If $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$ and $\omega$ is a cube root of unity,then $(\omega-1)(\omega-\alpha_1)(\omega-\alpha_2)(\omega-\alpha_3)(\omega-\alpha_4)+\omega$ is equal to

Let $\alpha$ and $\beta$ be the roots of $x^2 - \sqrt{2}x + 1 = 0$. Then the value of $\alpha^{50} + \beta^{50}$ is:

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 the complex number $a$ is such that $|a|=1$ and $\arg (a)=\theta$,then the roots of the equation $\left(\frac{1+i z}{1-i z}\right)^4=a$ are $z=$

The real part of $z$ that satisfies $i z^4+1=0$ is

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are