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

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== Problem ==
 
== Problem ==
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The equation <math>z^6+z^3+1=0</math> has [[complex root]]s with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>.
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The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>.
  
 
== Solution 1 ==
 
== Solution 1 ==
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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> (the ninth degree [[roots of unity]]). 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>.
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We shall introduce another factor to make the equation easier to solve. 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> (the ninth degree [[roots of unity]]). 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>.
  
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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>.
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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>. (When we multiplied by <math>r^3 - 1</math> at the beginning, we introduced some extraneous solutions, and the solution with <math>120^\circ</math> was one of them.) This leaves <math>\boxed{\theta=160}</math>.
  
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Note: From above, notice that <math>z^6+z^3+1 = \frac {r^9-1}{r^3-1}</math>. 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 <math>\boxed{\theta=160}</math>.
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== 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.
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We can write them as <math>z^3 = \cos 240^\circ + i\sin 240^\circ</math> and <math>z^3 = \cos 120^\circ + i\sin 120^\circ</math>.  
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So we can use [[De Moivre's theorem]] (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above!
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For <math>\cos 240^\circ + i\sin 240</math> we have <math>(\cos 240^\circ + i\sin 240^\circ)^{1/3}</math> <math>\Rightarrow</math> <math>\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,</math> and <math>\cos 320^\circ + i\sin320^\circ.</math>
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Similarly for <math>(\cos 120^\circ + i\sin 120^\circ)^{1/3}</math>, we have <math>\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,</math> and <math>\cos 280^\circ + i\sin 280^\circ.</math>
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The only argument out of all these roots that fits the description is <math>\theta = \boxed{160}</math>
  
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== Solution 2 ==
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Note: We can add <math>120</math> to the angles of the previous solutions to get new solutions because De Moivre's formula says that <math>(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta</math> and <math>\frac{360}{3} = 120</math>. ~programmeruser
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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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~ blueballoon
  
 
== See also ==
 
== See also ==

Latest revision as of 18:05, 7 February 2023

Problem

The equation $z^6+z^3+1=0$ 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. 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 $r^6+r^3+1=3$. (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\circ$ was one of them.) This leaves $\boxed{\theta=160}$.

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. We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$. So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! For $\cos 240^\circ + i\sin 240$ we have $(\cos 240^\circ + i\sin 240^\circ)^{1/3}$ $\Rightarrow$ $\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,$ and $\cos 320^\circ + i\sin320^\circ.$ Similarly for $(\cos 120^\circ + i\sin 120^\circ)^{1/3}$, we have $\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,$ and $\cos 280^\circ + i\sin 280^\circ.$ The only argument out of all these roots that fits the description is $\theta = \boxed{160}$

Note: We can add $120$ to the angles of the previous solutions to get new solutions because De Moivre's formula says that $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta$ and $\frac{360}{3} = 120$. ~programmeruser

~ blueballoon

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AIME Problems and Solutions
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