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Demoivre Theorem






Powers and Roots Made Simple


Computing (1+i)100(1 + i)^{100} through repeated multiplication would fill pages with tedious algebra. Yet the answer emerges in seconds using De Moivre's theorem — a formula that transforms exponentiation into elementary arithmetic on angles and lengths. This same principle runs in reverse to extract roots, revealing that every nonzero complex number has exactly nn distinct nn-th roots, arranged in perfect geometric symmetry around the origin.



Introduction to De Moivre's Theorem

Raising complex numbers to powers in algebraic form quickly becomes unmanageable. Squaring z=a+biz = a + bi requires expanding (a+bi)2=a2+2abi+b2i2=(a2b2)+2abi(a + bi)^2 = a^2 + 2abi + b^2i^2 = (a^2 - b^2) + 2abi. Cubing demands multiplying this result by a+bia + bi again. By the time we reach z10z^{10} or z100z^{100}, the algebraic approach drowns in computation.

The trigonometric form offers rescue. When z=rcisθz = r\text{cis}\theta, multiplication becomes elegant: multiply moduli, add arguments. Squaring gives z2=r2cis(2θ)z^2 = r^2\text{cis}(2\theta). Cubing gives z3=r3cis(3θ)z^3 = r^3\text{cis}(3\theta). The pattern suggests a general rule — and De Moivre's theorem confirms it.

Abraham de Moivre, a French mathematician living in England, discovered this relationship in the early eighteenth century. Though he never stated the theorem in its modern form, his work on probability and trigonometric identities laid the foundation. The theorem bearing his name connects complex exponentiation to the simple operation of multiplying an angle by an integer.

De Moivre's theorem stands among the most useful results in complex analysis. It reduces high powers to trivial calculations, enables extraction of roots, derives trigonometric identities, and appears throughout physics and engineering wherever oscillations and rotations arise.

De Moivre's Formula

The theorem states a precise relationship between powers and angles. For any integer nn:

(cosθ+isinθ)n=cos(nθ)+isin(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)


Raise cisθ\text{cis}\theta to the nn-th power, and the result is cis(nθ)\text{cis}(n\theta). The angle simply multiplies by nn. No expansion, no collecting terms, no tracking powers of ii — just multiply the argument.

In the condensed notation: (cisθ)n=cis(nθ)(\text{cis}\theta)^n = \text{cis}(n\theta).

The formula extends naturally to complex numbers with arbitrary modulus. If z=rcisθz = r\text{cis}\theta, then:

zn=(rcisθ)n=rncis(nθ)z^n = (r\text{cis}\theta)^n = r^n\text{cis}(n\theta)


The modulus raises to the nn-th power while the argument multiplies by nn. Both operations are elementary. The number 2cis(30°)2\text{cis}(30°) raised to the sixth power becomes 26cis(180°)=64cis(180°)=642^6\text{cis}(180°) = 64\text{cis}(180°) = -64. What would require extensive algebra in rectangular form reduces to mental arithmetic in trigonometric form.

The formula works for negative integers as well. Since zn=1/znz^{-n} = 1/z^n, we have:

zn=rncis(nθ)z^{-n} = r^{-n}\text{cis}(-n\theta)


Negative exponents invert the modulus and negate the angle multiplier, maintaining the same structural simplicity.

Proof of De Moivre's Theorem

Mathematical induction establishes De Moivre's theorem for positive integers. The proof builds from the multiplication rule for trigonometric form: when two complex numbers multiply, their moduli multiply and their arguments add.

The base case n=1n = 1 holds trivially: (cosθ+isinθ)1=cos(1θ)+isin(1θ)(\cos\theta + i\sin\theta)^1 = \cos(1 \cdot \theta) + i\sin(1 \cdot \theta).

For the inductive step, assume the formula holds for n=kn = k:

(cosθ+isinθ)k=cos(kθ)+isin(kθ)(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)


We must prove it holds for n=k+1n = k + 1. Multiply both sides by cosθ+isinθ\cos\theta + i\sin\theta:

(cosθ+isinθ)k+1=(cos(kθ)+isin(kθ))(cosθ+isinθ)(\cos\theta + i\sin\theta)^{k+1} = (\cos(k\theta) + i\sin(k\theta))(\cos\theta + i\sin\theta)


The right side is a product of two numbers in trigonometric form. By the multiplication rule, we add arguments:

=cos(kθ+θ)+isin(kθ+θ)=cos((k+1)θ)+isin((k+1)θ)= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos((k+1)\theta) + i\sin((k+1)\theta)


This completes the induction. The theorem holds for all positive integers.

