I just attempted to do the question below, but it seems that even after seeing the answer I'm not sure I understand the motivation for the solution.

Let $\alpha =\cos\left(\frac{2\pi}{7}\right)+i\sin\left(\frac{2\pi}{7}\right)$. Show that $\alpha$ is a root of the equation $z^7-1=0$ and express the remaining roots in terms of $\alpha$.

Up to now it's ok, just need to express the solutions as: $\alpha, \alpha^2, \alpha^3,\dots$

The number $\alpha +\alpha^2+\alpha^4$ is a root of a quadratic equation: $$z^2+Az+B=0$$ where $A,B \in \mathbb{R}$. By guessing the other root find the numerical value of $A$ and $B$.

Show that: $$ \cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{8\pi}{7}\right)=-\frac{1}{2} $$ and evaluate: $$ \sin\left(\frac{2\pi}{7}\right)+\sin\left(\frac{4\pi}{7}\right)+\sin\left(\frac{8\pi}{7}\right)$$

This is the part that I'm not sure about, since when they say 'guess' I'm sure that it must be relatively obvious what they intend you to guess, but I can't see anything beyond trying to guess something similar to what was given.

I guessed the other root to be $\alpha^{-1}+\alpha^{-2}+\alpha^{-4}$ with the idea that it would make proving the cosine identity relatively simple (once we know $A$); but I'm not sure how to find $A$ and $B$. The solution gives the other root to be $\alpha^3+\alpha^5+\alpha^6$. I just can't see the specific motivation for doing that, unless you wanted to include all the roots of $z^7-1=0$ into the two roots of the quadratic $z^2+Az+B=0$, but I'm not sure whether theres a relation between the two. Is there something that I'm missing in the question that would make the guess more obvious?

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    $\begingroup$ If the root is complex then the complex conjugate must also be a root of the equation. $\endgroup$ – Rammus May 21 '15 at 18:47
  • $\begingroup$ Complex roots to real polynomial equations come in conjugate pairs. The sum of the two roots is $-b/a$, the difference of the two roots is $\sqrt{b^2-4ac}/a$. (Please check my algebra, I did it in my head.) Since $a=1$, I think should do it for you. $\endgroup$ – Ian May 21 '15 at 18:48
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    $\begingroup$ Should the $\sin(8\pi/6)$ in your last equation perhaps be $\sin(8\pi/7)$? Cheers! $\endgroup$ – Robert Lewis May 21 '15 at 18:52
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    $\begingroup$ Isn't it $\,\dfrac{8\pi}7\,$ rather than $\,\dfrac{8\pi}6\,$? $\endgroup$ – Bernard May 21 '15 at 18:52
  • $\begingroup$ And likewise for $\cos(8\pi/6)$? $\endgroup$ – Robert Lewis May 21 '15 at 18:54


Two conjugates complex numbers $\,z_0\,$ and $\,\overline{z}_0\,$ are roots of the quadratic equation: $$z^2-(z_0+\overline{z}_0)z+z_0\overline{z}_0.$$

Note that $z_0+\overline{z}_0=2\operatorname{Re}(z_0)$. Also, what is the minimal polynomial of $\alpha$?

