Let $$z_1$$ and $$z_2$$ be any two non-zero complex numbers such that $$3|z_1| = 4|z_2|$$. If $$z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1}$$ then maximum value of $$|z|$$ is:

Let us denote the two non-zero complex numbers by $$z_1$$ and $$z_2$$ and write them in polar form:

$$z_1 = r_1e^{i\theta_1}, \qquad z_2 = r_2e^{i\theta_2},$$

where $$r_1 = |z_1| \;,\; r_2 = |z_2| \;,\; \theta_1,\theta_2 \in \mathbb R.$$

The given condition $$3|z_1| = 4|z_2|$$ translates to

$$3r_1 = 4r_2 \quad\Longrightarrow\quad \frac{r_1}{r_2} = \frac{4}{3}.$$

We introduce the useful ratio

$$w = \frac{z_1}{z_2} = \frac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}.$$

Because $$\dfrac{r_1}{r_2} = \dfrac{4}{3},$$ the modulus of $$w$$ is fixed:

$$|w|=\frac{4}{3}.$$

Write the required complex number $$z$$ in terms of $$w$$:

$$z \;=\; \frac{3z_1}{2z_2} \;+\; \frac{2z_2}{3z_1} \;=\; \frac{3}{2}\frac{z_1}{z_2} + \frac{2}{3}\frac{z_2}{z_1} \;=\; \frac{3}{2}w + \frac{2}{3}\frac{1}{w}.$$

Substituting $$|w|=\dfrac{4}{3}$$, let

$$w = \frac{4}{3}e^{i\theta}\quad\text{for some real }\theta.$$

Then

$$\frac{1}{w} = \frac{3}{4}e^{-i\theta}.$$

We now compute each term of $$z$$:

$$\frac{3}{2}w = \frac{3}{2}\left(\frac{4}{3}e^{i\theta}\right) \;=\; 2e^{i\theta},$$

$$\frac{2}{3}\cdot\frac{1}{w} = \frac{2}{3}\left(\frac{3}{4}e^{-i\theta}\right) \;=\; \frac{1}{2}e^{-i\theta}.$$

So we have

$$z = 2e^{i\theta} + \frac12\,e^{-i\theta}.$$

Using Euler’s formula $$e^{i\theta}=\cos\theta + i\sin\theta$$ and $$e^{-i\theta}=\cos\theta - i\sin\theta,$$ we obtain

$$z = 2(\cos\theta + i\sin\theta) \;+\; \frac12(\cos\theta - i\sin\theta) = \left(2+\frac12\right)\cos\theta \;+\; i\left(2-\frac12\right)\sin\theta = \frac52\cos\theta \;+\; i\,\frac32\sin\theta.$$

Hence the modulus of $$z$$ is

$$|z| = \sqrt{\left(\frac52\cos\theta\right)^2 + \left(\frac32\sin\theta\right)^2} = \sqrt{\frac{25}{4}\cos^2\theta + \frac{9}{4}\sin^2\theta} = \frac12\sqrt{25\cos^2\theta + 9\sin^2\theta}.$$

Define

$$f(\theta)=25\cos^2\theta + 9\sin^2\theta.$$

Since $$\cos^2\theta = 1-\sin^2\theta,$$ we can write

$$f(\theta)=25(1-\sin^2\theta) + 9\sin^2\theta = 25 - 16\sin^2\theta.$$

The term $$\sin^2\theta$$ ranges between 0 and 1. Therefore $$f(\theta)$$ attains its maximum when $$\sin^2\theta=0,$$ i.e. when $$\theta=0,\,\pi,\,2\pi,\dots$$ giving

$$f_{\text{max}} = 25.$$

Substituting back, the maximum modulus is

$$|z|_{\text{max}} = \frac12\sqrt{25} = \frac12\cdot5 = \frac52.$$

Hence, the correct answer is Option C.