Solve the equation |eiθ − 1| = 2 for θ (0 ≤ θ < 2π), and verify the solution geometrically

To solve the equation |eiθ − 1| = 2 for θ (0 ≤ θ < 2π), we first need to understand that eiθ is a complex number in the form of a+bi, where a = cos(θ) and b = sin(θ).

Step 1: Rewrite the equation The equation |eiθ − 1| = 2 can be rewritten as |cos(θ) + i*sin(θ) - 1| = 2.

Step 2: Simplify the equation This simplifies to |(cos(θ) - 1) + i*sin(θ)| = 2.

Step 3: Apply the absolute value to a complex number The absolute value of a complex number a+bi is √(a² + b²). So, we can rewrite the equation as √((cos(θ) - 1)² + sin²(θ)) = 2.

Step 4: Simplify the equation This simplifies to (cos²(θ) - 2cos(θ) + 1 + sin²(θ)) = 4.

Step 5: Use the Pythagorean identity We know that cos²(θ) + sin²(θ) = 1. So, the equation simplifies to -2cos(θ) + 2 = 4.

Step 6: Solve for cos(θ) Solving for cos(θ), we get cos(θ) = -1.

Step 7: Solve for θ The solutions for θ in the interval [0, 2π) are θ = π.

Geometric verification: The complex number eiθ is represented as a point on the unit circle in the complex plane. The point corresponding to θ = π is (-1,0). The distance from (-1,0) to (1,0) is 2, which verifies our solution.