Problem:
Let P be the product of those roots of z6+z4+z3+z2+1=0 that have positive imaginary part, and suppose that P=r(cosθ∘+isinθ∘), where 0<r and 0≤θ<360. Find θ.
Solution:
Divide both sides of the given equation by z3, which gives z3+z+1+z−1+z−3=0. This takes the form w3−2w+1=0, where w=z+z−1. Factor the cubic polynomial to obtain (w−1)(w2+w−1)=0. Now replace w by z+z−1 and multiply both sides of the equation by z3. This yields
0=(z2−z+1)(z4+z3+z2+z+1)=z+1z3+1⋅z−1z5−1
It follows that the six values for z are the fifth roots of 1 and the cube roots of −1, with the exception of 1 and −1. These roots may be written in polar form cosϕ∘+isinϕ∘, where ϕ takes on the following values: 72,144,216,288,60,300. The roots with positive imaginary part have ϕ-values 72,144, and 60. The product of these roots is cosθ∘+isinθ∘, where θ=72+144+60=276.
Note: This solution illustrates a general method for solving symmetric equations of degree 2n, by reducing them to equations of degree n. In this example, it is even possible to find non-trigonometric formulas for the roots, by repeated use of the quadratic formula. In particular, the roots of w2+w−1=0 are w=21(−1±5), and the four z-values from the ensuing equation z+z−1=w are fifth roots of 1. They are z=41(−1−5±i10−25) and z=41(−1+5±i10+25). These formulas for fifth roots imply the ruler-and-compass constructibility of a regular pentagon.