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Topos Uranos
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The complex exponential and Euler’s formula

Summary

This lesson defines the exponential of a complex number. It starts from two observations: the complex number cos⁡θ+isin⁡θ\cos \theta + i \sin \theta turns sums of angles into products, as a power turns sums of exponents into products, and the exponential series, evaluated at an imaginary number, splits into the series of the cosine and of the sine. Both lead to Euler’s formula, eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \theta, and to the definition ea+bi=ea(cos⁡b+isin⁡b)e^{a + bi} = e^a \left( \cos b + i \sin b \right). The lesson proves that the complex exponential turns sums into products, that its modulus is eae^a and its argument bb, that it never vanishes and that it is periodic with period 2πi2\pi i; it writes every complex number in the exponential form reiθr e^{i\theta}; it expresses the cosine and the sine by means of exponentials, extends them to the whole plane and discovers the hyperbolic functions in them. It ends with the complex logarithm, which has infinitely many values, and with powers with complex exponents, among which iii^i turns out to be a real number.

Learning objectives

  1. Explain why the law of exponents and the exponential series lead to Euler’s formula, distinguish that heuristic calculation from a proof and extend the definition of the exponential to every complex exponent.
  2. Prove the properties of the complex exponential (exponential of a sum, modulus, argument and periodicity) and operate with the exponential form of complex numbers.
  3. Express the cosine and the sine by means of exponentials, extend them to the complex numbers, decide which trigonometric identities remain valid in C\mathbb{C} and relate them to the hyperbolic functions.
  4. Solve the equation ew=ze^w = z, define the principal logarithm, compute powers with complex exponents and explain why they have, in general, infinitely many values.

Euler’s formula

In the lesson on the polar form of complex numbers it was proved that, when two complex numbers are multiplied, the moduli multiply and the arguments add. For complex numbers of modulus 11, the theorem says that

(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 E(θ)E(\theta) denotes the complex number cos⁡θ+isin⁡θ\cos \theta + i \sin \theta, the equality above is written E(α)E(β)=E(α+β)E(\alpha) E(\beta) = E(\alpha + \beta), and moreover E(0)=1E(0) = 1. Now, turning sums into products is the characteristic property of powers, ax+y=axaya^{x + y} = a^x a^y, studied in the lesson of the course Algebra and Geometry I on the exponential function; and De Moivre’s formula, E(θ)n=E(nθ)E(\theta)^n = E(n\theta), proved in the lesson on De Moivre’s formula, is the law (ax)n=anx\left( a^x \right)^n = a^{nx}. Everything suggests, therefore, that E(θ)E(\theta) is a power with an exponent proportional to θ\theta. The exponential series indicates which one.

The exponential series with an imaginary exponent

In the Course I lesson on the number e it was proved that ee is the sum of the series of the reciprocals of the factorials. More generally, for every real xx the following three equalities hold, in which the sums of infinitely many terms mean, as there, the value that the partial sums approach:

ex=∑k=0∞xkk!,cos⁡x=∑m=0∞(−1)mx2m(2m)!,sin⁡x=∑m=0∞(−1)mx2m+1(2m+1)!e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}, \qquad \cos x = \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{(2m)!}, \qquad \sin x = \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m + 1}}{(2m + 1)!}

These three series are admitted here without proof: they are obtained with Taylor’s theorem, which is studied in the course Calculus II, where it is also proved that they converge for every real xx; the convergence of the first one for every complex xx belongs to the study of series and of complex variables. The exponential series uses only sums, products and divisions, which make sense in C\mathbb{C}, and it can be evaluated at x=iθx = i\theta. Now, the symbol eiθe^{i\theta} does not yet have any meaning: a power with an imaginary exponent is not a product of equal factors, nor the value of the real exponential. The calculation in the example below is therefore not a proof that eiθe^{i\theta} equals anything, but a motivation: it shows that the sum of the series at iθi\theta is cos⁡θ+isin⁡θ\cos \theta + i \sin \theta, so that, if the series is to go on defining the exponential for imaginary exponents, there is no other possible choice.

