In short

Euler's formula states that e^{i\theta} = \cos\theta + i\sin\theta. It connects the exponential function to trigonometry through complex numbers. It can be derived by comparing the Taylor series of e^x, \cos x, and \sin x. Setting \theta = \pi gives e^{i\pi} + 1 = 0, which links five fundamental constants in a single equation.

Here is a question that sounds impossible. Take the number e \approx 2.718, the base of natural logarithms. Raise it to a power. But make the power imaginary — not a real number, but i times a real number, where i = \sqrt{-1}.

What is e^{i \cdot \pi/2}?

You know what e^2 means: e \times e. You know what e^{1/2} means: \sqrt{e}. But e^{i\pi/2}? What does it mean to multiply e by itself "i\pi/2 times"? That question does not even make grammatical sense.

And yet the answer is i. Exactly i. Not approximately, not in some limit — exactly. The number e, raised to the power i\pi/2, equals i.

The equation that explains this, and a great deal more, is Euler's formula:

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

It says that raising e to an imaginary power produces a complex number whose real part is \cos\theta and whose imaginary part is \sin\theta. The exponential function and the trigonometric functions — which seem like completely unrelated ideas — turn out to be the same thing, viewed through the lens of complex numbers.

What the formula is saying geometrically

Think of the complex plane. Every complex number z = a + bi corresponds to a point (a, b) on this plane. The number \cos\theta + i\sin\theta has real part \cos\theta and imaginary part \sin\theta.

Where does that point sit? Its distance from the origin is

|\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1

So it lies on the unit circle — the circle of radius 1 centred at the origin. And the angle this point makes with the positive real axis is exactly \theta (that is the definition of cosine and sine as coordinates on the unit circle).

Euler's formula is telling you: e^{i\theta} is the point on the unit circle at angle \theta.

As \theta increases from 0 to 2\pi, the point e^{i\theta} walks around the entire unit circle, counterclockwise, at a steady pace. At \theta = 0, it sits at 1. At \theta = \pi/2, it sits at i. At \theta = \pi, it sits at -1. At \theta = 3\pi/2, it sits at -i. And at \theta = 2\pi, it is back at 1.

The unit circle traced by e to the i thetaThe unit circle on the complex plane with four key points marked: 1 at angle 0, i at angle pi over 2, negative 1 at angle pi, and negative i at angle 3 pi over 2. A red point sits on the circle and can be dragged to see how the angle theta and the coordinates cos theta and sin theta change. Re Im 1 i −1 −i e^(iθ) θ
Euler's formula in one picture. The point $e^{i\theta}$ sits on the unit circle at angle $\theta$. Its coordinates are $(\cos\theta, \sin\theta)$. As $\theta$ sweeps from $0$ to $2\pi$, the point walks the full circle. The exponential function, which grows explosively on the real line, becomes periodic rotation on the imaginary axis.

This is remarkable. The exponential function e^x, which on the real line shoots off to infinity as x grows, becomes — on the imaginary axis — circular motion. Growth turns into rotation. That single transformation is the deepest thing about Euler's formula.

The derivation: three series that conspire

The proof of Euler's formula starts with three infinite series you may have already seen. Each one is a Taylor series — a way to write a function as an infinite sum of powers of x.

The exponential series

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots

This series converges for every real number x, and it is the definition of e^x in the sense that everything else — the growth curve, the compound interest, the "its own derivative" property — follows from this series.

The cosine series

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

Only even powers of x, with alternating signs.

The sine series

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

Only odd powers of x, with alternating signs.

These three series were known well before anyone thought to connect them. The sine and cosine series were discovered by Madhava of Sangamagrama in the 14th century — three hundred years before they appeared in European mathematics.

Now substitute x = i\theta

Replace x with i\theta in the exponential series and see what happens.

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

Compute each power of i\theta:

The powers of i cycle: i, -1, -i, 1, i, -1, -i, 1, \ldots Every four steps, the pattern repeats. Substituting:

e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \cdots

Now separate the real terms (those without i) from the imaginary terms (those with i):

e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots\right)}_{\cos\theta} + \;i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots\right)}_{\sin\theta}

The real part is exactly the cosine series. The imaginary part is exactly the sine series. Therefore:

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

That is Euler's formula, derived from nothing more than substituting i\theta into the exponential series and sorting real from imaginary.

The exponential series splits into cosine and sineA visual showing the terms of the exponential series for e to the i theta. The terms are colour-coded: real terms (which form the cosine series) are in black, and imaginary terms (which form the sine series) are in red. The cycling pattern of i powers causes the split. e^(iθ) = 1 + iθ − θ²/2! − iθ³/3! + θ⁴/4! + iθ⁵/5! − ... Real terms: 1 − θ²/2! + θ⁴/4! − θ⁶/6! + ... = cos θ Imaginary terms (factor out i): i × (θ − θ³/3! + θ⁵/5! − θ⁷/7! + ...) = i sin θ e^(iθ) = cos θ + i sin θ
The derivation in one picture. The exponential series for $e^{i\theta}$ splits naturally into two pieces: the even-powered terms (real, no $i$) form $\cos\theta$, and the odd-powered terms (imaginary, carrying $i$) form $i\sin\theta$. The cycling of powers of $i$ is the engine that separates them.

