Finding Values

Geometry Level 5

There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $|z| = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ .

The answer is 840.

4 solutions

Otto Bretscher
Sep 18, 2015

Entertaining, charming problem! Thanks!

We will measure angles in degrees throughout as suggested by the author.

Taking a look at the unit circle, we realize that the only way the difference of the unit vectors $z^{28}$ and $z^8$ can be 1 is $z^{28}=e^{60i}$ and $z^8=e^{120i}$ , or the conjugates thereof. The equation $z^8=e^{120i}$ holds for $\theta=\frac{120}{8}+\frac{360}{8}k=15+45k$ , and $z^{28}=e^{60i}$ holds for $\theta=\frac{15}{7}+\frac{90}{7}k$ . The common solutions are of the form $\theta=15+90k$ , that is $\theta=15,105,195,285$ , as well as the conjugates, $\theta=75,165,255,345$ . The answer we seek is $75+165+255+345=\boxed{840}$

Moderator note:

Nice observation that $z^{28} - z^8 = 1$ gives us a lot of information about this root of unity.

Nice Solution!

This problem is from 2001 AIME II Problem 14.

Alan Yan - 5 years, 8 months ago

Hobart Pao
Oct 26, 2015

I immediately recognized this problem. It was on my honors precalc test last year...

Your honors precalc is way more difficult than the one offered at our school then. This problem would have for sure stumped all of us.

Xuming Liang - 5 years, 7 months ago

No, my teacher just loved trolling us. Basically what he did is the following: During class, he projected the pdf of the textbook in word document form on the screen and just scrolled down. The homeworks were two very easy standard textbook problems. The tests had 10 very easy multiple choice questions but 3 really challenging problems (like this one) at the end worth 10 points each. If you couldn't solve it (which most people couldn't), you just lose 1/10 points. If you even get the right idea, you get full credit. So yeah :P

Hobart Pao - 5 years, 7 months ago

Lu Chee Ket
Oct 20, 2015

75 + 165 + 255 + 345 = 840.

P = 0.258819045102521 and Q = 0.965925826289068 of plus and minus for all but only the following are wanted:

0.258819045102521+0.965925826289068i

-0.965925826289068+0.258819045102521i

-0.258819045102521-0.965925826289068i

0.965925826289068-0.258819045102521i

A trap of 15.

Abdelhamid Saadi
Sep 22, 2015

Let's try a proof in Nicolas Bourbaki style:

Lemme 1:

a complex $z$ which satisfies $|z| = 1$ and $|z - 1| = 1$ is a solution of $x^2 - x + 1 = 0$

Proof: $z = z(\bar z - 1) (z - 1) = (z\bar z - z) (z - 1) =(1 - z)(z - 1)$ Then $z^2 - z + 1 = 0$

Lemme 2:

a complex $z$ which satisfies $|z| = 1$ and $z^{7} - z^{2} - 1 = 0$ is a solution of $x^2 - x + 1 = 0$

Proof : $1 = |z^{7} - z^{2}| = |z^2||z^5 - 1| = |z^5 - 1|$ Since $|z^5| = 1$ from Lemme 1, we have $z^5$ is solution of $x^2 - x + 1 = 0$ .

Then $z^{10} - z^{5} + 1 = 0$ $z^3 = z^3(z^7 - z^2) = z^{10} - z^5 = -1$ Since $z \neq 0$ , $z$ is a solution of $x^2 - x + 1 = 0$

Problem:

Let consider now a complex number that satisfy both $z^{28} - z^{8} - 1 = 0$ and $|z| = 1$

This can be written : ${(z^{4})}^{7} - {(z^{4})}^{2} - 1 = 0$ and $|z^{4}| = 1$

From Lemme 2, $z^{4}$ is a solution of $x^2 - x + 1 = 0$

Then $z^{4}=e^{60^{o} i}$ or $z^{4}=e^{300^{o} i}$ with an Abuse of notation .

This gives $\theta=15,105,195,285$ for the first equation

And $\theta=75,165,255,345$ for the second

The answer we seek is $75+165+255+345=\boxed{840}$

Finding Values

The answer is 840.

4 solutions

Moderator note:

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