<p>If \(z = re^{i\theta}\), then prove that \(|e^{iz}| = e^{-r\sin\theta}\).</p>
Step-by-Step Solution
Key Concept: Express e^(iz) by substituting z = re^(iθ), then separate the result into real and imaginary parts to find the modulus using |e^(a+ib)| = e^a.
<p><strong>Step 1:</strong> Substitute z = re^(iθ) into e^(iz).</p><p>e^(iz) = e^(i·re^(iθ)) = e^(ire^(iθ))</p><p><strong>Step 2:</strong> Expand e^(iθ) = cos θ + i sin θ.</p><p>e^(iz) = e^(ir(cos θ + i sin θ)) = e^(ir cos θ - r sin θ)</p><p><strong>Step 3:</strong> Separate into real and imaginary parts.</p><p>e^(iz) = e^(-r sin θ + ir cos θ) = e^(-r sin θ) · e^(ir cos θ)</p><p><strong>Step 4:</strong> Apply the modulus formula for complex exponentials.</p><p>|e^(iz)| = |e^(-r sin θ)| · |e^(ir cos θ)| = e^(-r sin θ) · 1 = e^(-r sin θ)</p><p>Since |e^(a+ib)| = e^a and |e^(iθ)| = 1 for real θ.</p><p>∴ |e^(iz)| = e^(-r sin θ) ✓</p>
Correct Answer: Proof