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Euler's identity

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Euler’s identity is one of my favorite equations.

eiπ+1=0.e^{i\pi} + 1 = 0.

It contains five constants, each exactly once: the additive identity 00, the multiplicative identity 11, the circle constant π\pi, the base of the natural exponential ee, and the imaginary unit i=1i = \sqrt{-1}. They are joined by one addition and one exponentiation, with no other symbols.

The identity is the θ=π\theta = \pi case of Euler’s formula

eiθ=cosθ+isinθ.e^{i\theta} = \cos\theta + i\sin\theta.

For a real exponent, exe^{x} is a positive real number. For a purely imaginary exponent, eiθe^{i\theta} has modulus 11 and argument θ\theta, so it is the point on the unit circle at angle θ\theta. At θ=π\theta = \pi that point is 1-1, which is the identity. This post derives Euler’s formula three ways, geometrically, as a limit, and from the power series, then reads off what each constant contributes.

Multiplying by i is a quarter turn

Put a complex number a+bia + bi at the point (a,b)(a, b) in the plane. Addition is then ordinary vector addition. Multiplication has a geometric description too.

Write each number in terms of its modulus rr (its distance from the origin) and its argument ϕ\phi (its angle from the positive real axis), so z=r(cosϕ+isinϕ)z = r(\cos\phi + i\sin\phi). Multiplying two such numbers and using the angle-addition formulas for sine and cosine gives

r1(cosα+isinα)r2(cosβ+isinβ)=r1r2(cos(α+β)+isin(α+β)).r_1(\cos\alpha + i\sin\alpha)\cdot r_2(\cos\beta + i\sin\beta) = r_1 r_2\bigl(\cos(\alpha + \beta) + i\sin(\alpha + \beta)\bigr).

So multiplying two complex numbers multiplies their lengths and adds their angles.

Now take the special case of multiplying by ii. Its modulus is 11 and its argument is a quarter turn, 9090^\circ. So multiplying any point by ii leaves its length unchanged and rotates it a quarter turn counterclockwise. Check it on the powers of ii: starting from 11 at (1,0)(1, 0), one multiplication gives ii at (0,1)(0, 1), the next gives i2=1i^2 = -1 at (1,0)(-1, 0), then i3=ii^3 = -i at (0,1)(0, -1), then i4=1i^4 = 1, home again after four quarter turns. The imaginary unit is the 9090^\circ rotation of the plane, written as a number.

The complex exponential traces the unit circle

Define z(θ)=eiθz(\theta) = e^{i\theta} and follow the point as θ\theta increases.

The exponential is the function that is its own derivative. With the constant ii sitting in the exponent, the chain rule gives

ddθeiθ=ieiθ.\frac{d}{d\theta}\, e^{i\theta} = i\, e^{i\theta}.

Read that as a statement about motion: the velocity z(θ)z'(\theta) is the position z(θ)z(\theta) multiplied by ii, that is, the position turned a quarter turn. Two things follow.

So eiθe^{i\theta} travels around the unit circle at unit speed. After parameter θ\theta it has covered arc length θ\theta, which on the unit circle is an angle of θ\theta radians (angle measured so that a full turn is 2π2\pi). Its coordinates are therefore (cosθ,sinθ)(\cos\theta, \sin\theta), which is Euler’s formula read straight off the motion:

eiθ=cosθ+isinθ.e^{i\theta} = \cos\theta + i\sin\theta.

Drag θ\theta below and watch eiθe^{i\theta} move around the circle, its projections on the two axes tracing cosine and sine. The red velocity arrow stays exactly a quarter turn ahead of the radius. At θ=π\theta = \pi the point sits at 1-1:

θ = 1.05
cos θ = 0.50sin θ = 0.87e = 0.50 + 0.87i

The compound-interest limit

A second derivation comes from continuous compounding.

Money at annual rate rr, compounded nn times, multiplies by (1+r/n)n(1 + r/n)^n over the year, and as the compounding gets finer this tends to ere^r. Each of the nn steps multiplies by 1+r/n1 + r/n, a number just larger than 11, so the balance grows.

