Official solution
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 \[\theta \].