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Difference between revisions of "1996 AIME Problems/Problem 11"

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(this cannot be right .... *double checks* *triple checks* ... the answer is never 0 ...)
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Let <math>\mathrm {P}</math> be the product of the roots of <math>z^6+z^4+z^3+z^2+1=0</math> that have an imaginary part, and suppose that <math>\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})</math>, where <math>0<r</math> and <math>0\leq \theta <360</math>. Find <math>\theta</math>.
 
Let <math>\mathrm {P}</math> be the product of the roots of <math>z^6+z^4+z^3+z^2+1=0</math> that have an imaginary part, and suppose that <math>\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})</math>, where <math>0<r</math> and <math>0\leq \theta <360</math>. Find <math>\theta</math>.
  
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__TOC__
 
== Solution ==
 
== Solution ==
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{{solution}}
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=== Solution 1 ===
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<cmath>\begin{eqnarray*}
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0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\
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0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1}\\
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0 &=& \frac{(z^2-z+1)(z^5-1)}{z-1}
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\end{eqnarray*}</cmath>
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Thus <math>z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis} \frac{360 k}{5}, k = 1, 2, \ldots 4</math>, or <math>z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis} 60, 300</math>. Therefore all of the roots are complex, and the answer is <math>\boxed{0}</math>.
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=== Solution 2 ===
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== See also ==
 
== See also ==
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*[[1996 AIME Problems]]
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{{AIME box|year=1996|num-b=10|num-a=12}}
  
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{{AIME box|year=1996|num-b=10|num-a=12}}
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[[Category:Intermediate Complex Numbers Problems]]

Revision as of 20:00, 27 November 2007

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 $\boxed{0}$.

Solution 2

See also

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
All AIME Problems and Solutions
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