2020 CIME I Problems/Problem 6

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.

Solution 3

We have $\frac{z^{850}+z^{350}+1}{3}=0$ . Geometrically speaking, this means that the centroid with vertices on $z^{850}, z^{350}, 1$ is the origin. However, because the circumcenter is also the origin, the only configuration that works is if all three points form an equilateral triangle. Thus, $z^{850}, z^{350}$ are the primitive 3rd roots of unity in some order. Let $y=z^50$ , then because $17$ and $7$ are relatively prime, there is one solution of $y$ for each permutation, so $2$ in total. Thus, there are $2*50=100$ solutions.

2020 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2020 CIME I Problems/Problem 6

Contents

Problem 6

Solution 1

Solution 2

Solution 3

See also