Solve this following

$3\left|z_{1}\right|=4\left|z_{2}\right|$

$\Rightarrow \frac{\left|z_{1}\right|}{\left|z_{2}\right|}=\frac{4}{3}$

$\Rightarrow \frac{\left|3 z_{1}\right|}{\left|2 z_{2}\right|}=2$

Let $\frac{3 \mathrm{z}_{1}}{2 \mathrm{z}_{2}}=\mathrm{a}=2 \cos \theta+2 \mathrm{i} \sin \theta$

$\mathrm{z}=\frac{3 \mathrm{z}_{1}}{2 \mathrm{z}_{2}}+\frac{2 \mathrm{z}_{2}}{3 \mathrm{z}_{1}}=\mathrm{a}+\frac{1}{\mathrm{a}}$

$=\frac{5}{2} \cos \theta+\frac{3}{2} i \sin \theta$

Now all options are incorrect

Remark :

There is a misprint in the problem actual problem should be :

"Let $z_{1}$ and $z_{2}$ be any non-zero complex number such that $3\left|z_{1}\right|=2\left|z_{2}\right|$.

If $\mathrm{z}=\frac{3 \mathrm{z}_{1}}{2 \mathrm{z}_{2}}+\frac{2 \mathrm{z}_{2}}{3 \mathrm{z}_{1}}$, then"

Given

$3\left|z_{1}\right|=2\left|z_{2}\right|$

Now $\left|\frac{3 z_{1}}{2 z_{2}}\right|=1$

Let $\frac{3 z_{1}}{2 z_{2}}=a=\cos \theta+i \sin \theta$

$\mathrm{z}=\frac{3 \mathrm{z}_{1}}{2 \mathrm{z}_{2}}+\frac{2 \mathrm{z}_{2}}{3 \mathrm{z}_{1}}$

$=a+\frac{1}{a}=2 \cos \theta$

$\therefore \quad \operatorname{Im}(z)=0$

Now option (4) is correct.