De Moivre's Theorem

Let $\alpha , \beta$ be the roots of the quadratic equation $x^2+\sqrt{6}x+3=0$. Then $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$ is equal to

Let $\alpha , \beta$  be the roots of the equation $x^2-\sqrt{2x}+2=0$ Then ${\alpha }^{14}+{\beta }^{14}$ is equal to

If $x^2-\sqrt{2}x+2=0$ has roots ${\alpha }^{14}+{\beta }^{14}$ is:

${\left(\frac{-1-i}{\sqrt{2}}\right)}^{100}$ equals -

${\left(\cos \theta +i\sin \theta \right)}^n$ is equal to

If ${\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)}^m=1,2022<m<2029$, then $m=$

One of the values of ${\left(-32i\right)}^{\frac{2}{5}}$ is

If $x^6={\left(4-3i\right)}^5$, then the product of all of its roots is (where $\theta =-{\tan }^{-1}\left(\frac{3}{4}\right)$)

Value of ${\left(\frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta }\right)}^8+{\left(\frac{1+\cos \theta -i\sin \theta }{1+\cos \theta +i\sin \theta }\right)}^{16}$ is

${\left(\sin \theta -i\cos \theta \right)}^3$ is equal to

If $x+\frac{1}{x}=\sqrt{2}$ then $x^{29}+\frac{1}{x^{29}}=$

If $\left|z\right|=1 \text{and} z\ne -1,$ then ${\left(\frac{1+z}{1+\overline{z}}\right)}^4+ {\left(\frac{1+\overline{z}}{1+z}\right)}^4$ is

What is the product of all the values of $(1-i{)}^{\frac{2}{5}}=$

If $n$ is a positive integer, then $(1+i\sqrt{3}{)}^n+(1-i\sqrt{3}{)}^n=$

If $Z+\frac{1}{Z}=2\sin \left(6°\right)$, then the value of $Z^{1000}+\frac{1}{Z^{1000}}$ is equal to

The value of $\sqrt[4]{-1}$ is 

If $z={\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^5+{\left(1+i\right)}^2+{\left(1-i\right)}^2$, then  (where $i=\sqrt{-1}$)

Consider a regular $10-$gon with its vertices on the unit circle. With one vertex fixed, draw straight lines to the other $9$ vertices. Call them ${\mathrm{L}}_1, {\mathrm{L}}_2,\dots ,{\mathrm{L}}_9$ and denote their lengths by $l_1,l_2,\dots ,l_9$ respectively. Then the product $l_1{l}_2\dots l_9$ is

If $n$ is an integer and $z=cis\theta$, then $\frac{z^{2n}-1}{z^{2n}+1}=$