If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ where $i = \sqrt{-1}$,then $(1 + iz + z^5 + iz^8)^9$ is equal to:

$(r, \theta)$ 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) = $

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

The value of $\sum_{n=0}^{\infty}\left(\frac{2 i}{3}\right)^n$ is

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:

If $(2+i)$ and $(\sqrt{5}-2i)$ are the roots of the equation $(x^{2}+ax+b)(x^{2}+cx+d)=0$ where $a, b, c$ and $d$ are real constants,then the product of all the roots of the equation is