The complex number $\frac{{1 + 2i}}{{1 - i}}$ lies in which quadrant of the complex plane

If $i = \sqrt { - 1} $, then $1 + {i^2} + {i^3} - {i^6} + {i^8}$ is equal to

${(1 + i)^{10}}$, where ${i^2} = - 1,$ is equal to

Let $S$ be the set of all $(\alpha, \beta), \pi<\alpha, \beta<2 \pi$, for which the complex number $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real. Let $Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$ . Then $\sum_{(\alpha, \beta) \in s }\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to.

The values of $x$ and $y$ for which the numbers $3 + i{x^2}y$ and ${x^2} + y + 4i$ are conjugate complex can be

The value of ${(1 + i)^6} + {(1 - i)^6}$ is