AIME 1984 P8

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Problem Statement

The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.

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Solution

$$\text{let } y = z^3 \implies y^2 + y + 1 = 0$$ $$\text{Using quadratic formula: } z^3 = y = \dfrac{-1 \pm i\sqrt{3}}{2}$$ $$\text{If } z^3 = \dfrac{-1}{2} + \dfrac{\sqrt{3}}{2}i = \cos 120^{\circ} + i\sin 120^{\circ}$$ $$z = \cos \dfrac{1}{3}(120 + 360k) + i\sin \dfrac{1}{3}(120 + 360k), k \in \mathbb{Z}$$ $$\implies \theta = \dfrac{1}{3}(120 + 360k) = 40 + 120k$$ $$\text{Since } 90 < \theta < 180 \implies \theta = 160$$ $$\text{If } z^3 = \dfrac{-1}{2} - \dfrac{\sqrt{3}}{2}i = \cos 240^{\circ} + i\sin 240^{\circ}$$ $$z = \cos \dfrac{1}{3}(240 + 360k) + i\sin \dfrac{1}{3}(240 + 360k), k \in \mathbb{Z}$$ $$\implies \theta = \dfrac{1}{3}(240 + 360k) = 80 + 120k$$ $$\text{Since } 90 < \theta < 180 \implies \text{No solution for } \theta$$ $$\therefore \theta = 160^{\circ}$$