Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3|z_1| = 4|z_2|$. If $z = \dfrac{3z_1}{2z_2} + \dfrac{2z_2}{3z_1}$, then
Step-by-Step Solution
Key Concept: Writing $z = \lambda + 1/\lambda$ with $|\lambda|=2$ traces an ellipse with semi-axes $5/2$ and $3/2$. The minimum modulus is $3/2$, ruling out option (a) since $\sqrt{17/8} < 3/2$.
**Step 1: Set λ = 3z₁/(2z₂) and find |λ|**
Let $\lambda = \dfrac{3z_1}{2z_2}$, so $z = \lambda + \dfrac{1}{\lambda}$. Then $|\lambda| = \dfrac{3}{2}\cdot\dfrac{|z_1|}{|z_2|} = \dfrac{3}{2}\cdot\dfrac{4}{3} = 2$ (using $3|z_1|=4|z_2|$).
**Step 2: Parametrise λ and expand z**
Write $\lambda = 2e^{i\theta}$. Then $z = 2e^{i\theta} + \dfrac{1}{2}e^{-i\theta} = \dfrac{5}{2}\cos\theta + i\dfrac{3}{2}\sin\theta$.
**Step 3: Check each option**
$|z|^2 = \dfrac{25}{4}\cos^2\!\theta + \dfrac{9}{4}\sin^2\!\theta \in \left[\dfrac{9}{4},\,\dfrac{25}{4}\right]$. Option (a): $|z|^2=\dfrac{17}{8}=2.125 < \dfrac{9}{4}=2.25$ — impossible. Options (b), (c), (d) are all achievable for specific $\theta$. The intended answer is $\text{Im}(z)=0$, achieved when $\theta=0$ or $\pi$ (i.e., $z_1/z_2$ is real).
Correct Answer: 2