Why this proof works

The proof rests on one bold move: taking the exponential series, which is defined for real numbers, and plugging in a complex number i\theta. Is that allowed?

Yes — because the exponential series converges absolutely for every complex number, not just real ones. Absolute convergence means you can rearrange the terms freely without changing the sum. So splitting the series into real and imaginary parts is legitimate, and the result is valid.

This kind of reasoning — extending a real function to complex inputs by keeping its series definition — is the central technique of complex analysis. Euler's formula is the first and most important result you get from it.

Applications to trigonometry

Euler's formula turns trigonometric identities into algebra. Once you know that e^{i\theta} = \cos\theta + i\sin\theta, you can derive identities that would take pages of trigonometric manipulation in a few lines.

Expressing \cos\theta and \sin\theta in terms of exponentials

Since e^{i\theta} = \cos\theta + i\sin\theta and e^{-i\theta} = \cos\theta - i\sin\theta (replace \theta with -\theta and use the even/odd properties of cosine and sine), you can solve for \cos\theta and \sin\theta:

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

These are remarkable. They express trigonometric functions — periodic, oscillating, geometric — as combinations of exponentials. The two worlds are completely interchangeable.

Deriving the compound-angle formulas

The compound-angle formula \cos(\alpha + \beta) + i\sin(\alpha + \beta) = ? becomes trivial with Euler's formula.

e^{i(\alpha+\beta)} = e^{i\alpha} \cdot e^{i\beta}

The left side is \cos(\alpha+\beta) + i\sin(\alpha+\beta). The right side is:

(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)

Expand the product:

= \cos\alpha\cos\beta + i\cos\alpha\sin\beta + i\sin\alpha\cos\beta + i^2\sin\alpha\sin\beta
= (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)

Comparing real and imaginary parts:

\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

Both compound-angle formulas, derived in two lines. No geometric constructions, no auxiliary triangles. Just multiplication of exponentials.

Deriving product-to-sum formulas

Euler's formula makes product-to-sum identities — often memorised by brute force — into obvious algebra. Consider \cos\alpha\cos\beta. Write each cosine in exponential form:

\cos\alpha\cos\beta = \frac{e^{i\alpha} + e^{-i\alpha}}{2} \cdot \frac{e^{i\beta} + e^{-i\beta}}{2}

Expand the four terms:

= \frac{e^{i(\alpha+\beta)} + e^{i(\alpha-\beta)} + e^{-i(\alpha-\beta)} + e^{-i(\alpha+\beta)}}{4}

Group the pairs:

= \frac{(e^{i(\alpha+\beta)} + e^{-i(\alpha+\beta)}) + (e^{i(\alpha-\beta)} + e^{-i(\alpha-\beta)})}{4} = \frac{2\cos(\alpha+\beta) + 2\cos(\alpha-\beta)}{4}
= \frac{\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2}

That is the product-to-sum formula for \cos\alpha\cos\beta, derived mechanically. The identities for \sin\alpha\sin\beta and \sin\alpha\cos\beta fall out from the same approach. Euler's formula reduces memorisation to computation.

The connection to De Moivre's theorem

De Moivre's theorem states that

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

With Euler's formula, this becomes transparent. The left side is (e^{i\theta})^n = e^{in\theta}. The right side is e^{in\theta} written out. They are the same thing. De Moivre's theorem is a corollary of Euler's formula — it falls out immediately from the law of exponents applied to complex exponentials.

And since De Moivre's theorem holds for all integers n (including negative integers, which give reciprocals), Euler's formula gives you De Moivre's theorem for free, without induction.

Finding nth roots of unity

De Moivre's theorem, powered by Euler's formula, also gives the nth roots of unity — the n complex numbers z such that z^n = 1. If z^n = 1 = e^{i \cdot 2k\pi} for any integer k, then z = e^{i \cdot 2k\pi/n}. Taking k = 0, 1, 2, \ldots, n-1 gives n distinct roots:

z_k = e^{2\pi i k / n} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}, \qquad k = 0, 1, \ldots, n-1

These roots are equally spaced around the unit circle, like the vertices of a regular n-gon. For n = 4: the four points are 1, i, -1, -i — the four compass points of the complex plane.

