Skip to content
Topos Uranos
English

De Moivre’s formula and powers of complex numbers

Summary

This lesson proves De Moivre’s formula, according to which the power with integer exponent nn of a complex number of modulus 11 and argument θ\theta has argument nθn\theta, and applies it in five directions. It computes powers with large exponents without expanding products and decides when a power is real or equals 11; it recovers the multiple-angle formulas by comparing the formula with the binomial theorem; it expresses the powers of the cosine and of the sine as sums of cosines of multiple angles, an operation called linearization; it computes the sums of sines and of cosines in arithmetic progression as a geometric sum of complex numbers, and it obtains sums of binomial coefficients as the real and imaginary parts of the powers of 1+i1 + i.

Learning objectives

  1. Prove De Moivre’s formula for every integer exponent and interpret the powers of a complex number geometrically.
  2. Compute powers of complex numbers by means of the polar form and determine when they are real or equal to 11.
  3. Derive the multiple-angle formulas and linearize the powers of the sine and of the cosine.
  4. Compute trigonometric sums and sums of binomial coefficients as the real and imaginary parts of sums and of powers of complex numbers.

De Moivre’s formula

In the lesson on the polar form of complex numbers, every nonzero complex number was written as z=r(cos⁡θ+isin⁡θ)z = r \left( \cos \theta + i \sin \theta \right), where r=∣z∣r = \abs{z} is its modulus and θ\theta is one of its arguments, and it was proved, with the formulas for the sine and the cosine of a sum, that in a product the moduli are multiplied and the arguments are added. For two complex numbers of modulus 11, that product theorem reads as follows:

(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)=cos⁡(α+β)+isin⁡(α+β)\left( \cos \alpha + i \sin \alpha \right) \left( \cos \beta + i \sin \beta \right) = \cos(\alpha + \beta) + i \sin(\alpha + \beta)

If the two factors are equal, the product is a square, with argument 2θ2\theta; if it is multiplied once more by the same factor, the result is a cube, with argument 3θ3\theta. When the product is repeated, the argument is added to itself as many times as there are factors, so that raising to a power reduces to multiplying an angle by an integer. This fact, which Abraham de Moivre used in an equivalent form at the beginning of the eighteenth century and which Euler wrote in its present form in 1748, is the subject of this lesson.

TheoremDe Moivre’s formula

For every real number θ\theta and every integer nn, the following equality holds.

(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)\left( \cos \theta + i \sin \theta \right)^n = \cos(n\theta) + i \sin(n\theta)

The proof treats three cases: the exponent 00, the natural exponents, by induction, and the negative exponents, by means of the inverse. It uses the principle of induction in the form of the lesson of the course Algebra and Geometry I on summations, progressions and induction, and the powers with integer exponent from the lesson on the complex numbers, where z0=1z^0 = 1 and z−m=1zmz^{-m} = \frac{1}{z^m}. The figure follows the proof on the circle of radius 11, with a small angle θ\theta and n=3n = 3.

Proof

  1. (cos⁡θ+isin⁡θ)0=1=cos⁡0+isin⁡0\left( \cos \theta + i \sin \theta \right)^0 = \resaltar{1} = \resaltar{\cos 0 + i \sin 0}

    For n=0n = 0, the left-hand side equals 11 by definition, and so does the right-hand side, because cos⁡0=1\cos 0 = 1 and sin⁡0=0\sin 0 = 0.

  2. (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)\dato{h}{\left( \cos \theta + i \sin \theta \right)^n = \cos(n\theta) + i \sin(n\theta)}

    For n=1n = 1, the equality is evident. Induction hypothesis: we assume that it holds for a natural number nn.

  3. (cos⁡θ+isin⁡θ)n+1=(cos⁡θ+isin⁡θ)n(cos⁡θ+isin⁡θ)=(cos⁡(nθ)+isin⁡(nθ))(cos⁡θ+isin⁡θ)\left( \cos \theta + i \sin \theta \right)^{n + 1} = \resaltar{\left( \cos \theta + i \sin \theta \right)^n} \left( \cos \theta + i \sin \theta \right) = \resaltar{\left( \cos(n\theta) + i \sin(n\theta) \right)} \left( \cos \theta + i \sin \theta \right)

    The power that comes after is the product of the preceding one by the base, and the preceding one is replaced by the induction hypothesis.

