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(cosΞΈβ+isinΞΈβ), where 0<r and 0β€ΞΈ<360. Find ΞΈ.
Solution:
Divide both sides of the given equation by z3, which gives z3+z+1+zβ1+zβ3=0. This takes the form w3β2w+1=0, where w=z+zβ1. Factor the cubic polynomial to obtain (wβ1)(w2+wβ1)=0. Now replace w by z+zβ1 and multiply both sides of the equation by z3. This yields
It follows that the six values for z are the fifth roots of 1 and the cube roots of β1, with the exception of 1 and β1. These roots may be written in polar form cosΟβ+isinΟβ, where Ο takes on the following values: 72,144,216,288,60,300. The roots with positive imaginary part have Ο-values 72,144, and 60. The product of these roots is cosΞΈβ+isinΞΈβ, where ΞΈ=72+144+60=276β.
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+wβ1=0 are w=21β(β1Β±5β), and the four z-values from the ensuing equation z+zβ1=w are fifth roots of 1. They are z=41β(β1β5βΒ±i10β25ββ) and z=41β(β1+5βΒ±i10+25ββ). These formulas for fifth roots imply the ruler-and-compass constructibility of a regular pentagon.