Euler's Formula — Moving Particle
Interactive decomposition of Figure 1.8 from Visual Complex Analysis (Tristan Needham, Oxford University Press, 2023), Section 1.2.2. If is the position of a particle in the complex plane, then its velocity is always perpendicular to its position vector. The particle traces the unit circle, and after time t = θ it has traveled distance θ along the arc, giving .
Velocity Vectors at Key Angles
Each figure shows the position vector Z(θ) = eiθ (blue arrow) and the velocity vector V = iZ (orange arrow) drawn from the tip of Z. The velocity is always the position rotated by π/2 (90°), keeping the particle on the unit circle.
Animated: Particle Tracing the Unit Circle
Watch the particle move along . The blue arrow is the position vector Z(t), the orange arrow is the velocity V = iZ. The dashed arc shows the path traced so far. After one full revolution (t = 2π), the particle returns to its starting point.
The Argument
Step 1. The exponential function ex is its own derivative: . This is a defining property — if df/dx = f and f(0) = 1, then f(x) = ex.
Step 2. Extend this to imaginary exponents by insisting the property still holds for k = i: . Here t is time, and Z(t) = eit is the position of a particle in the complex plane.
Step 3. The derivative dZ/dt is the velocity V. Since V = iZ, multiplying by i rotates the position vector by π/2. The velocity is always perpendicular to the position.
Step 4. Starting at Z(0) = 1 with velocity i (straight up), the particle moves upward. A moment later, its new velocity is perpendicular to its new position. The particle traces the unit circle.
Step 5. Since |Z(t)| = 1 always, the speed |V| = |Z| = 1 is constant. After time t = θ, the particle has traveled distance θ along the circle, landing at angle θ. Therefore: .
Reference: Needham, Tristan. Visual Complex Analysis, Section 1.2.2 "Moving Particle Argument", Figure 1.8. Oxford University Press, 2023. 25th Anniversary Edition.