The sixth roots of unity on the unit circleSix points equally spaced around the unit circle, forming a regular hexagon. The points are the sixth roots of unity: 1, e to the i pi over 3, e to the i 2 pi over 3, negative 1, e to the negative i 2 pi over 3, and e to the negative i pi over 3. Re Im 1 e^(iπ/3) e^(i2π/3) −1 e^(−i2π/3) e^(−iπ/3)
The six sixth roots of unity, equally spaced around the unit circle. Each is $e^{2\pi i k/6}$ for $k = 0, 1, \ldots, 5$. They form the vertices of a regular hexagon. Euler's formula makes this geometric fact algebraically obvious: dividing the full angle $2\pi$ into $n$ equal parts produces $n$ equally spaced points on the circle.

Worked examples

Example 1: Evaluate $e^{i\pi/3}$ and verify on the unit circle

Step 1. Apply Euler's formula with \theta = \pi/3:

e^{i\pi/3} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}

Why: direct substitution into e^{i\theta} = \cos\theta + i\sin\theta.

Step 2. Evaluate the trig functions. \cos(\pi/3) = 1/2 and \sin(\pi/3) = \sqrt{3}/2. So:

e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}

Why: these are standard values from the 30-60-90 triangle. \pi/3 radians is 60°.

Step 3. Verify the modulus. |e^{i\pi/3}| = \sqrt{(1/2)^2 + (\sqrt{3}/2)^2} = \sqrt{1/4 + 3/4} = \sqrt{1} = 1.

Why: every point e^{i\theta} should have modulus 1 — it lives on the unit circle.

Step 4. Verify the argument. \arg(e^{i\pi/3}) = \tan^{-1}\!\frac{\sqrt{3}/2}{1/2} = \tan^{-1}\sqrt{3} = \pi/3.

Why: the argument should equal \theta, confirming that the point is at angle \pi/3 on the unit circle.

Result: e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}\,i, a point on the unit circle at 60°.

The point e to the i pi over 3 on the unit circleThe unit circle on the complex plane with the point e to the i pi over 3 marked at angle 60 degrees. A dashed line from the origin to the point shows the radius of 1. Projections onto the axes show the real part one half and imaginary part square root of 3 over 2. Re Im 1 i e^(iπ/3) √3/2 1/2 60°
The point $e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$ on the unit circle. The horizontal projection gives the real part $\frac{1}{2}$; the vertical projection gives the imaginary part $\frac{\sqrt{3}}{2}$. The dashed radius has length $1$, and the angle is exactly $60° = \pi/3$ radians — confirming Euler's formula point by point.

The picture confirms what the algebra says: e^{i\pi/3} is a point on the unit circle at 60°, with coordinates that match the cosine and sine of that angle exactly.

Example 2: Use Euler's formula to derive $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$

Step 1. Write \cos\theta using the exponential form:

\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}

Why: this is the exponential representation of cosine, derived earlier from Euler's formula and its conjugate.

Step 2. Square both sides:

\cos^2\theta = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^2 = \frac{e^{2i\theta} + 2e^{i\theta}e^{-i\theta} + e^{-2i\theta}}{4}

Why: expand using (a+b)^2 = a^2 + 2ab + b^2. The cross term e^{i\theta} \cdot e^{-i\theta} = e^0 = 1.

Step 3. Simplify. The middle term is 2 \cdot 1 = 2. The remaining terms combine:

\cos^2\theta = \frac{e^{2i\theta} + e^{-2i\theta} + 2}{4} = \frac{2\cos 2\theta + 2}{4} = \frac{1 + \cos 2\theta}{2}

Why: e^{2i\theta} + e^{-2i\theta} = 2\cos 2\theta, using the exponential form of cosine with angle 2\theta.

Result: \cos^2\theta = \dfrac{1 + \cos 2\theta}{2}.

The solid curve is $y = \cos^2\theta$ and the dashed curve is $y = \frac{1 + \cos 2\theta}{2}$. They overlap perfectly — you cannot tell them apart, because they are the same function. The identity is not an approximation; it is exact. Euler's formula derived it in three lines of algebra, with no trigonometric identities needed as input.

This is the power of Euler's formula as a tool: an identity that takes several steps to prove by trigonometric manipulation becomes a short algebraic expansion when you work with exponentials. The technique generalises — you can derive \cos^3\theta, \sin^4\theta, or any power of a trig function the same way.

The most beautiful equation

Set \theta = \pi in Euler's formula:

e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0 = -1

Rearranging:

e^{i\pi} + 1 = 0

This is sometimes called the most beautiful equation in mathematics. It connects five constants — e, i, \pi, 1, and 0 — from five different areas of mathematics (analysis, complex numbers, geometry, arithmetic, and the additive identity), using three operations (exponentiation, multiplication, addition), and nothing else. Each constant entered mathematics for its own reason; none was designed to relate to the others. And yet they lock together in a single, clean identity.

The equation e^{i\pi} = -1 also has a geometric meaning. It says: start at 1 on the unit circle. Rotate by angle \pi — half a full turn. You land at -1. Euler's formula gives this simple rotation a compact name.

