomials — it expresses \cos(3\theta) purely in terms of \cos\theta.

The triple-angle formula derived from De Moivre's theoremA flow diagram showing the derivation. At the top, the expression cos theta plus i sin theta cubed. An arrow labelled De Moivre leads to cos 3 theta plus i sin 3 theta on the left. Another arrow labelled binomial expansion leads to the expanded form on the right. Equating real parts gives the cos 3 theta formula; equating imaginary parts gives the sin 3 theta formula. (cos θ + i sin θ)³ De Moivre Binomial cos 3θ + i sin 3θ (c³ − 3cs²) + i(3c²s − s³) equate real and imaginary parts cos 3θ = cos³θ − 3 cos θ sin²θ sin 3θ = 3 cos²θ sin θ − sin³θ The same method works for any positive integer n.
The triple-angle formulas in one picture. Expand $(\cos\theta + i\sin\theta)^3$ two ways — De Moivre on the left, binomial theorem on the right — then equate real and imaginary parts. The method generalises: for $n = 4$, you get the quadruple-angle formulas; for $n = 5$, the quintuple-angle formulas; and so on.

Interactive: powers on the unit circle

Drag the red point on the unit circle below. The readouts show the angle \theta and the angles 2\theta and 3\theta (modulo 2\pi). Watch how the second and third powers jump ahead on the circle — each power multiplies the angle, wrapping around when it exceeds \pi.

Interactive De Moivre explorer on the unit circleA complex plane with a unit circle. A red draggable point sits on the curve x squared plus y squared equals 1. Readouts show the angle theta, 2 theta, and 3 theta updating as the reader drags the point around the circle. ReIm 1−1 i−i drag the red point around the circle
Drag the red point around the unit circle. The readouts show $\theta$, $2\theta$, $3\theta$, and $\operatorname{Re}(z^3) = \cos^3\theta - 3\cos\theta\sin^2\theta$. As $\theta$ increases slowly, $3\theta$ races ahead three times faster — visually demonstrating how De Moivre's theorem multiplies the angle.

Worked examples

Example 1: Compute $(1 + i)^8$

A power of a complex number with clean modulus and argument. De Moivre's theorem avoids expanding eight nested products.

Step 1. Convert to polar form.

r = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}
\theta = \operatorname{Arg}(1 + i) = \frac{\pi}{4}

So 1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right).

Why: the point (1, 1) sits in the first quadrant, at 45° from the real axis and distance \sqrt{2} from the origin.

Step 2. Apply De Moivre's theorem with n = 8.

(1 + i)^8 = (\sqrt{2})^8 \left(\cos\frac{8\pi}{4} + i\sin\frac{8\pi}{4}\right)

Why: the modulus gets raised to the 8th power and the argument gets multiplied by 8. This is the entire computation — two operations instead of seven multiplications.

Step 3. Simplify.

(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16
\frac{8\pi}{4} = 2\pi
\cos(2\pi) = 1, \qquad \sin(2\pi) = 0
(1 + i)^8 = 16(1 + 0i) = 16

Why: after eight quarter-turns of \pi/4 each, the angle completes exactly two full rotations and returns to 0 (equivalently 2\pi). The number lands back on the positive real axis, and the modulus 16 tells you how far along it.

Result: (1 + i)^8 = 16.

The eighth power of 1 plus i shown as two full rotations ending at 16 on the real axisA complex plane with a spiral path starting from the point 1 plus i on a small circle. Each squaring doubles the angle and squares the distance. After eight multiplications, the point lands at 16 on the positive real axis. The initial point, the intermediate squares, and the final point are marked. ReIm 0 1 2 4 8 16 1 + i (n=1) 2i (n=2) −4 (n=4) (1+i)⁸ = 16
Successive powers of $1 + i$ spiral outward from the origin. After eight multiplications, the angle has made two full turns ($8 \times \pi/4 = 2\pi$) and the modulus has grown to $(\sqrt{2})^8 = 16$. The result is $16$ — a real number, sitting on the positive real axis. De Moivre's theorem predicted this in one line.

The answer is a real number. That is no coincidence — eight quarter-turns make two full circles, so the imaginary part vanishes. Without De Moivre's theorem, you would need to expand (1+i)^8 by repeated squaring: (1+i)^2 = 2i, (2i)^2 = -4, (-4)^2 = 16. The theorem tells you the answer instantly and tells you why it is real.

Example 2: Express $\cos(4\theta)$ in terms of $\cos\theta$ alone

The De Moivre plus binomial method applied to derive a multiple-angle formula.

Step 1. Write De Moivre's theorem for n = 4.

(\cos\theta + i\sin\theta)^4 = \cos(4\theta) + i\sin(4\theta)

Why: the left side will be expanded using the binomial theorem, then the real part gives \cos(4\theta).

