Euler's Formula

⚡ Key Takeaways

Euler's formula states that for any real number $\theta$ , $e^{i\theta} = \cos\theta + i\sin\theta$ . It binds exponential growth to circular motion, expressing complex exponentials in terms of trigonometric functions. It is foundational across complex analysis, signal processing, differential equations, and physics. Discovered by Leonhard Euler around 1740, published in his Introductio in analysin infinitorum (1748).

💡 The Core Insight

Multiplying by $e^{i\theta}$ is the same as rotating by angle $\theta$ . The exponential function, defined by the property that it equals its own derivative, when fed an imaginary argument, produces a quantity whose "growth" is always perpendicular to its current value. Perpendicular growth means circular motion — and circular motion is precisely sine and cosine. This is why a single exponential captures both.

DefinitionEuler's Formula

For any real number $\theta$ (measured in radians), the complex exponential satisfies:

1

\quad e^{i\theta} = \cos\theta + i\sin\theta

Derivation hint: Compare the Taylor series of $e^{i\theta}$ with those of $\cos\theta$ and $i\sin\theta$ . Here $e \approx 2.71828$ is Euler's number, $i$ is the imaginary unit with $i^2 = -1$ , and $\cos\theta$ , $\sin\theta$ are the standard trigonometric functions. The quantity $e^{i\theta}$ is a complex number of modulus $1$ , lying on the unit circle at angle $\theta$ from the positive real axis.

TheoremEuler's Identity

Setting $\theta = \pi$ in Euler's formula yields:

2

\quad e^{i\pi} + 1 = 0

Derivation hint: Substitute $\theta = \pi$ so that $\cos\pi = -1$ and $\sin\pi = 0$ . This identity connects $e$ , $i$ , $\pi$ , $1$ , and $0$ through the operations of addition, multiplication, and exponentiation, each appearing exactly once.

DefinitionInverse Relations

Euler's formula can be inverted to express the trigonometric functions in terms of complex exponentials:

3

\quad \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Derivation hint: Add and subtract $e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$ . These are indispensable for evaluating integrals, simplifying trigonometric identities, and analyzing waveforms.

ProofSeries Derivation

We derive Euler's formula by comparing power series, all of which converge absolutely for every complex argument.

Step 1. The exponential function has the Maclaurin series, valid for any complex $z$ :

4

\quad e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots

Derivation hint: This is the defining Taylor expansion of $e^z$ about $z=0$ ; it converges everywhere.

Step 2. Substitute $z = i\theta$ and use the cyclic powers of $i$ : $i^0=1$ , $i^1=i$ , $i^2=-1$ , $i^3=-i$ , $i^4=1$ , repeating with period $4$ :

5

\quad e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots

Derivation hint: Replace each $i^n$ by its value and track the alternating signs introduced by $i^2 = -1$ .

Step 3. Separate real and imaginary parts:

6

\quad e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)

Derivation hint: Group terms with no factor of $i$ separately from those carrying a factor of $i$ .

Step 4. Recognize the two grouped series as the Maclaurin expansions of $\cos\theta$ and $\sin\theta$ :

7

\quad \cos\theta = \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}, \qquad \sin\theta = \sum_{n=0}^\infty \frac{(-1)^n\theta^{2n+1}}{(2n+1)!}

Derivation hint: These are the standard Taylor series for cosine and sine about $\theta = 0$ . Therefore $e^{i\theta} = \cos\theta + i\sin\theta$ . $\blacksquare$

∎

ProofDifferential Equation Derivation

An alternative route avoids series and clarifies the geometric meaning.

Step 1. Define $f(\theta) = \cos\theta + i\sin\theta$ . Differentiate:

8

\quad f'(\theta) = -\sin\theta + i\cos\theta = i(\cos\theta + i\sin\theta) = i\,f(\theta)

Derivation hint: Factor out $i$ and use $i^2 = -1$ so that $i\cdot i\sin\theta = -\sin\theta$ .

Step 2. The function $f$ satisfies the initial value problem $f'(\theta) = i\,f(\theta)$ with $f(0) = 1$ . The unique solution of $g' = i g$ , $g(0)=1$ , is $g(\theta) = e^{i\theta}$ .

