Question:

<p><strong>Step 1:</strong> Multiply numerator and denominator by conjugate of denominator:</p><p>$\dfrac{3}{2+e^{i\theta}} = \dfrac{3(2-e^{-i\theta})}{(2+e^{i\theta})(2-e^{-i\theta})}$</p><p><strong>Step 2:</strong> Simplify denominator: $(2+e^{i\theta})(2-e^{-i\theta}) = 4 + 2(e^{i\theta} - e^{-i\theta}) + 1 = 5 + 4i\sin\theta$</p><p><strong>Step 3:</strong> Let $e^{i\theta} = \cos\theta + i\sin\theta$. Rationalize:</p><p>$\dfrac{3}{2+\cos\theta + i\sin\theta} = \dfrac{3(2+\cos\theta - i\sin\theta)}{(2+\cos\theta)^2 + \sin^2\theta}$</p><p><strong>Step 4:</strong> Denominator becomes: $4 + 4\cos\theta + \cos^2\theta + \sin^2\theta = 5 + 4\cos\theta$</p><p><strong>Step 5:</strong> Therefore: $\dfrac{3(2+\cos\theta)}{5+4\cos\theta} + i\dfrac{-3\sin\theta}{5+4\cos\theta} = ax + iby$</p><p><strong>Step 6:</strong> This gives $x = \dfrac{3(2+\cos\theta)}{5+4\cos\theta}$ and $y = \dfrac{-3\sin\theta}{5+4\cos\theta}$</p><p><strong>Step 7:</strong> Eliminate θ: From the parametric equations, $(x - \frac{3}{2})^2 + y^2 = (\frac{3}{4})^2$</p><p>∴ The locus is a <strong>circle</strong> with center $(\frac{3}{2}, 0)$ and radius $\frac{3}{4}$</p>