$(1-i \sqrt{3})^{2025}=$

If $\alpha$ is a root of $z^2-z+1=0$,then $\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^2+\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^3+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^4+\left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^5=$

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 $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$ where $x$ and $y$ are real,then the ordered pair $(x, y)$ is

If $|Z|=2$, $Z_1=\frac{Z}{2} e^{i \alpha}$ and $\theta$ is the $\operatorname{amp}(Z)$, then $\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}}=$

One of the $15^{\text{th}}$ roots of $-1$ is