Step 3. By uniqueness of solutions to first-order linear ODEs, $f(\theta) = e^{i\theta}$ , establishing the formula. $\blacksquare$

∎

LE

Leonhard Euler

1707–1783

The Swiss mathematician Leonhard Euler was the most prolific mathematician in history, contributing foundational work to analysis, number theory, mechanics, and graph theory. He introduced the modern notation for $e$ , $i$ , the function notation $f(x)$ , and the Greek letter $\pi$ for the circle constant. Euler discovered his eponymous formula while studying the logarithms of complex and negative numbers, settling a long debate (with Leibniz and Johann Bernoulli) over the meaning of $\ln(-1)$ . Remarkably, he produced much of his greatest work after becoming totally blind in 1771, dictating to assistants from memory.

🧠 Intuition

Think of $e^{i\theta}$ as a particle moving on the unit circle. At $\theta=0$ it sits at the point $1$ on the real axis. As $\theta$ increases, the differential equation $f' = if$ says its velocity equals $i$ times its position — and multiplying a complex number by $i$ rotates it $90°$ counterclockwise. So the velocity is always perpendicular to the radius, which is the hallmark of uniform circular motion at unit speed. After traveling an arc length $\theta$ , the particle is at angle $\theta$ , with coordinates $(\cos\theta, \sin\theta)$ . The real exponential $e^t$ grows because its velocity points outward (parallel to position); the imaginary exponential $e^{i\theta}$ circles because its velocity points sideways (perpendicular to position). Euler's formula is the precise statement of this geometric fact.

Figure

The complex exponential

e^{i\theta}

traces the unit circle; its real and imaginary parts are

\cos\theta

and

\sin\theta

.

ExampleComputing a Complex Power

Given: Evaluate $(1+i)^8$ using Euler's formula.

Step 1. Write $1+i$ in polar form. Its modulus is $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ , and its argument is $\theta = \arctan(1/1) = \pi/4$ . Thus

9

\quad 1+i = \sqrt{2}\,e^{i\pi/4}

Derivation hint: Use $r = \sqrt{a^2+b^2}$ and $\theta = \arctan(b/a)$ for $a+bi$ in the first quadrant.

Step 2. Raise to the 8th power using $(re^{i\theta})^n = r^n e^{in\theta}$ :

10

\quad (1+i)^8 = (\sqrt{2})^8\, e^{i\cdot 8\pi/4} = 16\, e^{i2\pi}

Derivation hint: $(\sqrt 2)^8 = 2^4 = 16$ and $8\cdot\pi/4 = 2\pi$ .

Step 3. Since $e^{i2\pi} = \cos 2\pi + i\sin 2\pi = 1$ :

\boxed{(1+i)^8 = 16}

ExampleIntegrating with the Exponential Form

Given: Evaluate $\displaystyle\int_0^{\pi} e^{x}\cos x \, dx$ using $\cos x = \tfrac12(e^{ix}+e^{-ix})$ .

Step 1. Note that $e^x\cos x = \operatorname{Re}\!\left(e^{x}e^{ix}\right) = \operatorname{Re}\!\left(e^{(1+i)x}\right)$ .

Step 2. Integrate the complex exponential:

11

\quad \int_0^\pi e^{(1+i)x}\,dx = \left[\frac{e^{(1+i)x}}{1+i}\right]_0^\pi = \frac{e^{(1+i)\pi} - 1}{1+i}

Derivation hint: Treat $1+i$ as a constant; the antiderivative of $e^{kx}$ is $e^{kx}/k$ .

Step 3. Evaluate $e^{(1+i)\pi} = e^{\pi}e^{i\pi} = e^\pi(-1) = -e^\pi$ . So the integral is

12

\quad \frac{-e^\pi - 1}{1+i} = \frac{-(e^\pi+1)}{1+i}\cdot\frac{1-i}{1-i} = \frac{-(e^\pi+1)(1-i)}{2}

Derivation hint: Multiply numerator and denominator by the conjugate $1-i$ , since $(1+i)(1-i)=2$ .

