Euler's Formula

The most important formula connecting exponential and trigonometric functions. Derived by substituting iθ into the Taylor series of eˣ and separating real and imaginary parts. The special case θ = π gives Euler's identity: e^(iπ) + 1 = 0.

Three functions — $e^x$, $\cos\theta$, $\sin\theta$ — look entirely unrelated when first encountered. Euler's formula reveals they are the same object, viewed from different angles.

The Taylor Series of $e^x$

For any real number $x$:

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

This series converges for all $x \in \mathbb{R}$. We now ask: what happens if we substitute $x = i\theta$, where $\theta$ is real?

Substituting $x = i\theta$

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

Each power of $i\theta$ simplifies using the cycle $i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$:

$$(i\theta)^1 = i\theta, \quad (i\theta)^2 = -\theta^2, \quad (i\theta)^3 = -i\theta^3, \quad (i\theta)^4 = \theta^4, \quad (i\theta)^5 = i\theta^5, \quad \ldots$$

Substituting:

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

Separating Real and Imaginary Parts

Collect terms without $i$ (real part) and terms with $i$ (imaginary part):

$$e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots\right)}_{\text{real part}} + \; i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\text{imaginary part}}$$

These two series are exactly the Taylor expansions of $\cos\theta$ and $\sin\theta$:

$$\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \qquad \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots$$

Therefore:

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

Euler's Identity

Setting $\theta = \pi$:

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

Five fundamental constants — $e$, $i$, $\pi$, $1$, $0$ — in a single equation. This is not a coincidence or a trick. It is the direct consequence of the series expansion above.

What the Formula Is Really Saying

Writing $z = e^{i\theta}$ means $|z| = 1$ and $\arg(z) = \theta$. The exponential $e^{i\theta}$ is simply the point on the unit circle at angle $\theta$. As $\theta$ increases from $0$ to $2\pi$, $e^{i\theta}$ traces the unit circle exactly once anticlockwise.

This is why multiplying any complex number $r e^{i\phi}$ by $e^{i\alpha}$ rotates it by $\alpha$ — you are adding angles in the exponent.