Key Concepts and Formulas

• Modulus Properties: For complex numbers $z_1$ and $z_2$ and a scalar $c$, we have $|cz_1| = |c||z_1|$ and $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$.
• Polar Form: A complex number $z$ can be written as $z = r(\cos\theta + i\sin\theta)$, where $r = |z|$ and $\theta = \arg(z)$.
• Sum of a Complex Number and its Reciprocal: If $w = R(\cos\theta + i\sin\theta)$, then $w + \frac{1}{w} = \left(R + \frac{1}{R}\right)\cos\theta + i\left(R - \frac{1}{R}\right)\sin\theta$.

Step-by-Step Solution

Step 1: Simplify the Expression and Use the Given Condition We are given $3|z_1| = 4|z_2|$, which implies $\frac{|z_1|}{|z_2|} = \frac{4}{3}$. We want to simplify $z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1}$. Let $k = \frac{3z_1}{2z_2}$. Then $z = k + \frac{1}{k}$. This substitution simplifies the expression and sets up the use of the polar form.

Step 2: Find the Modulus of k We need to find $|k|$ to use the polar form. $|k| = \left| \frac{3z_1}{2z_2} \right| = \frac{|3z_1|}{|2z_2|} = \frac{3|z_1|}{2|z_2|} = \frac{3}{2} \cdot \frac{|z_1|}{|z_2|} = \frac{3}{2} \cdot \frac{4}{3} = 2$ Thus, $|k| = 2$.

Step 3: Express z in Terms of the Argument of k Let $\theta = \arg(k)$. Then $k = 2(\cos\theta + i\sin\theta)$. Therefore, $z = k + \frac{1}{k} = 2(\cos\theta + i\sin\theta) + \frac{1}{2}(\cos(-\theta) + i\sin(-\theta))$ $z = 2(\cos\theta + i\sin\theta) + \frac{1}{2}(\cos\theta - i\sin\theta)$ $z = \left(2 + \frac{1}{2}\right)\cos\theta + i\left(2 - \frac{1}{2}\right)\sin\theta = \frac{5}{2}\cos\theta + i\frac{3}{2}\sin\theta$

Step 4: Analyze the Options We have $z = \frac{5}{2}\cos\theta + i\frac{3}{2}\sin\theta$.

• (A) $\text{Im}(z) = 0$: This implies $\frac{3}{2}\sin\theta = 0$, so $\sin\theta = 0$. This means $\theta = n\pi$ for some integer $n$. In this case, $k = \frac{3z_1}{2z_2}$ is real, and $z$ is also real.
• (B) $|z| = \sqrt{\frac{17}{2}}$: $|z|^2 = \frac{25}{4}\cos^2\theta + \frac{9}{4}\sin^2\theta = \frac{17}{2}$. This gives $25\cos^2\theta + 9\sin^2\theta = 34$, or $25\cos^2\theta + 9(1 - \cos^2\theta) = 34$, which simplifies to $16\cos^2\theta = 25$, so $\cos^2\theta = \frac{25}{16}$. This is impossible since $|\cos\theta| \le 1$.
• (C) $|z| = \frac{1}{2}\sqrt{9 + 16\cos^2\theta}$: $|z|^2 = \frac{25}{4}\cos^2\theta + \frac{9}{4}\sin^2\theta = \frac{25}{4}\cos^2\theta + \frac{9}{4}(1 - \cos^2\theta) = \frac{16}{4}\cos^2\theta + \frac{9}{4} = \frac{1}{4}(16\cos^2\theta + 9)$. Thus $|z| = \frac{1}{2}\sqrt{16\cos^2\theta + 9}$.
• (D) Re(z) $= 0$: This implies $\frac{5}{2}\cos\theta = 0$, so $\cos\theta = 0$. This means $\theta = \frac{\pi}{2} + n\pi$ for some integer $n$. In this case, $z$ is purely imaginary.

Since the correct answer is (A), it means that we should consider the case where $\text{Im}(z) = 0$.

Common Mistakes & Tips

• Remember the modulus properties correctly.
• When dealing with expressions involving a complex number and its reciprocal, consider using the polar form.
• Be careful with trigonometric identities and simplifications.

Summary

We were given a complex number $z$ in terms of $z_1$ and $z_2$, with a condition on their moduli. By introducing a new complex number $k$, we simplified the expression for $z$ and used the polar form to analyze the real and imaginary parts. By setting $\text{Im}(z)=0$, we see that option (A) is possible.

Final Answer The final answer is $\boxed{A}$.