If $x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$,where $\omega$ is a complex cube root of unity,then $xyz$ is equal to

If ${z_1} = a + ib$ and ${z_2} = c + id$ are complex numbers such that $|{z_1}| = |{z_2}| = 1$ and $R({z_1}\overline {{z_2}} ) = 0,$ then the pair of complex numbers ${w_1} = a + ic$ and ${w_2} = b + id$ satisfies

The product of the real roots of the equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$,given that one of the roots is $3 + i\sqrt{6}$,is:

If a complex number $z$ is such that $(7+i)(z+\bar{z})-(4+i)(z-\bar{z})+116i=0$,then $z\bar{z}=$

If $z_{1}, z_{2}, z_{3}$ are complex numbers such that $|z_{1}|=|z_{2}|=|z_{3}|=|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=1$, then $|z_{1}+z_{2}+z_{3}|$ is

$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$,then the set of possible values of $a$ is