Let $\mathrm {P}$ be the product of the [roots](https://artofproblemsolving.com/wiki/index.php/Root) of $z^6+z^4+z^3+z^2+1=0$ that have a positive [imaginary](https://artofproblemsolving.com/wiki/index.php/Imaginary) part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$, where $0<r$ and $0\leq \theta <360$. Find $\theta$.
\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}

Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$,
or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 300$
(see [cis](https://artofproblemsolving.com/wiki/index.php/Cis)).
Discarding the roots with negative imaginary parts (leaving us with $\mathrm{cis} \theta,\ 0 < \theta < 180$), we are left with $\mathrm{cis}\ 60, 72, 144$; their product is $P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}$.