  • $\begingroup$ So if $z_0=\alpha+\alpha^2+\alpha^4$ then surely they complex conjugate would be $\alpha^{-1}+\alpha^{-2}+\alpha^{-4}$ since $\overline{\alpha}=\cos(\frac{2\pi}{7})-i \sin(\frac{2\pi}{7})$. Which is what I'd imagined it would have to be to make the imaginary parts cancel when you add the two roots, or like you say $z_0+\overline{z_0}=2 \text{Re}(z_0)$. But how would you find $A$ without using the sum of the cosines? Since $A=1$ and $B=2$. $\endgroup$ – Jay May 21 '15 at 19:18
  • $\begingroup$ What is the minimal polynomial of $\alpha$? Also the conjugates are positive powers of $\alpha$. $\endgroup$ – Bernard May 21 '15 at 19:24
  • $\begingroup$ I'm not sure what you mean by minimal polynomial, I haven't heard of it before. I'm not sure what you mean by they have to be positive powers, since I thought if you had the root: $\alpha+\alpha^2+\alpha^4$ which is equal to: $\cos(2\pi/7)+i\sin(2\pi/7)+\cos(4\pi/7)+i\sin(4\pi/7)+\dots$ the conjugate would be: $\cos(2\pi/7)-i\sin(2\pi/7)+\cos(4\pi/7)-i\sin(4\pi/7)+\dots$ which is $\alpha^{-1}+\alpha^{-2}+\alpha^{-4}$? Obviously this is wrong but I'm not sure what else would give the conjugates. Thanks $\endgroup$ – Jay May 21 '15 at 19:38
  • $\begingroup$ That is perfectly right, but is not useful here. For algebraic numbers such as $\alpha$, it is the polynomial of minimal degree of which $\alpha$ is aroot. To find it, you have to factor $z^7-1$. $\endgroup$ – Bernard May 21 '15 at 19:42
  • $\begingroup$ so it would be $(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1)=0$ $\endgroup$ – Jay May 21 '15 at 19:46

Very Engaging Problem!!! I find this one to be 'mos 'def significant because it brings together a collection of essential facts about roots of unity into one example which can be stated in more or less elementary terms. For this reason I have written out my answer in terms of first principles, explaining each step as I go along, even at risk of re-deriving a few basic facts.

It is not necessary to employ guesswork here; indeed, all is unravelled logically from the stated hypothesis, as follows:

First off, recall that for a real polynomial $p(z) \in \Bbb R[z]$, $\lambda \in \Bbb C$ is a zero of $p(z)$ if and only if $\bar \lambda \in \Bbb C$ is also;

$p(\lambda) = 0 \Leftrightarrow p(\bar \lambda) = 0; \tag{1}$

(1) may be seen by writing

$p(z) = \sum_0^n p_i z^i, \tag{2}$

with each $p_i \in \Bbb R$; then

$\overline{p(\lambda)} = \overline{\sum_0^n p_i \lambda^i} = \sum_0^n \overline{p_i \lambda^i} = \sum_0^n \bar p_i \bar \lambda^i = p(\bar \lambda) \tag{3}$

since $\bar p_i = p_i \in \Bbb R$, $1 \le i \le n$. (1) then is an immediate consequence of (3). Note we have only used basic properties of complex conjugation here, viz. $\overline{a + b} = \bar a + \bar b$, $\overline{ab} = \bar a \bar b$, and $\bar a = a$ for $a \in \Bbb R$.

It follows from the above that if $\alpha + \alpha^2 + \alpha^4$ is a root of the real quadratic polynomial

$q(x) = z^2 + Az + B \in \Bbb R[z], \tag{4}$

the other zero must indeed be

$\overline{\alpha + \alpha^2 + \alpha^4} = \bar \alpha + \bar \alpha^2 + \bar \alpha^4. \tag{5}$

We further massage (5); observe that for any unimodular complex number $\beta$,

$\beta \bar \beta = \vert \beta \vert^2 = 1, \tag{6}$


$\bar \beta = \beta^{-1}; \tag{6}$

exploiting this property in (5) yields

$\overline{\alpha + \alpha^2 + \alpha^4} = \alpha^{-1} + \alpha^{-2} + \alpha^{-4}. \tag{7}$

Next, we have

$\alpha^k \alpha^{7 - k} = \alpha^{k + (7 - k)} = \alpha^7 = 1 \tag{8}$

for any $k \in \Bbb Z$; thus

$\alpha^{-k} = (\alpha^k)^{-1} = \alpha^{7 - k}; \tag{9}$

now (7) becomes

$\overline{\alpha + \alpha^2 + \alpha^4} = \alpha^6 + \alpha^5 + \alpha^3; \tag{10}$

(7) presents the other root of $q(z)$ solely in terms of powers of $\alpha$.

