Convert the complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ into polar form.

If $z_1, z_2, z_3$ are the roots of the equation $z^3 - z^2(4 + 3i) + z(3 + 8i) - 5i = 0$, then $Re(z_1) + Re(z_2) + Re(z_3)$ is -

If $z = x - iy$ and $z^{1/3} = p + iq$ $(x, y, p, q \in R)$, then $\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)}$ is equal to

If the set $\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\}$ is equal to the interval $(\alpha, \beta]$,then $24(\beta-\alpha)$ 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 $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$,then $2 x+4 y=$