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| | __TOC__ | | __TOC__ |
| − | == Solution ==
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| − | === Solution 1 ===
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| − | To satisfy <math>z^{28} - z^{8} - 1 = 0</math>, <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math> and <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>.
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| − | Since <math>\mid z \mid = 1</math>, <math>z</math> is on the [[unit circle]] centered at the origin in the [[complex plane]].
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| − | Since <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math>, <math>z^{28}</math> and <math>z^8</math> have the same <math>y</math> coordinate. Since <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>, <math>z^{28}</math> is <math>1</math> unit to the right of <math>z^{8}</math>. It is easy to see that the only possibilities are <math>(z^{28},z^{8})=(\text{cis}\,(60),\text{cis}\,(120))</math> or <math>(\text{cis}\,{(300)},\text{cis}\,{(240)})</math>.
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| − | <center><asy>
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| − | pathpen = black+linewidth(0.7); pen l = linewidth(0.6);
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| − | D(unitcircle); D((-1.5,0)--(1.5,0),l,Arrows(5)); D((0,-1.5)--(0,1.5),l,Arrows(5));
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| − | D(D(expi(pi/3))--D(expi(2*pi/3)),EndArrow(3)); D(D(expi(4*pi/3)) -- D(expi(5*pi/3)),BeginArrow(3));
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| − | MP("1",(0.5,0));MP("1",(0,3^.5/2),SE);MP("\mathrm{cis}60",expi(1*pi/3),NE);MP("\mathrm{cis}120",expi(2*pi/3),NW);MP("\mathrm{cis}240",expi(4*pi/3),SW);MP("\mathrm{cis}300",expi(5*pi/3),SE);
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| − | </asy></center>
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| − | For the first possibility:
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| − | <cmath>
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| − | \begin{align*}
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| − | z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(60) \Rightarrow 28\theta \equiv 60 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{90} \\
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| − | z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(120) \Rightarrow 8\theta \equiv 120 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{45} \end{align*}</cmath>
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| − | Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.
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| − | For the second possibility:
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| − | <cmath> \begin{align*}
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| − | z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(300) \Rightarrow 28\theta \equiv 300 \pmod{360} &\Rightarrow \theta \equiv 75 \pmod{90} \\
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| − | z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(240) \Rightarrow 8\theta \equiv 240 \pmod{360} &\Rightarrow \theta \equiv 30 \pmod{45} \end{align*}</cmath>
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| − | Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.
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| − | Therefore <math>(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)</math> and <math>\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}</math>.
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| − |
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| − | === Solution 2 ===
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| − | Rearrange the given equation as <math>z^8\left(z^{20}-1\right) = 1</math>; the magnitudes of both sides must be equal, so
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| − | <cmath>\left|z^8\left(z^{20}-1\right)\right| = \left|z^{20}-1\right| = \left| 1 \right| = 1</cmath>
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| − | Thus the distance between <math>z^{20} = \text{cis}\, 20\theta </math> and <math>(1,0)</math> on the coordinate plane is <math>1</math>. By the distance formula,
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| − | <cmath>1 = \sqrt{(\cos 20\theta - 1)^2 + \sin ^2 20\theta} = \sqrt{2 - 2 \cos 20\theta} \Longrightarrow \cos 20\theta = \frac 12</cmath>
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| − | And <math>20\theta = 60, 300 + 360n</math>, while <math>z^{20} - 1 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i - 1 = \text{cis}\,(120,240)</math>. Thus <math>z^8 = \frac{1}{z^{20}-1} = \text{cis}^{-1}\, \{120,240\} = \text{cis}\,\{240,120\}</math>. We thus have <math>20\theta = 60 + 360n</math> and <math>8\theta = 240 + 360n</math> or <math>20\theta = 300 + 360n</math> and <math>8\theta = 120 + 360n</math>. From here, follow the above solution.
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| | == See also == | | == See also == |
| | {{AIME box|year=2001|n=II|num-b=13|num-a=15}} | | {{AIME box|year=2001|n=II|num-b=13|num-a=15}} |
complex numbers that satisfy both 
and 
. These numbers have the form 
, where 
and angles are measured in degrees. Find the value of 
.