ExampleThe exponential series at an imaginary number

Let θ\theta be real. We substitute x=iθx = i\theta in the exponential series and separate the terms of even index from those of odd index, using the cycle of the powers of ii from the Course I lesson on the complex numbers. The figure draws, for θ=π\theta = \pi, the partial sums SnS_n as a polygonal path: each term is the one before it rotated a quarter turn and multiplied by θk\frac{\theta}{k}.

Solution

  1. ∑k=0∞(iθ)kk!=1+iθ+(iθ)22!+(iθ)33!+⋯\dato{s}{\sum_{k=0}^{\infty} \frac{(i\theta)^k}{k!}} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \cdots

    We replace xx by iθi\theta in the series, without giving its sum any name yet. The first two partial sums are S0=1S_0 = 1 and S1=1+iθS_1 = 1 + i\theta.

  2. (iθ)2m=(−1)mθ2m,(iθ)2m+1=(−1)miθ2m+1\resaltar{(i\theta)^{2m}} = \dato{p}{\resaltar{(-1)^m \theta^{2m}}}, \qquad \resaltar{(i\theta)^{2m + 1}} = \dato{q}{\resaltar{(-1)^m i \theta^{2m + 1}}}

    The powers of ii with even exponent are (−1)m(-1)^m, because i2m=(i2)mi^{2m} = \left( i^2 \right)^m; those with odd exponent are obtained by multiplying by ii. The terms of even index are real and those of odd index, purely imaginary.

  3. S2N+1=∑m=0N(−1)mθ2m(2m)!+i∑m=0N(−1)mθ2m+1(2m+1)!\resaltar{S_{2N + 1}} = \dato{r}{\resaltar{\sum_{m=0}^{N} \frac{(-1)^m \theta^{2m}}{(2m)!} + i \sum_{m=0}^{N} \frac{(-1)^m \theta^{2m + 1}}{(2m + 1)!}}}

    In the partial sum S2N+1S_{2N + 1}, which has finitely many terms, we gather those of even index on one side and those of odd index on the other, and we factor ii out of the latter: the partial sums of the series of the cosine and of the sine result.

  4. ∑k=0∞(iθ)kk!=∑m=0∞(−1)mθ2m(2m)!+i∑m=0∞(−1)mθ2m+1(2m+1)!=cos⁡θ+isin⁡θ\sum_{k=0}^{\infty} \frac{(i\theta)^k}{k!} = \resaltar{\sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m}}{(2m)!}} + i \resaltar{\sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m + 1}}{(2m + 1)!}} = \resaltar{\cos \theta} + i \resaltar{\sin \theta}

    The two real sums approach the cosine and the sine of θ\theta, by the admitted series; and S2NS_{2N} differs from S2N+1S_{2N + 1} by a term that approaches 00. The sum of the series is the complex number of modulus 11 and argument θ\theta.

  5. Sn=∑k=0n(iπ)kk!→cos⁡π+isin⁡π=−1\resaltar{S_n = \sum_{k=0}^{n} \frac{(i\pi)^k}{k!}} \rightarrow \resaltar{\cos \pi + i \sin \pi = -1}

    With θ=π\theta = \pi, the polygonal path of the partial sums turns in a spiral and approaches −1-1: the sum S12S_{12} is less than one thousandth away from −1-1.

The calculation above, which rests on the admitted series, determines the sum of the series; what it cannot do is prove that this sum is “ee raised to the power iθi\theta”, because this expression has no prior meaning. It is therefore adopted as a definition.

The imaginary exponential

For every real number θ\theta, the imaginary exponential eiθe^{i\theta} is the complex number of modulus 11 and argument θ\theta given by Euler’s formula, which Leonhard Euler published in 1748.
eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \theta

With θ=0\theta = 0 one gets ei0=1e^{i0} = 1, as the law of exponents requires. The properties of the polar form and De Moivre’s formula now translate into the laws of powers, which no longer have to be remembered as rules about angles.

