1996 AIME Problems/Problem 11

Problem

Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have an imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ .

Solution

Solution 1

$\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1}\\ 0 &=& \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}$

Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis} \frac{360 k}{5}, k = 1, 2, \ldots 4$ , or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis} 60, 300$ . Therefore all of the roots are complex, and the answer is $\boxed{0}$ .

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1996 AIME Problems/Problem 11

Problem

Contents

Solution

Solution 1

Solution 2

See also