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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>.
+
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.  
 +
We can write them as <math>z^3 = cos240 + sin240i</math> and <math>z^3 = cos120 + sin120i</math>. 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 them!
 +
For <math>cos240 + sin240i</math> we have <math>(cos240 + sin240i)^{1/3}</math> <math>\Rightarrow</math> <math>cos80 + sin80i, cos200 + sin200i,</math> and <math>cos320 + sin320i.</math>
 +
Similarly for <math>(cos120 + sin120i)^{1/3}</math>, we have <math>cos40 + sin40i, cos160 + sin160i,</math> and <math>cos280 + sin280i.</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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~ blueballoon
  
 
== See also ==
 
== See also ==

Revision as of 13:13, 3 May 2021

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 $z^3 = cos240 + sin240i$ and $z^3 = cos120 + sin120i$. 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 them! For $cos240 + sin240i$ we have $(cos240 + sin240i)^{1/3}$ $\Rightarrow$ $cos80 + sin80i, cos200 + sin200i,$ and $cos320 + sin320i.$ Similarly for $(cos120 + sin120i)^{1/3}$, we have $cos40 + sin40i, cos160 + sin160i,$ and $cos280 + sin280i.$ The only argument out of all these roots that fits the description is $\theta = \boxed{160}$

~ blueballoon

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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