$\frac{3 + 2i\sin \theta}{1 - 2i\sin \theta}$ will be purely imaginary,if $\theta = $ [Where $n$ is an integer]

If $\alpha$ and $\beta$ are the roots of the equation $2z^2 - 3z - 2i = 0$,where $i = \sqrt{-1}$,then $16 \cdot \operatorname{Re}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right)$ is equal to

For a complex number $z$, let $\operatorname{Re}(z)$ denote the real part of $z$. Let $S$ be the set of all complex numbers $z$ satisfying $z^4 - |z|^4 = 4iz^2$, where $i = \sqrt{-1}$. Then the minimum possible value of $|z_1 - z_2|^2$, where $z_1, z_2 \in S$ with $\operatorname{Re}(z_1) > 0$ and $\operatorname{Re}(z_2) < 0$, is:

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is

Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.

If $z_1=x_1+i y_1$, $z_2=x_2+i y_2$, $z_3=x_1+\frac{i x_2}{2}$, and $z_4=2 y_1+i y_2$ are complex numbers such that $|z_1|=1$, $|z_2|=2$, and $\operatorname{Re}(z_1 \bar{z}_2)=0$, then: