<p>If \(\text{Im}\left(\dfrac{z-1}{e^{\theta i}} + \dfrac{e^{\theta i}}{z-1}\right) = 0\), then find the locus of \(z\).</p>
Step-by-Step Solution
Key Concept: When the imaginary part of a complex expression equals zero, the expression is purely real. Use the property that w + w̄ = 2Re(w) and recognize that e^(θi)/(z-1) is the conjugate of (z-1)/e^(θi) when |z-1| = 1.
<p><strong>Step 1:</strong> Let w = (z-1)/e^(θi). Then the expression becomes w + e^(θi)/(z-1) = w + 1/w̄ (after recognizing the structure).</p><p><strong>Step 2:</strong> For Im(w + 1/w̄) = 0, the expression must be purely real. Write w = re^(φi) where r = |z-1|. Then 1/w̄ = (1/r)e^(-φi).</p><p><strong>Step 3:</strong> The expression becomes re^(φi) + (1/r)e^(-φi). For this to be real: Im[re^(φi) + (1/r)e^(-φi)] = r·sin(φ) - (1/r)·sin(φ) = 0.</p><p><strong>Step 4:</strong> Factoring: sin(φ)[r - 1/r] = 0. Either sin(φ) = 0 (giving a line through z=1) or r = 1/r, which means r² = 1, so r = 1 (taking positive value).</p><p><strong>Step 5:</strong> Since the problem involves arbitrary θ and geometric consistency, r = 1 is the valid locus, meaning |z - 1| = 1.</p><p>∴ <strong>Answer: Circle having center at (1 + i0) and radius 1</strong></p>
Correct Answer: Circle having center at 1 + i0 and radius 1