The equation $z^6+z^3+1=0$ has one complex root with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine $\theta,$ in degrees.
To solve for the complex roots of this equation, I need to rewrite it in a form that allows me to use the roots of unity. I notice that the equation is a quadratic in terms of $z^3,$ so I can factor it as $(z^3)^2+z^3+1=0.$ Then I can use the quadratic formula to find the values of $z^3.$ I get $z^3=\frac{-1\pm\sqrt{-3}}{2}.$ To find the complex roots of this equation, I need to write these values of $z^3$ in polar form, using the modulus and the argument. The modulus of $z^3$ is $\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1.$ The argument of $z^3$ is the angle that the complex number makes with the positive real axis. For $z^3=\frac{-1+\sqrt{-3}}{2},$ the argument is $120^\circ,$ since it is in the second quadrant and has a slope of $\sqrt{3}.$ For $z^3=\frac{-1-\sqrt{-3}}{2},$ the argument is $240^\circ,$ since it is in the third quadrant and has a slope of $-\sqrt{3}.$ To find the values of $z,$ I need to take the cube roots of $z^3.$ This means that the modulus of $z$ is the cube root of the modulus of $z^3,$ which is $1^{1/3}=1.$ The argument of $z$ is one-third of the argument of $z^3,$ since multiplying complex numbers adds their arguments. So, for $z^3=1e^{120^\circ i},$ I get $z=1e^{40^\circ i}, 1e^{160^\circ i}, 1e^{280^\circ i}.$ For $z^3=1e^{240^\circ i},$ I get $z=1e^{80^\circ i}, 1e^{200^\circ i}, 1e^{320^\circ i}.$ These are the six complex roots of the equation, and they are equally spaced around the unit circle. The problem asks for the root with argument between $90^\circ$ and $180^\circ.$ The only root that satisfies this condition is $z=1e^{160^\circ i}.$ Therefore, $\theta=160^\circ.$ 