Extension to negative integers follows from the reciprocal. For n>0n > 0:

zn=1zn=1rncis(nθ)=1rncis(nθ)=rncis(nθ)z^{-n} = \frac{1}{z^n} = \frac{1}{r^n\text{cis}(n\theta)} = \frac{1}{r^n} \cdot \text{cis}(-n\theta) = r^{-n}\text{cis}(-n\theta)


The formula zn=rncis(nθ)z^n = r^n\text{cis}(n\theta) thus holds for all integers nn, positive, negative, or zero.

Applying De Moivre's Theorem to Powers

Computing high powers follows a three-step procedure. First convert to trigonometric form, then apply De Moivre's formula, finally convert back to algebraic form if required.

Consider (1+i)10(1 + i)^{10}.

Step 1: Convert 1+i1 + i to trigonometric form. The modulus is 1+i=12+12=2|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}. The argument satisfies tanθ=1/1=1\tan\theta = 1/1 = 1 with the point in the first quadrant, giving θ=45°=π/4\theta = 45° = \pi/4. Thus 1+i=2cis(45°)1 + i = \sqrt{2}\text{cis}(45°).

Step 2: Apply De Moivre's theorem.

(1+i)10=(2)10cis(10×45°)=25cis(450°)=32cis(450°)(1 + i)^{10} = (\sqrt{2})^{10}\text{cis}(10 \times 45°) = 2^5\text{cis}(450°) = 32\text{cis}(450°)


Since 450°=360°+90°450° = 360° + 90°, the angle reduces to 90°90°:

=32cis(90°)=32(cos90°+isin90°)=32(0+i)=32i= 32\text{cis}(90°) = 32(\cos 90° + i\sin 90°) = 32(0 + i) = 32i


Step 3: The answer 32i32i is already in algebraic form.

Verification through direct calculation would require multiplying 1+i1 + i by itself ten times — tedious and error-prone. De Moivre's theorem completes the same computation in a few lines.

Another example: find (3i)6(\sqrt{3} - i)^6. The modulus is 3+1=2\sqrt{3 + 1} = 2. The argument, with a=3>0a = \sqrt{3} > 0 and b=1<0b = -1 < 0, lies in the fourth quadrant: θ=30°\theta = -30°. So 3i=2cis(30°)\sqrt{3} - i = 2\text{cis}(-30°), and:

(3i)6=26cis(180°)=64cis(180°)=64(1)=64(\sqrt{3} - i)^6 = 2^6\text{cis}(-180°) = 64\text{cis}(-180°) = 64(-1) = -64

Introduction to nn-th Roots

De Moivre's theorem runs in reverse to solve equations of the form zn=wz^n = w. Given a complex number ww, what values of zz satisfy this equation? The answer reveals surprising multiplicity: every nonzero complex number has exactly nn distinct nn-th roots.

Real number experience suggests uniqueness. The equation x3=8x^3 = 8 has one real solution: x=2x = 2. But in C\mathbb{C}, the equation z3=8z^3 = 8 has three solutions. The cube root of 88 is not just 22 but three different complex numbers, each of which cubes to 88.

Where do these extra roots come from? The key lies in angle periodicity. Adding 360°360° (or 2π2\pi radians) to any argument produces the same complex number: cis(θ)=cis(θ+360°)\text{cis}(\theta) = \text{cis}(\theta + 360°). When we write w=Rcisϕw = R\text{cis}\phi, we could equally write w=Rcis(ϕ+360°)w = R\text{cis}(\phi + 360°) or w=Rcis(ϕ+720°)w = R\text{cis}(\phi + 720°).

Now suppose zn=wz^n = w with z=rcisαz = r\text{cis}\alpha. By De Moivre's theorem, rncis(nα)=Rcisϕr^n\text{cis}(n\alpha) = R\text{cis}\phi. Matching moduli gives r=R1/nr = R^{1/n}. Matching arguments gives nα=ϕ+360°kn\alpha = \phi + 360°k for some integer kk, so α=ϕ+360°kn\alpha = \frac{\phi + 360°k}{n}.

Different values of kk yield different angles α\alpha. As kk runs from 00 to n1n-1, we obtain nn distinct arguments. Beyond k=n1k = n-1, the angles repeat modulo 360°360°. Exactly nn roots exist, no more, no fewer.

The Formula for nn-th Roots

Let w=Rcisϕw = R\text{cis}\phi be any nonzero complex number. The nn-th roots of ww — all solutions to zn=wz^n = w — are given by:

zk=R1/ncis(ϕ+2πkn)for k=0,1,2,,n1z_k = R^{1/n}\text{cis}\left(\frac{\phi + 2\pi k}{n}\right) \quad \text{for } k = 0, 1, 2, \ldots, n-1


In degree measure:

zk=R1/ncis(ϕ+360°kn)for k=0,1,2,,n1z_k = R^{1/n}\text{cis}\left(\frac{\phi + 360°k}{n}\right) \quad \text{for } k = 0, 1, 2, \ldots, n-1


Each root has the same modulus: zk=R1/n|z_k| = R^{1/n}, the positive real nn-th root of the original modulus. The roots differ only in their arguments.