Step 4. Take the real part:

\boxed{\int_0^\pi e^x\cos x\,dx = \frac{-(e^\pi+1)}{2} = -\frac{e^\pi+1}{2} \approx -12.07}

⚠️ Common mistake

Euler's formula does not say $e^{i\theta}$ is "imaginary." It is a complex number with both real part $\cos\theta$ and imaginary part $\sin\theta$ . Its modulus is always $1$ : $|e^{i\theta}| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$ . Also, $\theta$ must be in radians, not degrees — the series for $\sin$ and $\cos$ , and the derivative identities, only hold in radian measure.

⚠️ Common mistake

The complex exponential is not one-to-one. Because $e^{i(\theta + 2\pi k)} = e^{i\theta}$ for any integer $k$ , the function is periodic with period $2\pi$ . Consequently the complex logarithm is multivalued, and naïve manipulations like " $\ln(e^{i\theta}) = i\theta$ " require choosing a branch.

Counterexample

A tempting but false manipulation: "Since $e^{i\pi} = -1$ , squaring gives $e^{2\pi i} = 1$ , so taking square roots, $e^{\pi i} = e^0 = 1$ , hence $-1 = 1$ ." The error is that the complex square root (like the logarithm) is multivalued: $\sqrt{e^{2\pi i}}$ has two values, $+1$ and $-1$ , and the rule $(e^z)^{1/2} = e^{z/2}$ fails as a single-valued identity. This illustrates why branch cuts of the complex logarithm cannot be ignored.

connection

Euler's formula is the keystone of Complex Analysis, where it underlies the polar representation $z = re^{i\theta}$ and De Moivre's Theorem, $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ . It generalizes to The Complex Exponential Function $e^{z}=e^{x}(\cos y + i\sin y)$ for $z = x+iy$ . The inverse relations [3] are the seed of the Fourier Series and Fourier Transform, which decompose signals into complex exponentials. In differential equations, it provides the complex eigenfunctions $e^{\lambda t}$ that solve Linear Ordinary Differential Equations with constant coefficients. Geometrically, it links to Rotation Matrices via the isomorphism between $e^{i\theta}$ and the $2\times 2$ rotation by $\theta$ .

🔬 Where this appears

⏳ History

- 1714: Roger Cotes publishes the logarithmic form

\ln(\cos\theta + i\sin\theta) = i\theta

, an early equivalent statement, though buried in his work on integration.

- 1722: Abraham de Moivre states the power identity

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

, implicit in his earlier work since 1707.

- 1740: Euler, in correspondence and unpublished notes, recognizes that

e^{i\theta}

and

\cos\theta + i\sin\theta

satisfy the same differential equation, deriving the formula.

- 1741–1743: Euler resolves the dispute over

\ln(-1)

, showing it equals

i\pi

, by treating logarithms as multivalued — a direct consequence of his formula.

- 1748: Euler publishes the formula explicitly in Introductio in analysin infinitorum, presenting both the series and exponential viewpoints.

- 1797–1806: Caspar Wessel and Jean-Robert Argand develop the geometric interpretation of complex numbers as points in a plane, giving Euler's formula its rotational meaning.

- 1831: Carl Friedrich Gauss popularizes the complex plane and the term "complex number," cementing the geometric reading of

e^{i\theta}

.

- 1822: Joseph Fourier's Théorie analytique de la chaleur uses sinusoidal decompositions, later recast in the complex-exponential language built on Euler's formula.

open-problem

While Euler's formula itself is fully proven, its extensions touch deep and partly open territory. The multivaluedness of the complex logarithm, $\log z = \ln|z| + i(\arg z + 2\pi k)$ , demands the machinery of Riemann surfaces to handle rigorously; choosing branch cuts consistently across analytic continuations remains a subtle practical issue. Generalizations to noncommutative settings — the matrix exponential $e^{A}$ and Lie group exponentials — obey $e^{A}e^{B} = e^{A+B}$ only when $A$ and $B$ commute; the general case is governed by the Baker–Campbell–Hausdorff formula, whose full structure and convergence in infinite-dimensional algebras is still actively studied. Questions about the transcendence of values like $e^{i\theta}$ for algebraic $\theta$ connect to the Gelfond–Schneider theorem and open problems in transcendental number theory.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also