A complex equation

The equation is $z^6 + z^3 + 1 = 0$ .

Let $z^3 = x$ .

The equation now becomes $x^2 + x + 1 = 0$ .

This equation has roots $w , w^2$ where $w$ is a cube root of unity.

$\Rightarrow z^3 = w^{n}$ or $z^3=w^{2n}$ where $n$ is not a multiple of 3.

$\Rightarrow z = w^{\frac n 3}$ or $z =w^{\frac{2n}{3}}$ .

$\Rightarrow r e^{i\theta} = e^{i \frac {2 n \pi}{9}}$ $\quad$ or $\quad r e^{i\theta} = e^{i \frac {4 n \pi}{9}}$ where n is not a multiple of 3.

The only possible value according to the given conditions is $\boxed{160^\circ}$ .

$Solving~the~quadratic~in~z^3, \\z^3=\dfrac{-1\pm \sqrt3 i} 2\\ =\large e^{i*(\frac{2\pi} 3 +2\pi n ) }\\ \therefore~\large z= e^{i*\frac {2\pi} 3 *(\frac 1 3 + n ) } .\\ \therefore~\theta=2\pi *(\frac{\frac 1 3 + n} 3 )\\ If~n=0,~~\theta=\frac{2\pi } 9^c=40^o\\ If~n=1,~~\theta=\frac{20\pi } 9^c=160^o\\ \therefore \theta=~~~~~~~~\Large \color{#D61F06}{160^o}$

Substitute $x=z^3$ , which gives $x^2+x+1$ . This is a well-known quotient, namely $\frac{x^3-1}{x-1}$ . We can substitute back our $z$ and get the equation $\frac{z^9-1}{z^3-1}=0$ The next step comes from identifying roots of unity. Whenever $z^n-r$ is a factor in an expression that equals zero, then there are $n$ equally-spaced points on the complex plane. In this case, $z^9-1$ tells us that there should be 9 points, including the point $(1,0)$ , equally spaced apart. With $360^{\circ}$ in a circle, this gives $40$ degrees between each successive root of unity, so all arguments are multiples of 40. But beware! The denominator, $z^3-1$ , creates a domain restriction on $\textit{cubic}$ roots of unity. Hence, we can't include the 9th roots of unity that are also cubic roots of unity. The cubic roots of unity, naturally, occur at $0^{\circ}$ , $120^{\circ}$ , and $240^{\circ}$ , multiples of 120. The only multiple of 40 in the range 90 to 180 which is NOT a multiple of 120 is $\boxed{160}$ .