$z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta} $ $\times \frac{1+2 i \sin \theta}{1+2 i \sin \theta}$

$z=\frac{\left(3-4 \sin ^{2} \theta\right)+8 i \sin \theta}{1+4 \sin ^{2} \theta}$

For purely imaginary real part should be zero. i.e. $3-4 \sin ^{2} \theta=0$

ie. $\sin \theta=\pm \frac{\sqrt{3}}{2}$

$\theta=-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3},$ Sum of all values is $-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{2 \pi}{3}$