If $Z_1 = \sqrt{3} + i \sqrt{3}$ and $Z_2 = \sqrt{3} + i$,and $\left(\frac{Z_1}{Z_2}\right)^{50} = x + iy$,then the point $(x, y)$ lies in

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2} + (2i - 1) = 0$. Then,the value of $|\alpha^{8} + \beta^{8}|$ is equal to

Let $S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$. Then $\sum_{z \in S} |z + \sqrt{3}i|^2$ is equal to:

If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg(z) - \arg(\omega) = \frac{3 \pi}{2}$,then $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$ is:
(Here $\arg(z)$ denotes the principal argument of complex number $z$)

$\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \times \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \times \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \times \ldots \infty =$

If $\sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta)$ where $-\pi < \theta < \pi$,then find the value of $r^2(\sec \theta+3 \operatorname{cosec}^2 \theta)$.