If $1, \omega, \omega^2$ are the cube roots of unity, then

$$ 1\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+2\left(3+\frac{1}{\omega}\right)\left(3+\frac{1}{\omega^2}\right) +3\left(4+\frac{1}{\omega}\right)\left(4+\frac{1}{\omega^2}\right)+\ldots 10 \text { terms }= $$

$$ (1+\sqrt{3} i)^6-(\sqrt{3}+i)^6= $$

If $z=x+i y$ and $x^2+y^2=1$, then $\frac{1+x+i y}{1+x-i y}=$

If $x^6=(\sqrt{3}-i)^5$, then the product of all of its roots is