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

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== Solution 2 ==
 
== Solution 2 ==
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Note that the substitution <math>y=z^3</math> simplifies this to <math>y^2+y+1</math>. Simply applying the quadratic formula gives roots <math>y_{1,2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have angles of 120 and 240, respectively. This means <math>arg(z) = \frac{120,240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>.
+
Note that the substitution <math>y=z^3</math> simplifies this to <math>y^2+y+1</math>. Simply applying the quadratic formula gives roots <math>y_{1,2}=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have angles of 120 and 240, respectively. This means <math>arg(z) = \frac{120,240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>.
 +
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=7|num-a=9}}
 
{{AIME box|year=1984|num-b=7|num-a=9}}
  
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 07:11, 30 June 2012

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 (Problems • Answer Key • Resources)
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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