The complex numbers $\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ (where $i = \sqrt{-1}$) are conjugate to each other for:

Let $S$ be the set of all $(\alpha, \beta)$ such that $\pi < \alpha, \beta < 2\pi$,for which the complex number $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$ for $(\alpha, \beta) \in S$. Then $\sum_{(\alpha, \beta) \in S} \left(i Z_{\alpha \beta} + \frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to:

If $z_1$ and $z_2$ are complex numbers such that $\frac{2 z_1}{3 z_2}$ is a purely imaginary number, then the value of $\left|\frac{z_1-z_2}{z_1+z_2}\right|$ is

Among the statements:
$(S1) :$ The set $\{z \in \mathbb{C} - \{-i\} : |z|=1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\}$ contains exactly two elements,and
$(S2) :$ The set $\{z \in \mathbb{C} - \{-1\} : |z|=1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\}$ contains infinitely many elements.

If $\sqrt{-3-4 i}=re^{i \theta}$,then $r^2 \tan \theta=$

Sum of the squares of the imaginary roots of the equation $z^8-20z^4+64=0$ is