The complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ is equal to $.....$

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 $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is a real number, then $\theta$ will be (where $I$ is the set of integers):

Assertion $(A)$: If $z$ is a complex number such that $|z| \geq 3$,then the least value of $|z + \frac{3}{z}|$ is $1$.
Reason $(R)$: $|z_1 - z_2| \leq |z_1| + |z_2|$,for any two complex numbers $z_1, z_2$.
The correct option among the following is:

$(r, \theta)$ denotes $r(\cos \theta + i \sin \theta)$. If $x = (1, \alpha)$,$y = (1, \beta)$,$z = (1, \gamma)$ and $x + y + z = 0$,then $\sum \cos (2\alpha - \beta - \gamma) = $

If $(x-iy)^{1/3} = 2-i\sqrt{3}$ and the point $z = (x, y)$ lies on the line $\frac{x}{2} + \frac{y}{\sqrt{3}} = k$,then $k =$