<p>If \(z\) is a complex number of unit modulus and argument \(\theta\), then \(\arg\left(\dfrac{1+z}{1+\bar{z}}\right)\) equals</p>
Step-by-Step Solution
Key Concept: Since |z| = 1, we have z = e^(iθ) = cos(θ) + i·sin(θ). The key is to express (1+z)/(1+z̄) in terms of θ and recognize that the argument depends only on the imaginary part of the numerator and denominator after simplification.
<p><strong>Step 1:</strong> Since |z| = 1 with arg(z) = θ, write z = cos(θ) + i·sin(θ) and z̄ = cos(θ) - i·sin(θ)</p><p><strong>Step 2:</strong> Compute the numerator: 1 + z = (1 + cos(θ)) + i·sin(θ)</p><p><strong>Step 3:</strong> Compute the denominator: 1 + z̄ = (1 + cos(θ)) - i·sin(θ)</p><p><strong>Step 4:</strong> Form the quotient:</p><p>$$\frac{1+z}{1+z̄} = \frac{(1+\cos\theta) + i\sin\theta}{(1+\cos\theta) - i\sin\theta}$$</p><p><strong>Step 5:</strong> Multiply by conjugate of denominator:</p><p>$$= \frac{[(1+\cos\theta) + i\sin\theta]^2}{(1+\cos\theta)^2 + \sin^2\theta}$$</p><p><strong>Step 6:</strong> Simplify denominator: (1+cos(θ))² + sin²(θ) = 1 + 2cos(θ) + cos²(θ) + sin²(θ) = 2(1 + cos(θ))</p><p><strong>Step 7:</strong> Expand numerator: (1+cos(θ))² + 2i(1+cos(θ))sin(θ) - sin²(θ) = 2cos²(θ) + 2i(1+cos(θ))sin(θ)</p><p><strong>Step 8:</strong> Simplify the fraction:</p><p>$$= \frac{2\cos^2(\theta/2) + 2i·2\cos(\theta/2)\sin(\theta/2)·2\cos(\theta/2)}{4\cos^2(\theta/2)}$$</p><p><strong>Step 9:</strong> Using half-angle formulas: 1 + cos(θ) = 2cos²(θ/2) and sin(θ) = 2sin(θ/2)cos(θ/2), the result is:</p><p>$$\frac{1+z}{1+z̄} = e^{i\theta}$$</p><p><strong>Step 10:</strong> Therefore, arg$$\left(\frac{1+z}{1+z̄}\right) = \theta$$</p><p>∴ Answer: C (θ)</p>
Correct Answer: C