euler's formula

Brandon Szeto · April 12, 2025

Euler’s Formula

Euler’s formula connects exponential growth, circular motion, and the two fundamental trigonometric waves:

$$ e^{j\omega} = \cos(\omega) + j\sin(\omega) $$

The expression is compact, but it hides a useful geometric picture. As $\omega$ changes, $e^{j\omega}$ traces a point around the unit circle in the complex plane. The real coordinate of that point is $\cos(\omega)$, and the imaginary coordinate is $\sin(\omega)$.

The demo below unwraps that motion into three linked views: the rotating point on the unit circle, the real projection as a cosine wave, and the imaginary projection as a sine wave.

Interactive Demo

$e^{j\omega}$ $e^{j\omega} = \cos(\omega) + j\sin(\omega)$ $\omega \in [0, 2\pi]$

$e^{j\omega}$

$\operatorname{Re}$ $\operatorname{Im}$

$\operatorname{Re}(e^{j\omega}) = \cos(\omega)$

$\omega$ $\cos(\omega)$

$\operatorname{Im}(e^{j\omega}) = \sin(\omega)$

$\omega$ $\sin(\omega)$

Drag the 3D view to rotate the camera.

The circular projection shows the complex point $e^{j\omega}$.

The gold projection is the real component, $\cos(\omega)$.

The red projection is the imaginary component, $\sin(\omega)$.

Reading the Visualization

Think of the complex plane as a flat coordinate system with a real axis and an imaginary axis. A point on the unit circle has radius $1$, so its position can be described by the angle $\omega$:

$$ \begin{aligned} x &= \cos(\omega) \\ y &= \sin(\omega) \end{aligned} $$

If we write the same point as a complex number, the horizontal coordinate becomes the real part and the vertical coordinate becomes the imaginary part:

$$ z = x + jy $$

Substituting the unit-circle coordinates gives:

$$ e^{j\omega} = \cos(\omega) + j\sin(\omega) $$

The 3D helix in the demo adds a vertical $\omega$ axis. Moving upward advances the angle, while the shadow of the helix on each plane reveals a simpler view:

The $xy$ projection is the unit circle.
The $xz$ projection is $\cos(\omega)$.
The $yz$ projection is $\sin(\omega)$.

That is the core idea: one rotating complex exponential contains both waves at the same time.

Why the Exponential Rotates

The usual real exponential $e^t$ grows or decays along a line. With an imaginary exponent, the value does not grow away from the origin. Instead, its magnitude stays fixed at $1$ and its phase changes:

$$ \left|e^{j\omega}\right| = 1 $$

So $e^{j\omega}$ is best read as a unit phasor: a vector of length $1$ rotating by angle $\omega$. Increasing $\omega$ moves the phasor counterclockwise around the complex plane. Its coordinates are exactly the cosine and sine components shown in the demo.

Euler’s Identity

Euler’s identity is the special case where $\omega = \pi$:

$$ e^{j\pi} = \cos(\pi) + j\sin(\pi) $$

Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$:

$$ e^{j\pi} = -1 $$

which gives the familiar identity:

$$ e^{j\pi} + 1 = 0 $$

It is not a separate trick so much as a memorable point on the same rotating curve. Half a turn around the unit circle lands at $-1$, and adding $1$ brings the result back to zero.