  4. (cos⁡θ+isin⁡θ)n+1=(cos⁡(nθ)cos⁡θ−sin⁡(nθ)sin⁡θ)+i(sin⁡(nθ)cos⁡θ+cos⁡(nθ)sin⁡θ)=cos⁡((n+1)θ)+isin⁡((n+1)θ)\left( \cos \theta + i \sin \theta \right)^{n + 1} = \resaltar{\left( \cos(n\theta) \cos \theta - \sin(n\theta) \sin \theta \right)} + i \resaltar{\left( \sin(n\theta) \cos \theta + \cos(n\theta) \sin \theta \right)} = \dato{p}{\cos((n + 1)\theta) + i \sin((n + 1)\theta)}

    When the product is expanded, the real part and the imaginary part are the right-hand sides of the formulas for the cosine and the sine of a sum, from the lesson on the sum and difference formulas: this is the theorem on the product and quotient in polar form. The formula for n+1n + 1 is obtained, and by the principle of induction it holds for every natural number.

  5. (cos⁡θ+isin⁡θ)−1=cos⁡(−θ)+isin⁡(−θ)=cos⁡θ−isin⁡θ\left( \cos \theta + i \sin \theta \right)^{-1} = \dato{i}{\resaltar{\cos(-\theta) + i \sin(-\theta)}} = \resaltar{\cos \theta - i \sin \theta}

    By the product theorem, cos⁡θ+isin⁡θ\cos \theta + i \sin \theta multiplied by cos⁡(−θ)+isin⁡(−θ)\cos(-\theta) + i \sin(-\theta) gives cos⁡0+isin⁡0=1\cos 0 + i \sin 0 = 1: the second factor is therefore the inverse of the first. Since the cosine is even and the sine is odd, the inverse is the conjugate.

  6. (cos⁡θ+isin⁡θ)−m=(cos⁡(−θ)+isin⁡(−θ))m=cos⁡(−mθ)+isin⁡(−mθ)\left( \cos \theta + i \sin \theta \right)^{-m} = \resaltar{\left( \cos(-\theta) + i \sin(-\theta) \right)^m} = \resaltar{\cos(-m\theta) + i \sin(-m\theta)}

    If n=−mn = -m, with mm natural, the power with exponent −m-m is the mmth power of the inverse, by the laws of exponents. The natural case is applied to the angle −θ-\theta.

The formula extends to all nonzero complex numbers, because the power of a product is the product of the powers, also for negative exponents: the inverse of a product is the product of the inverses.

TheoremPowers in polar form

If z=r(cos⁡θ+isin⁡θ)z = r \left( \cos \theta + i \sin \theta \right) is a nonzero complex number, then for every integer nn we have zn=rn(cos⁡(nθ)+isin⁡(nθ))z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right): the modulus of znz^n is rnr^n, and nθn\theta is an argument of znz^n. The real factor is separated from the factor of modulus 11, and De Moivre’s formula is applied to the latter. The figure follows the proof with a complex number of modulus slightly greater than 11 and n=3n = 3, and ends with z=3+i=2(cos⁡π6+isin⁡π6)z = \sqrt{3} + i = 2\left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) and its first three powers.

Proof

  1. zn=(r(cos⁡θ+isin⁡θ))n=rn(cos⁡θ+isin⁡θ)nz^n = \left( r \left( \cos \theta + i \sin \theta \right) \right)^n = \dato{p}{\resaltar{r^n} \left( \cos \theta + i \sin \theta \right)^n}

    The power of a product is the product of the powers: for natural exponents, by the commutativity and the associativity of the product, and for negative exponents because the inverse of a product is the product of the inverses.

  2. zn=rn(cos⁡θ+isin⁡θ)n=rn(cos⁡(nθ)+isin⁡(nθ))z^n = r^n \resaltar{\left( \cos \theta + i \sin \theta \right)^n} = \dato{f}{r^n \resaltar{\left( \cos(n\theta) + i \sin(n\theta) \right)}}

    De Moivre’s formula, valid for every integer nn, is applied to the second factor.

