Difference between revisions of "2020 CIME I Problems/Problem 6"

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Find the number of complex numbers <math>z</math> satisfying <math>|z|=1</math> and <math>z^{850}+z^{350}+1=0</math>.
 
Find the number of complex numbers <math>z</math> satisfying <math>|z|=1</math> and <math>z^{850}+z^{350}+1=0</math>.
  
==Solution==
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==Solution 1==
{{solution}}
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We reduce the problem to <math>z^{17}+z^7+1</math>, remembering to multiply the final product by 50. We need the imaginary parts of the numbers <math>z^{17},z^7</math> to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form <math>cis(15x)</math> (this holds true because we are only looking for solutions with a magnitude of <math>1</math>). We also need the real parts to sum to <math>-1</math>. We check all the multiples of 15 that result in <math>cis(x)</math> being negative, and find that only two work(or alternatively, if you are good, you can guess that only <math>120</math> and <math>240</math> work). The answer is then <math>100</math>.
  
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==Solution 2==
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Let <math>f(w)=w^{850}+w^{350}+1</math> and suppose <math>z</math> is such that <math>f(z)=0</math> and <math>|z|=1</math>. Note that <math>z</math> is not real (by descartes rule of signs), thus <math>\overline{z}=|z|^2/z=1/z</math> is also a root of <math>f</math>. It follows that <math>z^{850}f(1/z)=z^{850}+z^{500}+1=0</math>. Subtracting we have <cmath>0=z^{500}-z^{350}=z^{350}(z^{150}-1)=z^{350}(z^{50}-1)(z^{100}+z^{50}+1)</cmath>
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Now <math>z^{350}\neq 0</math> else <math>f(z)=1\neq 0</math>, and <math>z^{50}\neq 1</math> else <math>f(z)=3\neq0</math>. Hence if <math>|z|=1</math> and <math>f(z)=0</math> then we must have <math>z^{100}+z^{50}+1=0</math>. Conversely, if <math>z</math> satisfies <math>z^{100}+z^{50}+1=0</math> then <math>z^{150}=1</math> so that <math>|z|=1</math> and <math>z^{850}+z^{350}+1=z^{100}+z^{50}+1=0</math>. Therefore <math>z</math> satisfies <math>f(z)=0</math> and <math>|z|=1</math> if and only if <math>z^{100}+z^{50}+1=0</math>. Note that this equation has 100 solutions (by the fundamental theorem of algebra) lying on the unit circle. Furthermore they are distinct since the solutions of <math>z^{150}=1</math> are distinct and <math>z^{100}+z^{50}+1</math> is a factor of <math>z^{150}-1</math>. Therefore the answer is 100.
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==See also==
 
{{CIME box|year=2020|n=I|num-b=5|num-a=7}}
 
{{CIME box|year=2020|n=I|num-b=5|num-a=7}}
 
{{MAC Notice}}
 
{{MAC Notice}}

Latest revision as of 00:07, 30 December 2021

Problem 6

Find the number of complex numbers $z$ satisfying $|z|=1$ and $z^{850}+z^{350}+1=0$.

Solution 1

We reduce the problem to $z^{17}+z^7+1$, remembering to multiply the final product by 50. We need the imaginary parts of the numbers $z^{17},z^7$ to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form $cis(15x)$ (this holds true because we are only looking for solutions with a magnitude of $1$). We also need the real parts to sum to $-1$. We check all the multiples of 15 that result in $cis(x)$ being negative, and find that only two work(or alternatively, if you are good, you can guess that only $120$ and $240$ work). The answer is then $100$.

Solution 2

Let $f(w)=w^{850}+w^{350}+1$ and suppose $z$ is such that $f(z)=0$ and $|z|=1$. Note that $z$ is not real (by descartes rule of signs), thus $\overline{z}=|z|^2/z=1/z$ is also a root of $f$. It follows that $z^{850}f(1/z)=z^{850}+z^{500}+1=0$. Subtracting we have \[0=z^{500}-z^{350}=z^{350}(z^{150}-1)=z^{350}(z^{50}-1)(z^{100}+z^{50}+1)\] Now $z^{350}\neq 0$ else $f(z)=1\neq 0$, and $z^{50}\neq 1$ else $f(z)=3\neq0$. Hence if $|z|=1$ and $f(z)=0$ then we must have $z^{100}+z^{50}+1=0$. Conversely, if $z$ satisfies $z^{100}+z^{50}+1=0$ then $z^{150}=1$ so that $|z|=1$ and $z^{850}+z^{350}+1=z^{100}+z^{50}+1=0$. Therefore $z$ satisfies $f(z)=0$ and $|z|=1$ if and only if $z^{100}+z^{50}+1=0$. Note that this equation has 100 solutions (by the fundamental theorem of algebra) lying on the unit circle. Furthermore they are distinct since the solutions of $z^{150}=1$ are distinct and $z^{100}+z^{50}+1$ is a factor of $z^{150}-1$. Therefore the answer is 100.

See also

2020 CIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

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