Step 2. Expand the left side using the binomial theorem with c = \cos\theta, s = \sin\theta.

(c + is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4
= c^4 + 4c^3 s\,i + 6c^2 s^2\, i^2 + 4c s^3\, i^3 + s^4\, i^4
= c^4 + 4c^3 s\, i - 6c^2 s^2 - 4cs^3\, i + s^4

Why: recall i^2 = -1, i^3 = -i, i^4 = 1. Each power of i places the term in the real or imaginary group.

Step 3. Collect real and imaginary parts.

Real: c^4 - 6c^2 s^2 + s^4

Imaginary: 4c^3 s - 4cs^3 = 4cs(c^2 - s^2)

So \cos(4\theta) = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta.

Why: the real part of (\cos\theta + i\sin\theta)^4 equals \cos(4\theta) by De Moivre's theorem. Matching real parts is all that is needed.

Step 4. Replace \sin^2\theta with 1 - \cos^2\theta to express everything in terms of \cos\theta.

Let c = \cos\theta. Then s^2 = 1 - c^2.

\cos(4\theta) = c^4 - 6c^2(1 - c^2) + (1 - c^2)^2
= c^4 - 6c^2 + 6c^4 + 1 - 2c^2 + c^4
= 8c^4 - 8c^2 + 1
= 8\cos^4\theta - 8\cos^2\theta + 1

Why: expressing the formula purely in \cos\theta is useful for solving equations like \cos(4\theta) = k, which become polynomial equations in \cos\theta.

Result: \cos(4\theta) = 8\cos^4\theta - 8\cos^2\theta + 1.

Verification of the quadruple-angle formula by comparing cos 4 theta with the polynomial 8 cos-to-the-fourth minus 8 cos-squared plus 1A table showing three test values of theta: 0, pi over 6, and pi over 4. For each, the left column computes cos 4 theta directly, the right column evaluates 8 cos-to-the-fourth theta minus 8 cos-squared theta plus 1. Both columns match at each row, confirming the formula. θ cos 4θ 8c⁴ − 8c² + 1 0 cos 0 = 1 8 − 8 + 1 = 1 π/6 cos(2π/3) = −1/2 8(9/16) − 6 + 1 = −1/2 π/4 cos π = −1 8(1/4) − 4 + 1 = −1 All three rows match — the formula is confirmed.
A spot-check of the formula $\cos(4\theta) = 8\cos^4\theta - 8\cos^2\theta + 1$ at three values. At $\theta = 0$: both sides give $1$. At $\theta = \pi/6$: $\cos(2\pi/3) = -1/2$, and $8(3/4)^2 - 8(3/4) + 1 = 9/2 - 6 + 1 = -1/2$. At $\theta = \pi/4$: $\cos\pi = -1$, and $8(1/4) - 4 + 1 = -1$. The formula checks out.

The bonus formula \sin(4\theta) = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta comes from the imaginary part of the same expansion. De Moivre's theorem delivers both multiple-angle formulas simultaneously — a single computation replaces two separate derivations.

Common confusions

Going deeper

If you can state De Moivre's theorem, use it to compute powers, and derive multiple-angle formulas by separating real and imaginary parts of a binomial expansion, you have the core. The material below is for readers who want to see where the theorem leads.

Roots of unity

De Moivre's theorem runs in reverse to find roots. The equation z^n = 1 asks: which points on the unit circle, when raised to the nth power, land at 1? By De Moivre, if z = \cos\theta + i\sin\theta, then z^n = \cos(n\theta) + i\sin(n\theta) = 1 requires n\theta = 2k\pi for integer k, so \theta = 2k\pi/n. The n distinct roots are:

z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \qquad k = 0, 1, 2, \dots, n-1

These n points are equally spaced around the unit circle, forming a regular n-gon. The nth Roots of Unity article covers this in full.

Chebyshev polynomials

The formula \cos(n\theta) = T_n(\cos\theta), where T_n is a polynomial of degree n, defines the Chebyshev polynomials of the first kind. De Moivre's theorem is the tool that proves T_n exists and gives a method to compute it for any n. The first few are: T_0(x) = 1, T_1(x) = x, T_2(x) = 2x^2 - 1, T_3(x) = 4x^3 - 3x, T_4(x) = 8x^4 - 8x^2 + 1 (which you just derived in Example 2). These polynomials appear throughout numerical analysis, approximation theory, and signal processing — they are the "best" polynomial approximations in a precise sense.

Connection to the binomial series

For non-integer exponents, the binomial theorem becomes an infinite series, and so does De Moivre's formula. The correct generalisation uses the principal value of the complex logarithm and requires analytic continuation — machinery from complex analysis that goes well beyond JEE. The integer case, which is what you need, is completely rigorous as proved above.

Where this leads next

De Moivre's theorem is the gateway to several deeper topics in complex numbers and trigonometry.