Given, $$3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$$

$$ \Rightarrow $$ $${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$$

$$ \Rightarrow $$ $${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$$

As we know, for any compled number

$${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$$(cos$$\theta $$ + i sin$$\theta $$)

= 2(cos$$\theta $$ + i sin$$\theta $$)

$$ \therefore $$ $${{2{z_2}} \over {3{z_1}}}$$ = $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$$

= $${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$$

= $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$
Now, given

$$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$$

= 2(cos$$\theta $$ + i sin$$\theta $$) + $${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$$

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

So, |z| = $$\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta } $$

= $${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta } $$

z is neither purely real nor purely imaginary and |z| depends on $$\theta $$.