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There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?

This is one of the AMC problems from this year. I've been trying to solve it, but I couldn't and a peek at the answers (not recommended, I know) talked about Euler's theorem etc., which I haven't learnt yet. Is there an 'easier' way to solve this problem?

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    $\begingroup$ Hint: use geometry interpretation. $\endgroup$ – z100 Oct 6 '17 at 16:29
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    $\begingroup$ If $z^6$ is real, what could it be? $\endgroup$ – Lord Shark the Unknown Oct 6 '17 at 16:31
  • $\begingroup$ @z100 I agree with you best way to solve this problem is to look at it geometrically ...the unit circle $\endgroup$ – Isham Oct 6 '17 at 16:33
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Since $|z|=1$ for all $z$ such that $z^{24}=1,$ the only two real numbers that $z^{6}$ can possibly be are the ones with magnitude $1,$ specifically $1$ and $-1.$ Since there are $6$ solutions to $z^{6}=1$ and $6$ more to $z^{6}=-1,$ there must be $\boxed{12}$ total $z$ satisfying the conditions.


Alternatively, consider that $(z^6)^4=z^{24}=1.$ Now, if we set $z^6=w,$ then for every $w$ there are $6$ solutions in $z.$ Since the solutions for $w$ are $\pm 1, \pm i,$ half of the $24$ solutions for $z$ satisfy $w$ being a real number, so the answer is again $\boxed{12}.$

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    $\begingroup$ And finally, those soolutions are exactly the solutions of $z^{12}=1$. $\endgroup$ – z100 Oct 6 '17 at 17:01
  • $\begingroup$ I didn't see that! Nice! (@OP: this is because $z^6=-1$ and $z^6=1$ both give $z^{12}=1$ when squared) $\endgroup$ – pie314271 Oct 6 '17 at 17:02
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Let's say $w=z^6$. We know that $w^4=1$, so $w=\pm 1,\pm i$. Each of these four numbers has $6$ distinct sixth roots.

Does that help?

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    $\begingroup$ The question says $w=z^6$ is real. Does that help to count the values? $\endgroup$ – Professor Vector Oct 6 '17 at 16:34
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    $\begingroup$ @ProfessorVector Yes. Half of the solution of $z^{24}=1$ have values $z^6=\pm 1$ and half $z^6=\pm i$ $\endgroup$ – Raffaele Oct 6 '17 at 16:35
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The $n$th complex roots of $1$ are the points on the complex unit circle with angles of $\frac{2\pi k}n, k\in \Bbb N$. This looks something like this (for $n=7$):

enter image description here

Now imagine this picture with $n=24$, and $n=6$. Where do they overlap?

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If you know something a bit more general than state in the opening sentence, namely that for any nonzero $a\in\Bbb C$ and any integer $n>0$ the equation $z^n=a$ has exactly $n$ solutions, then this is easy.

The condition is really not about $z$, but about $z^6$, so it is convenient to view the $24$ solutions to $z^{24}$ as obtained as follows: find two solutions to $x^2=1$, then for each find two solutions to $y^2=x$, then for each find $6$ solutions to $z^6=y$. Simple substitution says that $z^{24}=1$ for each such $z$, and we must have obtained all solutions this way. Now clearly your solutions for $x$ were $x=1$ and $x=-1$; taking $x=1$ you got the same solutions for $y$, which are real, but for $x=-1$ there are no real solutions for $y$.

In conclusion, half of the $24$ triples $(x,y,z)$ found will have $y$ real.

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Hint:

$$z^{24}=1=e^{2m\pi i}$$ where $m$ is any integer

$$z=e^{2m\pi i/24}$$

$$z^6=e^{m\pi/2}=\cos\frac{m\pi}2+i\sin\frac{m\pi}2$$ where $0\le m\le3$

So, we need $$\dfrac{m\pi}2\equiv0\pmod\pi\iff2|m$$

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The solutions to $z^{24}=1$ are $z=\sqrt[24]{|1|}e^{(\arg(1)+2k\pi)i/24}$ where $k\in\{0,1,2,\ldots,23\}$.

So we are concerned with seeing when $z^6=e^{(2k\pi)i/4}=e^{k\pi i/2}$ is in $\Bbb R$ for $k\in\{0,1,2,\ldots,23\}$. This occurs when $\sin\left(\frac{k\pi }{2}\right)=0$.

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No, there isn't a smart/easy way to solve this problem in my opinion. Either you have seen a lecture on roots of unity, and then you know how to solve it, or you haven't, and then it's extremely difficult to come up with the right bits of theory during a competition.

For this reason we prefer to avoid this kind of problems in many low-level mathematical competitions. (Source: I write problems for the ItalianMO.)

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    $\begingroup$ Or you've learned something about the complex plane as a geometric approach, and can work the problem out easily in your head. I've never heard of "roots of unity" as such, but this was an easy question. $\endgroup$ – Wildcard Oct 7 '17 at 2:03
  • $\begingroup$ @Wildcard What were you taught about this "geometric" approach? Did anyone mention to you the idea of a regular polygon in the complex plane? Maybe you were given a lecture on roots of unity without using that name. $\endgroup$ – Federico Poloni Oct 7 '17 at 6:50
  • $\begingroup$ Nope. I just understand that multiplication by a complex number can be correlated to rotation and scaling in the complex plane. That, plus a little playing with it, is all you actually need. $\endgroup$ – Wildcard Oct 7 '17 at 6:53
  • $\begingroup$ @Wildcard Well, good for you in that case. That requires some talent in my view. Would you like to write your solution to this problem, since all the ones I can read now rely on the idea of roots of unity? $\endgroup$ – Federico Poloni Oct 7 '17 at 7:00
  • $\begingroup$ Here is the sentence I take issue with: "Either you have seen a lecture on roots of unity, and then you know how to solve it, or you haven't...." This is a false dichotomy. It is totally possible to derive knowledge about "roots of unity" without ever attending a lecture on the subject, nor even reading about it. The current answers are more or less how I would do it, though I might use fewer symbols. I do agree it's not everyone who will so derive the knowledge, though. :) $\endgroup$ – Wildcard Oct 7 '17 at 7:05

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