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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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The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. This means <math>arg(z) = \frac{120 \text{or} 240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>.
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The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. This means <math>arg(z) = \frac{120 \; \text{or} \;240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:04, 29 June 2013

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 $m$ (the ninth degree roots of unity). 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 $\boxed{\theta=160}$.

Note: From above, notice that $z^6+z^3+1 = \frac {r^9-1}{r^3-1}$. Therefore, the solutions are all of the ninth degree roots of unity that are not also the third degree roots of unity. Checking, we see that the only angle is $\boxed{\theta=160}$.

Solution 2

The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$. Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$, which have arguments of $120$ and $240,$ respectively. This means $arg(z) = \frac{120 \; \text{or} \;240}{3} + \frac{360n}{3}$, 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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