The first root (k=0k = 0) has argument ϕ/n\phi/n, the original angle divided by nn. Each subsequent root adds 360°/n360°/n to the argument. The nn roots space themselves evenly around the circle of radius R1/nR^{1/n}.

Consider finding the cube roots of 88. Write 8=8cis(0°)8 = 8\text{cis}(0°), so R=8R = 8 and ϕ=0°\phi = 0°. The formula gives:

zk=81/3cis(0°+360°k3)=2cis(120°k)z_k = 8^{1/3}\text{cis}\left(\frac{0° + 360°k}{3}\right) = 2\text{cis}(120°k)


For k=0k = 0: z0=2cis(0°)=2z_0 = 2\text{cis}(0°) = 2.
For k=1k = 1: z1=2cis(120°)=2(12+32i)=1+3iz_1 = 2\text{cis}(120°) = 2(-\frac{1}{2} + \frac{\sqrt{3}}{2}i) = -1 + \sqrt{3}i.
For k=2k = 2: z2=2cis(240°)=2(1232i)=13iz_2 = 2\text{cis}(240°) = 2(-\frac{1}{2} - \frac{\sqrt{3}}{2}i) = -1 - \sqrt{3}i.

Three roots: 22, 1+3i-1 + \sqrt{3}i, and 13i-1 - \sqrt{3}i. Each cubes to 88.

Geometric Interpretation of Roots

The nn-th roots of any complex number arrange themselves with perfect geometric regularity. Plotting them on the complex plane reveals a pattern of elegant symmetry.

All nn roots share the same modulus R1/nR^{1/n}, placing them on a circle centered at the origin with that radius. They do not cluster or scatter randomly — they distribute themselves at equal angular intervals around this circle.

The angular separation between consecutive roots is exactly 360°n\frac{360°}{n} (or 2πn\frac{2\pi}{n} radians). The cube roots of any number sit 120°120° apart. Fourth roots sit 90°90° apart. Fifth roots sit 72°72° apart. This uniform spacing holds regardless of which number we take roots of — only the radius and starting angle vary.

The roots form vertices of a regular nn-gon inscribed in the circle. Cube roots give an equilateral triangle. Fourth roots give a square. Fifth roots give a regular pentagon. The polygon may appear rotated depending on the argument of the original number, but its shape is always perfectly regular.

This geometric picture aids computation and verification. If one root is known, the others follow by rotating 360°/n360°/n around the origin. If roots fail to form a regular polygon, an error has occurred. The visual pattern provides both a computational shortcut and a consistency check.

The conjugate relationship appears when the original number ww is real and positive. In that case ϕ=0°\phi = 0°, and the roots distribute symmetrically about the real axis. Complex roots come in conjugate pairs, with the real axis serving as a line of mirror symmetry.

Roots of Unity

The nn-th roots of unity — solutions to zn=1z^n = 1 — form a special case of fundamental importance. Since 1=1cis(0°)1 = 1\text{cis}(0°), the formula simplifies:

zk=cis(360°kn)=cis(2πkn)for k=0,1,2,,n1z_k = \text{cis}\left(\frac{360°k}{n}\right) = \text{cis}\left(\frac{2\pi k}{n}\right) \quad \text{for } k = 0, 1, 2, \ldots, n-1


All roots of unity have modulus 11 — they lie on the unit circle. Their arguments are multiples of 360°n\frac{360°}{n}, dividing the circle into nn equal arcs.

The root z0=cis(0°)=1z_0 = \text{cis}(0°) = 1 always appears. Every integer power of 11 equals 11, making 11 an nn-th root of unity for every nn.

The fourth roots of unity illustrate the pattern: z0=1z_0 = 1, z1=iz_1 = i, z2=1z_2 = -1, z3=iz_3 = -i. These four points mark the cardinal directions on the unit circle, forming a square. Each satisfies z4=1z^4 = 1.

The sixth roots of unity are 11, cis(60°)\text{cis}(60°), cis(120°)\text{cis}(120°), 1-1, cis(240°)\text{cis}(240°), cis(300°)\text{cis}(300°). In algebraic form: 11, 12+32i\frac{1}{2} + \frac{\sqrt{3}}{2}i, 12+32i-\frac{1}{2} + \frac{\sqrt{3}}{2}i, 1-1, 1232i-\frac{1}{2} - \frac{\sqrt{3}}{2}i, 1232i\frac{1}{2} - \frac{\sqrt{3}}{2}i. They form a regular hexagon on the unit circle.

