<p>If <span>\(z\)</span> is a complex number of unit modulus and argument <span>\(\theta\)</span>, then <span>\(\arg\left(\dfrac{1+z}{1+\bar{z}}\right)\)</span> equals</p>
Step-by-Step Solution
Key Concept: Since |z| = 1, we have z = e^(iθ) = cos(θ) + i·sin(θ), and the key is to express (1+z)/(1+z̄) in a form where we can directly read off its argument using geometric interpretation or algebraic simplification.
<p><strong>Step 1:</strong> Since |z| = 1 and arg(z) = θ, write z = e^(iθ) = cos(θ) + i·sin(θ), and z̄ = cos(θ) - i·sin(θ).</p><p><strong>Step 2:</strong> Compute 1 + z = (1 + cos(θ)) + i·sin(θ).</p><p><strong>Step 3:</strong> Compute 1 + z̄ = (1 + cos(θ)) - i·sin(θ).</p><p><strong>Step 4:</strong> Form the ratio: $\frac{1+z}{1+z̄} = \frac{(1+\cos\theta) + i\sin\theta}{(1+\cos\theta) - i\sin\theta}$</p><p><strong>Step 5:</strong> Multiply numerator and denominator by the conjugate of the denominator:</p><p>$$\frac{1+z}{1+z̄} = \frac{[(1+\cos\theta) + i\sin\theta]^2}{(1+\cos\theta)^2 + \sin^2\theta}$$</p><p><strong>Step 6:</strong> Expand numerator: $(1+\cos\theta)^2 + 2i\sin\theta(1+\cos\theta) - \sin^2\theta = 2\cos\theta(1+\cos\theta) + 2i\sin\theta(1+\cos\theta) = 2(1+\cos\theta)[\cos\theta + i\sin\theta]$</p><p><strong>Step 7:</strong> The denominator simplifies to $2(1+\cos\theta)$, so: $$\frac{1+z}{1+z̄} = \cos\theta + i\sin\theta = e^{i\theta}$$</p><p><strong>Step 8:</strong> Therefore, arg$\left(\frac{1+z}{1+z̄}\right) = \theta$</p><p>∴ Answer: B (which is θ)</p>
Correct Answer: B