Solve: $i x^2 - 3 x - 2 i = 0$

If $x+\frac{1}{x}=2 \sin \alpha$ and $y+\frac{1}{y}=2 \cos \beta$,then $x^3 y^3+\frac{1}{x^3 y^3}=$

If $3+i$ and $2-\sqrt{3}$ are the roots of the equation $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{n} x^{n}$ where $a_0, a_1, \ldots, a_{n} \in \mathbb{Z}$,then the least value of $n$ and the value of $a_0$ are respectively:

$\cos (x + iy)$ is equal to

Let $A = \left\{ \frac{1967 + 1686 i \sin \theta}{7 - 3 i \cos \theta} : \theta \in R \right\}$. If $A$ contains exactly one positive integer $n$,then the value of $n$ is

If $\alpha+i \beta$ and $\gamma+i \delta$ are the roots of $x^2-(3-2 i) x-(2 i-2)=0$,where $i=\sqrt{-1}$,then $\alpha \gamma+\beta \delta$ is equal to :