Difference between revisions of "1984 AIME Problems/Problem 8"

(Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with [[absolute value]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math>.
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We shall introduce another factor to make the equation easier to solve. Consider <math>r^3-1</math>. If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with [[absolute value]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math>. Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>.
  
This reduces <math>\theta</math> to either <math>120^{\circ}</math> or <math>160^{\circ}</math>. But <math>\theta</math> can't be <math>120^{\circ}</math> because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\boxed{\theta=160}</math>.
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This reduces <math>\theta</math> to either <math>120^{\circ}</math> or <math>160^{\circ}</math>. But <math>\theta</math> can't be <math>120^{\circ}</math> because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^6+r^3+1=3</math>. This leaves <math>\boxed{\theta=160}</math>.
  
 
Also,
 
Also,

Revision as of 10:06, 6 March 2011

Problem

The equation $z^6+z^3+1$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.

Solution 1

We shall introduce another factor to make the equation easier to solve. Consider $r^3-1$. If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$. Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$.

This reduces $\theta$ to either $120^{\circ}$ or $160^{\circ}$. But $\theta$ can't be $120^{\circ}$ because if $r=\cos 120^\circ +i\sin 120^\circ$, then $r^6+r^3+1=3$. This leaves $\boxed{\theta=160}$.

Also,

From above, you notice that $z^6+z^3+1 = \frac {r^9-1}{r^3-1}$. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is $\boxed{\theta=160}$.

Solution 2

Note that the substitution $y=z^3$ simplifies this to $y^2+y+1$. Simply applying the quadratic formula gives roots $y_{1,2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$, which have angles of 120 and 240, respectively. This means $arg(z) = \frac{120,240}{3} + \frac{360n}{3}$, and the only one between 90 and 180 is $\boxed{\theta=160}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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