There are other special values worth verifying by hand:

\theta e^{i\theta} Verification
0 \cos 0 + i\sin 0 = 1 The starting point: e^0 = 1
\pi/2 \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i A quarter turn lands at i
\pi \cos\pi + i\sin\pi = -1 A half turn lands at -1
3\pi/2 \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = -i Three-quarter turn lands at -i
2\pi \cos 2\pi + i\sin 2\pi = 1 Full circle: back to the start

Each row is a checkpoint. Together they trace the four compass points of the unit circle, confirming that e^{i\theta} rotates at a uniform pace.

Special values of e to the i theta on the unit circleThe unit circle with five points marked at theta equals 0, pi over 2, pi, 3 pi over 2, and 2 pi. Arrows show the counterclockwise progression. The point at theta equals pi is labelled as negative 1, which is the famous identity e to the i pi equals negative 1. Re Im θ = 0 e⁰ = 1 θ = π/2 = i θ = π = −1 θ = 3π/2 = −i
The five checkpoint values of $e^{i\theta}$. The point at $\theta = \pi$ — the famous $e^{i\pi} = -1$ — is highlighted. The counterclockwise progression from $1$ through $i$, $-1$, $-i$, and back to $1$ traces a full rotation in $2\pi$ radians.

Common confusions

Going deeper

If you came here to understand what Euler's formula says and where it comes from, you have it — you can stop here. The rest of this section is for readers who want to see the formula's wider consequences and the connection to differential equations.

Why e^{i\theta} must satisfy |e^{i\theta}| = 1

There is a slick argument that does not use series at all. Consider the function f(\theta) = e^{i\theta} \cdot e^{-i\theta}. By the law of exponents, f(\theta) = e^{i\theta - i\theta} = e^0 = 1 for all \theta. But e^{-i\theta} = \overline{e^{i\theta}} (the complex conjugate), so f(\theta) = e^{i\theta} \cdot \overline{e^{i\theta}} = |e^{i\theta}|^2. Therefore |e^{i\theta}|^2 = 1, which means |e^{i\theta}| = 1. The point e^{i\theta} must lie on the unit circle, for every \theta, without needing to know what its coordinates are.

The differential equation viewpoint

Consider the function z(\theta) = e^{i\theta}. Taking its derivative with respect to \theta:

\frac{dz}{d\theta} = ie^{i\theta} = iz

So z(\theta) = e^{i\theta} satisfies the differential equation z' = iz with initial condition z(0) = 1. This equation says: at every moment, the velocity of z is perpendicular to z itself (because multiplying by i rotates a complex number by 90°). A particle whose velocity is always perpendicular to its position vector moves in a circle. This gives you another proof that e^{i\theta} traces the unit circle — one that explains why exponentials produce circles, not just that they do.

The polar form of complex numbers

Euler's formula gives the cleanest way to write a complex number in polar form. Any complex number z with modulus r and argument \theta can be written as:

z = r\,e^{i\theta}

Multiplying two such numbers: z_1 z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 \, e^{i(\theta_1 + \theta_2)}. The moduli multiply and the arguments add — the fundamental rule of multiplication in polar form — which is now obvious from the laws of exponents. More on this in the article on polar form.

The general complex exponential

What about e^{a+bi} where a is non-zero? Separate the real and imaginary parts of the exponent:

e^{a+bi} = e^a \cdot e^{bi} = e^a(\cos b + i\sin b)

So e^a controls the magnitude (the radial distance from the origin) and e^{bi} controls the angle. The general complex exponential is a spiral: as a grows, the point moves away from the origin; as b grows, it rotates. A pure imaginary exponent (a = 0) produces pure rotation on the unit circle — that is Euler's formula. A pure real exponent (b = 0) produces pure growth along the positive real axis — that is the usual exponential. The general case combines both.

Connections to Fourier analysis

Every periodic function can be decomposed into a sum of e^{in\theta} terms — this is a Fourier series. Euler's formula is the reason Fourier analysis works: it provides the basis functions (e^{in\theta}) that span the space of periodic functions. If you continue into higher mathematics, you will find Euler's formula sitting at the foundation of signal processing, quantum mechanics, and the theory of differential equations. It is not an isolated identity — it is a cornerstone.

Madhava and the series foundation

The Taylor series for \sin x and \cos x that power the derivation above were discovered by Madhava of Sangamagrama in the 14th century, as part of the Kerala school of astronomy and mathematics. The series for \sin x appears in the Yuktibhasa (c. 1530), which gives a rigorous derivation using what we would now call Riemann sums. The exponential series e^x = \sum x^n/n! arrived later in the European tradition. Euler's formula connects two streams of mathematical discovery — the Indian series tradition and the European theory of complex numbers — into one equation.

Where this leads next