Problem:
The equation z6+z3+1=0 has one complex root with argument (angle) θ between 90∘ and 180∘ in the complex plane. Determine the degree measure of θ.
Solution:
Let w=z3. Then the given equation reduces to
w2+w+1=0
whose solutions are 2(−1+i3) and 2(−1−i3), with arguments of 120∘ and 240∘, respectively. From these, one finds the following six values for the argument of z:
Clearly, only the second one of these, 160∘, is between 90∘ and 180∘.
OR
Multiplying the given equation by z3−1=0 yields
z9−1=0
whose solutions are the ninth roots of unity:
zn=cos(n⋅40∘)+isin(n⋅40∘),n=0,1,2,…,8
of these, only z3 and z4 are in the second quadrant. However, since the solutions of z9−1=0 are distinct, and since z3 is a solution of z3−1=0, it cannot be a solution of the original equation. It follows that the desired root is z4, with degree measure 160.