The primitive nn-th root of unity, denoted ω=cis(2πn)\omega = \text{cis}\left(\frac{2\pi}{n}\right), generates all others through its powers. Since zk=ωkz_k = \omega^k, the complete set of nn-th roots of unity is {1,ω,ω2,,ωn1}\{1, \omega, \omega^2, \ldots, \omega^{n-1}\}. This generator perspective connects roots of unity to group theory and number theory.

Properties of Roots of Unity

The nn-th roots of unity satisfy algebraic identities that reflect their geometric symmetry. These properties appear throughout mathematics, from polynomial factorization to Fourier analysis.

The sum of all nn-th roots of unity equals zero:

k=0n1zk=1+ω+ω2++ωn1=0\sum_{k=0}^{n-1} z_k = 1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0


Geometrically, the roots form a regular polygon centered at the origin. Their vector sum — placing all arrows tip to tail — returns to the starting point, yielding zero. Algebraically, this is the sum of a geometric series with ratio ω1\omega \neq 1: 1ωn1ω=111ω=0\frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0.

The product of all nn-th roots of unity follows a pattern:

k=0n1zk=(1)n+1\prod_{k=0}^{n-1} z_k = (-1)^{n+1}


For odd nn, the product is 11. For even nn, the product is 1-1. This connects to the constant term in the polynomial zn1z^n - 1, which factors as (zz0)(zz1)(zzn1)(z - z_0)(z - z_1)\cdots(z - z_{n-1}).

When n2n \geq 2, complex roots of unity come in conjugate pairs. If ωk\omega^k is a root, so is ωk=ωnk\overline{\omega^k} = \omega^{n-k}. The roots ω\omega and ωn1\omega^{n-1} are conjugates; ω2\omega^2 and ωn2\omega^{n-2} are conjugates; and so on. Only z=1z = 1 (always) and z=1z = -1 (when nn is even) sit on the real axis without conjugate partners among the roots.

These symmetries constrain the algebraic behavior of expressions involving roots of unity and simplify many calculations.

Worked Examples

Concrete examples consolidate the root-finding technique. Each follows the same pattern: express the target in trigonometric form, apply the formula, enumerate all nn roots.

Square roots of $i$: Solve z2=iz^2 = i.

Write i=1cis(90°)i = 1\text{cis}(90°). The roots are zk=11/2cis(90°+360°k2)z_k = 1^{1/2}\text{cis}\left(\frac{90° + 360°k}{2}\right) for k=0,1k = 0, 1.

z0=cis(45°)=22+22iz_0 = \text{cis}(45°) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i

z1=cis(225°)=2222iz_1 = \text{cis}(225°) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i

Two roots, opposite each other on the unit circle.

Cube roots of $-8$: Solve z3=8z^3 = -8.

Write 8=8cis(180°)-8 = 8\text{cis}(180°). The roots are zk=2cis(180°+360°k3)z_k = 2\text{cis}\left(\frac{180° + 360°k}{3}\right) for k=0,1,2k = 0, 1, 2.

z0=2cis(60°)=1+3iz_0 = 2\text{cis}(60°) = 1 + \sqrt{3}i

z1=2cis(180°)=2z_1 = 2\text{cis}(180°) = -2

z2=2cis(300°)=13iz_2 = 2\text{cis}(300°) = 1 - \sqrt{3}i

Three roots forming an equilateral triangle with radius 22.

Fourth roots of unity: Solve z4=1z^4 = 1.

Write 1=1cis(0°)1 = 1\text{cis}(0°). The roots are zk=cis(90°k)z_k = \text{cis}(90°k) for k=0,1,2,3k = 0, 1, 2, 3.

z0=1z_0 = 1, z1=iz_1 = i, z2=1z_2 = -1, z3=iz_3 = -i

A square on the unit circle at the four cardinal points.

Fifth roots of $32$: Solve z5=32z^5 = 32.

Write 32=32cis(0°)32 = 32\text{cis}(0°). The roots are zk=2cis(72°k)z_k = 2\text{cis}(72°k) for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

z0=2z_0 = 2, z1=2cis(72°)z_1 = 2\text{cis}(72°), z2=2cis(144°)z_2 = 2\text{cis}(144°), z3=2cis(216°)z_3 = 2\text{cis}(216°), z4=2cis(288°)z_4 = 2\text{cis}(288°)

A regular pentagon with radius 22, one vertex on the positive real axis.