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It is not my own homework and I forgot how to solve this kind of things. Anyway, the following are the statement of the homework and my attempts:

The homework:

Let $w=e^{2i\pi/7}$, $u=w+w^2+w^4$, and $v=w^3+w^5+w^6$.

1) Calculate $u+v$ and then write $u^2$ in function of $u$.

2) Prove that $\mathrm{Im}(u)>0$.

3) Calculate the sum: $\sin(2\pi/7)+\sin(4\pi/7)+\sin(6\pi/7)$.

My attempts:

1) $u+v=w+w^2+w^4+w^3+w^5+w^6=w(1+w+w^2+w^3+w^4+w^5)$.

Hence, $u+v=w\dfrac{1-w^6}{1-w}=\dfrac{w-w^7}{1-w}=\dfrac{w-1}{1-w}=-1$.

Since $u+v=-1$ then $u^2+uv=-u$, so let's calculate $uv$.

$uv=(w+w^2+w^4)(w^3+w^5+w^6)=2w^7+w^4(1+w+w^2+w^3+w^4+w^5)$.

Then, $uv=2+w^3\dfrac{w-w^7}{1-w}=2-w^3$.

Hence, $u^2=-u-2+w^3$.

I do not know how to prove that $\mathrm{Im}(u)>0$.

Chiba
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    It was downvoted because it hadn't shown any attempt, I reckon. You could have edited the old question, adding your attempt, instead of creating a new question. I think this one will be better received by people here. – Ivo Terek Sep 15 '14 at 23:18
  • Hint: $2i\text{Im}(z) = z - \overline{z}$. – Gyu Eun Lee Sep 15 '14 at 23:53

1 Answers1

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Yeah there are alot of jerks on this site paranoid about doing other peoples homework. Thats right I am talking about you.. So if you care more about rules and institutions and all that garbage rather than mathematics please signal by entering a $-1$.

But if you care about math read on...

Anyway $w^7=1$ and if you factor out $w-1$ you get

$$w^6+w^5+w^4+w^3+w^2+w+1=0$$

And if you look you see this is the same as $u+v+1=0$ so $u+v=-1$ is the first problem.

For the second

$$u^2=w+w^4+w^2+2w^5+2 w^3 +2 w^6=u+2v=-2-u.$$

The imaginary part of $u$ is $$\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7} +\sin\frac{6\pi}{7}$$ and since all these terms are positive the imaginary part is positive.

Finally to find the imaginary part of $u$ and we also have $uv=2$ as can be checked. So $u$ and $v$ are solutions to $$y^2+y+2=0$$ and this means that

$$y=\frac{-1\pm \sqrt{-7}}{2}$$ and from the positivity, we have

$$u=\frac{-1 + \sqrt{-7}}{2}$$ so the trig sum is equal to $\frac{\sqrt{7}}{2}$

For those that think this is a trivial homework problem this is an example of Gauss periods, used in solving cyclotomic equations and Gauss's fourth proof of quadratic reciprocity.

Now are you not glad you read this far, knowledge is far better than hate.

Rene Schipperus
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