  3. rn>0→∣zn∣=rn,arg⁡zn=nθ\resaltar{r^n > 0} \rightarrow \abs{z^n} = \resaltar{r^n}, \qquad \arg z^n = \resaltar{n\theta}

    Since r>0r > 0, the power rnr^n is a positive real number, also for negative nn. By the theorem on equality in polar form, the expression obtained is a polar form of znz^n: its modulus is rnr^n, and nθn\theta is one of its arguments.

  4. (3+i)3=23(cos⁡π2+isin⁡π2)=8i\left( \sqrt{3} + i \right)^3 = \resaltar{2^3} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = \resaltar{8i}

    With z=3+iz = \sqrt{3} + i, the powers have moduli 22, 44 and 88 and arguments π6\frac{\pi}{6}, π3\frac{\pi}{3} and π2\frac{\pi}{2}: the cube is a pure imaginary number.

It follows that the powers z,z2,z3,…z, z^2, z^3, \ldots have moduli in geometric progression with common ratio rr and arguments in arithmetic progression with common difference θ\theta, in the sense of the Course I lesson on summations and progressions. If r=1r = 1, they all lie on the circle of radius 11, each one rotated through an angle θ\theta with respect to the one before it; if r>1r > 1, they rotate and move away from the origin; if r<1r < 1, they rotate and move toward it. Moreover, since zk+1=z⋅zkz^{k + 1} = z \cdot z^k and the product by zz is a rotation through the angle θ\theta composed with a dilation by a factor rr, all the triangles with vertices 00, zkz^k and zk+1z^{k + 1} are similar to the one with vertices 00, 11 and zz, as was proved in the lesson on the polar form.

ExampleThe spiral of the powers

Let zz be the complex number of modulus 1110\frac{11}{10}, slightly greater than 11, and argument π4\frac{\pi}{4}. We compute some of its powers up to the eighth and describe the curve on which they lie.

Proof

  1. z=1110(cos⁡π4+isin⁡π4)=11220(1+i)\dato{z}{z = \frac{11}{10} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)} = \frac{11\sqrt{2}}{20} (1 + i)

    In polar form and in rectangular form, the number is as follows.

  2. z2=(1110)2(cos⁡π2+isin⁡π2)=121100iz^2 = \resaltar{\left( \frac{11}{10} \right)^2} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = \resaltar{\frac{121}{100} i}

    By the theorem on powers in polar form, the square has argument π2\frac{\pi}{2}: it is a pure imaginary number.

  3. z4=(1110)4(cos⁡π+isin⁡π)=−1464110000z^4 = \resaltar{\left( \frac{11}{10} \right)^4} \left( \cos \pi + i \sin \pi \right) = \resaltar{-\frac{14641}{10000}}

    The fourth power has argument π\pi: it is a negative real number.

  4. z8=(1110)8(cos⁡2π+isin⁡2π)=214358881100000000z^8 = \resaltar{\left( \frac{11}{10} \right)^8} \left( \cos 2\pi + i \sin 2\pi \right) = \resaltar{\frac{214358881}{100000000}}

    The eighth power has argument 2π2\pi: after a full turn it comes back to the positive real half-axis, at a distance from the origin slightly greater than 22.

  5. ∣zk+1−zk∣=∣zk(z−1)∣=∣z∣k∣z−1∣\abs{z^{k + 1} - z^k} = \resaltar{\abs{z^k \left( z - 1 \right)}} = \resaltar{\abs{z}^k \abs{z - 1}}

    The distance between two consecutive powers also grows in geometric progression with common ratio 1110\frac{11}{10}, by the theorem on the modulus of a product. The points lie on the curve which, in polar coordinates, studied in the lesson on polar coordinates and polar curves, has the equation r=(1110)4θπr = \left( \frac{11}{10} \right)^{\frac{4\theta}{\pi}}: a spiral, called logarithmic, which, as is proved with differential calculus, cuts every line through the origin always at the same angle.