Now make the rate imaginary, r=iθr = i\theta, so each step multiplies by 1+iθ/n1 + i\theta/n. That factor has modulus 1+θ2/n2\sqrt{1 + \theta^2/n^2}, barely more than 11, and argument arctan(θ/n)θ/n\arctan(\theta/n) \approx \theta/n. Each step barely stretches, but it does turn, by about θ/n\theta/n. Over nn steps the turning accumulates to θ\theta while the modulus tends to 11, since (1+θ2/n2)n/21(1 + \theta^2/n^2)^{n/2} \to 1. In the limit,

eiθ=limn(1+iθn) ⁣n,e^{i\theta} = \lim_{n\to\infty}\left(1 + \frac{i\theta}{n}\right)^{\!n},

a point on the unit circle at angle θ\theta.

The widget draws the partial products 1, (1+iθ/n), (1+iθ/n)2, 1,\ (1 + i\theta/n),\ (1 + i\theta/n)^2,\ \dots as a chain of segments. Because each factor stretches a little, the chain bulges just outside the circle and spirals; as nn grows the stretch per step shrinks and the corners settle onto the red arc, the open blue endpoint closing in on the true eiθe^{i\theta} in red. Slide nn up and watch the polygon flatten onto the circle:

n = 12θ = 3.14|(1 + /n)n| = 1.49total turn = 3.07

The readout total turn is narctan(θ/n)n\arctan(\theta/n), the angle actually accumulated by the nn steps; it climbs to θ\theta as nn grows, while the modulus (1+θ2/n2)n/2(1 + \theta^2/n^2)^{n/2} falls back to 11.

The series check

For a derivation with no picture at all, expand. The exponential series ex=k0xk/k!e^{x} = \sum_{k \ge 0} x^k/k! converges for every complex xx, and absolutely, so we may substitute x=iθx = i\theta and regroup the terms:

eiθ=k=0(iθ)kk!.e^{i\theta} = \sum_{k=0}^{\infty} \frac{(i\theta)^k}{k!}.

The powers of ii run through 1,i,1,i1, i, -1, -i with period four. The even powers k=2mk = 2m give i2m=(1)mi^{2m} = (-1)^m and land on the real axis; the odd powers k=2m+1k = 2m+1 give i2m+1=(1)mii^{2m+1} = (-1)^m i and land on the imaginary axis. Splitting the sum along that parity,

eiθ=m=0(1)mθ2m(2m)!cosθ  +  im=0(1)mθ2m+1(2m+1)!sinθ.e^{i\theta} = \underbrace{\sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m}}{(2m)!}}_{\cos\theta} \;+\; i\underbrace{\sum_{m=0}^{\infty} \frac{(-1)^m \theta^{2m+1}}{(2m+1)!}}_{\sin\theta}.

The two halves are exactly the Taylor series of cosine and sine. Same conclusion as the geometry, now as an algebraic identity. (Absolute convergence is what licenses the regrouping.)

The half-turn from 1 to −1

Set θ=π\theta = \pi. The point eiθe^{i\theta} has walked halfway around the unit circle, from 11 to 1-1:

eiπ=1,equivalentlyeiπ+1=0.e^{i\pi} = -1, \qquad\text{equivalently}\qquad e^{i\pi} + 1 = 0.

Now each constant contributes something specific, and none can be changed:

Change any one of them and the statement is false. The identity is a genuine relation among these five constants, not a coincidence of notation.

Applications

Euler’s formula is why complex numbers are the standard tool for periodic and rotational problems.

Three corollaries

A little history

Euler’s formula predates Euler’s compact statement of it. Roger Cotes published the equivalent logarithmic relation iθ=ln(cosθ+isinθ)i\theta = \ln(\cos\theta + i\sin\theta) in 1714, though he did not exponentiate it. Abraham de Moivre had the nn-th power identity that carries his name, in equivalent form, by the 1720s. Leonhard Euler stated eiθ=cosθ+isinθe^{i\theta} = \cos\theta + i\sin\theta in his 1748 Introductio in analysin infinitorum and put the exponential at the center of the subject. The compact eiπ+1=0e^{i\pi} + 1 = 0 is a later way of writing the θ=π\theta = \pi case. When David Wells polled readers of The Mathematical Intelligencer in 1990 for the most beautiful theorem in mathematics, Euler’s identity finished first.

Common misconceptions

References



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