If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$, where $i = \sqrt{-1}$, then $(z^{201} - i)^{8}$ is equal to:

Let $\alpha$ and $\beta$ be the roots of $x^2 + \omega x + \omega^2 = 0$,where $\omega$ is an imaginary cube root of unity. If $z = \alpha^9 + i\beta^9$,then the value of $|z|$ is:

If $\omega$ is a complex cube root of unity and $(1+\omega)^7=A+B \omega$,then the values of $A$ and $B$ are,respectively.

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

If $x^{2}+x+1=0$, then the value of $(x+\frac{1}{x})^{4}+(x^{2}+\frac{1}{x^{2}})^{4}+(x^{3}+\frac{1}{x^{3}})^{4}+\dots+(x^{25}+\frac{1}{x^{25}})^{4}$ is:

If $(\sqrt{3}+i)^{10}=a+bi$,where $a, b \in R$,then the values of $a$ and $b$ are respectively: