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>. | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | {{ | + | === Solution 1 === |
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
== See also == | == See also == | ||
− | + | {{AIME box|year=1996|num-b=10|num-a=12}} | |
− | + | [[Category:Intermediate Complex Numbers Problems]] |
Revision as of 20:00, 27 November 2007
Problem
Let be the product of the roots of
that have an imaginary part, and suppose that
, where
and
. Find
.
Solution
Solution 1
Thus , or
. Therefore all of the roots are complex, and the answer is
.
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 |