Since $\alpha^7= 1$, we may write

$(\alpha - 1)(\sum_0^6 \alpha^i) = \alpha^7 - 1 = 0; \tag{11}$

since $\alpha \ne 1$, this yields

$\sum_0^6 \alpha^i = 0 \tag{12}$


$\sum_1^6 \alpha^i = -1, \tag{13}$

thus, by (10)

$(\alpha + \alpha^2 + \alpha^4) + \overline{\alpha + \alpha^2 + \alpha^4} = -1, \tag{14}$

from which we immediately conclude

$\Re(\alpha + \alpha^2 + \alpha^4) = -\dfrac{1}{2}; \tag{15}$


$\alpha = e^{(2\pi i / 7)} = \cos \dfrac{2\pi}{7} + i \sin \dfrac{2\pi}{7}, \tag{16}$

we easily see, using de Moivre's formula

$(\cos \theta + i\sin \theta)^n = \cos n\theta + i \sin n\theta, \tag{17}$


$\cos \dfrac{2\pi}{7} + \cos \dfrac{4\pi}{7} + \cos \dfrac{8\pi}{7} = -\dfrac{1}{2}. \tag{18}$

As for $A$ and $B$, setting

$\mu = \alpha + \alpha^2 + \alpha^4, \tag{19}$

we see that

$q(z) = (z - \mu)(z - \bar \mu) = z^2 - (\mu + \bar \mu)z + \mu \bar \mu; \tag{20}$

comparing (20) with (4) we find

$A = -(\mu + \bar \mu), \; B = \mu \bar \mu; \tag{21}$

so (14) directly indicates

$A = 1. \tag{22}$

To find $B$, we use $B = \mu \bar \mu$, with $\bar \mu$ given by (10), and $\alpha^7 = 1$:

$B = \mu \bar \mu = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6)$ $= \alpha^4 + \alpha^6 + 1 + \alpha^5 + 1 + \alpha + 1 + \alpha^2 + \alpha^4 = 3 + \sum_1^6 \alpha^i = 3 - 1 = 2. \tag{23}$

Finally, the sum

$\sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} + \sin \dfrac{8\pi}{7} \tag{24}$

may be evaluated by recognizing it as the imaginary part of $\mu = \alpha + \alpha^2 + \alpha^4$, which is one of the two roots of

$q(z) = z^2 + Az + B = z^2 + z + 2; \tag{25}$

applying the quadratic formula to (25), we find the roots are

$\dfrac{1}{2}(-1 \pm \sqrt{1^2 - 4(2)}) = \dfrac{1}{2}(-1 \pm \sqrt{-7}) = -\dfrac{1}{2} \pm i\dfrac{\sqrt{7}}{2}; \tag{26}$

next, we observe that

$\sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} + \sin \dfrac{8\pi}{7} > 0, \tag{27}$

since we may write

$\sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} + \sin \dfrac{8\pi}{7} = \sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} + \sin (\pi + \dfrac{\pi}{7})$ $= \sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} - \sin \dfrac{\pi}{7} = \sin \dfrac{4\pi}{7} + (\sin \dfrac{2\pi}{7} - \sin \dfrac{\pi}{7}); \tag{28}$

inspecting (28), we note that

$\dfrac{\pi}{7}, \dfrac{2\pi}{7}, \dfrac{4\pi}{7} \in (0, \pi), \tag{29}$


$\sin \dfrac{\pi}{7}, \sin \dfrac{2\pi}{7}, \sin \dfrac{4\pi}{7} > 0; \tag{30}$


$\dfrac{\pi}{7}, \dfrac{2\pi}{7} \in (0, \dfrac{\pi}{2}); \tag{31}$

since $\sin$ is strictly monotonically increasing on $(0, \pi/2)$, it follows that

$\sin \dfrac{2\pi}{7} - \sin \dfrac{\pi}{7} > 0; \tag{32}$

using (30), (32) in (28) shows that (27) binds; thus $\sin (2\pi /7) + \sin (4\pi /7) + \sin (8\pi / 7)$ corresponds to the root of (25) with positive imaginary part; that is

$\sin \dfrac{2\pi}{7} + \sin \dfrac{4\pi}{7} + \sin \dfrac{8\pi}{7} = \dfrac{\sqrt{7}}{2}; \tag{33}$

who would have thought? Not me, certainly, until, that is, I worked through this problem!

Nota Bene: How to generalize?


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