TheoremThe imaginary exponential

For all reals θ\theta, α\alpha and β\beta and every integer nn, the complex number eiθe^{i\theta} has modulus 11; eiαeiβ=ei(α+β)e^{i\alpha} e^{i\beta} = e^{i(\alpha + \beta)} and (eiθ)n=einθ\left( e^{i\theta} \right)^n = e^{in\theta} hold; the inverse of eiθe^{i\theta} is its conjugate e−iθe^{-i\theta}, and eiα=eiβe^{i\alpha} = e^{i\beta} if and only if α−β\alpha - \beta is an integer multiple of 2π2\pi. Each statement is the translation of a result from earlier lessons. The figure shows the point eiθe^{i\theta}, which travels around the circle of radius 11 counterclockwise as θ\theta increases.

Proof

  1. ∣eiθ∣=cos⁡2θ+sin⁡2θ=1=1\abs{e^{i\theta}} = \sqrt{\resaltar{\cos^2 \theta + \sin^2 \theta}} = \sqrt{\resaltar{1}} = 1

    The modulus is 11 by the Pythagorean identity of the lesson on the fundamental trigonometric identities: the point eiθe^{i\theta} is the point of the circle of radius 11 that corresponds to the angle θ\theta.

  2. eiαeiβ=(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)=cos⁡(α+β)+isin⁡(α+β)=ei(α+β)e^{i\alpha} e^{i\beta} = \resaltar{\left( \cos \alpha + i \sin \alpha \right) \left( \cos \beta + i \sin \beta \right)} = \resaltar{\cos(\alpha + \beta) + i \sin(\alpha + \beta)} = \dato{p}{e^{i(\alpha + \beta)}}

    The product rule is the product theorem in polar form with moduli equal to 11: the arguments add.

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

    The power is De Moivre’s formula, valid for every integer nn.

  4. e−iθ=cos⁡θ−isin⁡θ=eiθ‾,eiθe−iθ=ei0=1e^{-i\theta} = \resaltar{\cos \theta - i \sin \theta} = \conj{e^{i\theta}}, \qquad e^{i\theta} e^{-i\theta} = \resaltar{e^{i0}} = 1

    Since the cosine is even and the sine odd, e−iθe^{-i\theta} is the conjugate of eiθe^{i\theta}, its mirror image in the real axis; and by the product law, with β=−α\beta = -\alpha, it is also its inverse.

  5. eiα=eiβ↔α−β=2kπ,k∈Ze^{i\alpha} = e^{i\beta} \leftrightarrow \resaltar{\alpha - \beta = 2k\pi}, \quad k \in \mathbb{Z}

    By the equality of complex numbers in polar form, two complex numbers of modulus 11 coincide if and only if their arguments differ by an integer multiple of 2π2\pi: after a full turn, the point passes through the same place again.

Euler’s identity

With θ=π\theta = \pi, Euler’s formula gives eiπ=cos⁡π+isin⁡π=−1e^{i\pi} = \cos \pi + i \sin \pi = -1, that is:

eiπ+1=0e^{i\pi} + 1 = 0

This equality, called Euler’s identity, brings together in a single relation the numbers 00 and 11, the neutral elements of addition and multiplication, the imaginary unit ii and the two fundamental constants of analysis, ee and π\pi. In the same way, eiπ2=ie^{i\frac{\pi}{2}} = i, e−iπ2=−ie^{-i\frac{\pi}{2}} = -i and e2πi=1e^{2\pi i} = 1. In particular, since i=eiπ2i = e^{i\frac{\pi}{2}}, the powers of ii are in=einπ2i^n = e^{in\frac{\pi}{2}}: the cycle of period 44 from the Course I lesson on the complex numbers is the periodicity of the imaginary exponential, because four quarter turns add up to 2π2\pi.