1996 AIME Problem 11

Question

Let

P

be the product of the roots of

z^6+z^4+z^3+z^2+1=0

that have positive imaginary part, and suppose that

P=r(\cos\theta^\circ+i\sin\theta^\circ),

where

r>0

and

0\leq\theta<360.

Find

\theta.

Answer

Divide by

z^3

and set

w=z+z^{-1}

Factor the resulting cubic in

w

and identify each value as

2\cos\phi

No root is zero. Dividing by

z^3

and setting

w=z+z^{-1}

gives

\begin{aligned}w^3-2w+1&=0,\\{}(w-1)(w^2+w-1)&=0.\end{aligned}

Its three roots are

\begin{aligned}1&=2\cos60^\circ,\\\frac{\sqrt5-1}{2}&=2\cos72^\circ,\\-\frac{1+\sqrt5}{2}&=2\cos144^\circ.\end{aligned}

For each value

w=2\cos\phi,

the corresponding roots of the original equation are

e^{i\phi}

and

e^{-i\phi}.

Thus the roots with positive imaginary part have arguments

60^\circ,

72^\circ,

and

144^\circ.

Their product has argument

60+72+144=276^\circ,

so

\theta=276.

Problem 11 in other years