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$\omega_1$ and $\omega_6$ are conjugate pairs. The same applies for for $\omega_2$ and $\omega_5$ and so on.

So $$\omega_1+\omega_2+\omega_4=a+ib$$ $$\omega_6+\omega_5+\omega_3=a-ib$$

$$\implies -1=2a$$ $$\implies a=-\frac 12$$

What should I do next ?

Somos
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Aditya
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4 Answers4

1

Hint: The quadratic residues modulo $7$ are precisely $1$, $2$ and $4$.

Alternatively, you could use the fact that $$\omega_k=\cos(\tfrac{2\pi}{7}k)+\sin(\tfrac{2pi}{7}k)i,$$ to note that $$\operatorname{Im}(\omega_1+\omega_2+\omega_3)=\sin(\tfrac{2pi}{7})+\sin(2\tfrac{2pi}{7})+\sin(4\tfrac{2pi}{7}).$$ Perhaps you could then aply some trigonometric identities if this expression isn't satisfactory.

Servaes
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1

Let $a=\frac{\pi}7$. Then, the roots are $w_k=e^{i2ka}$ and,

$$A=Im(\omega_1+\omega_2+\omega_4)=\sin2a+\sin4a+\sin8a$$

Evaluate

$$A^2 = \sin^22a+\sin^24a+\sin^28a + 2\sin2a\sin4a+2\sin4a\sin8a+2\sin8a\sin2a\tag 1$$

where,

$$\begin{array} && \sin^22a+\sin^24a+\sin^28a \\ & = \frac12(3-\cos4a-\cos8a-\cos16a)=\frac32+\frac12(\cos a+\cos3a+\cos5a) \\ & 2\sin2a\sin4a+2\sin4a\sin8a+2\sin8a\sin2a \\ & =(\cos2a-\cos6a)+(\cos4a-\cos12a)+(\cos6a-\cos10a) \\ & =(\cos2a-\cos6a)+(\cos4a-\cos2a)+(\cos6a-\cos4a) =0 \end{array}$$

Then, the expression (1) becomes

$$\begin{array} & A^2 & = \frac32+\frac12(\cos a+\cos3a+\cos5a) \\ & = \frac32+\frac1{2\sin a}(\cos a\sin a+\cos3a\sin a+\cos5a\sin a)\\ & = \frac32+\frac1{4\sin a}(\sin2a+\sin4a -\sin2a + \sin6a -\sin4a )\\ & = \frac32+\frac{\sin6a}{4\sin a}=\frac32+\frac{1}{4}=\frac74 \end{array}$$

Thus,

$$Im(\omega_1+\omega_2+\omega_4) = A = \frac{\sqrt7}2$$

Quanto
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Hint:

The roots of $$\dfrac{x^7-1}{x-1}=0$$ are $w_j, 1\le j\le6$

Also, $w_j\cdot w_k=w_{j+k}$

$(w_1+w_2+w_4)(w_3+w_5+w_6)$

$=w_4+w_5+w_7+w_6+w_7+w_9+w_7+w_8+w_{10}$

$=w_4+w_5+1+w_6+1+w_2+1+w_1+w_3=2+(\sum_{j=0}^6w_j)=2$

So, $w_1+w_2+w_4,w_3+w_5+w_6$ are the roots of $$t^2-t+2=0$$

Again $$\sin\dfrac{2\pi}7+\sin\dfrac{4\pi}7+\sin\dfrac{8\pi}7=2\sin\dfrac{5\pi}7\cos\dfrac{3\pi}7+\sin\dfrac{4\pi}7>0$$

lab bhattacharjee
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$$ \begin{aligned} (\omega_{1}+\omega_{2}+\omega_{4})^{2}&=(\omega_{1}+\omega_{2}+\omega_{4})+2(\omega_{3}+\omega_{5}+\omega_{6})\\ (\omega_{1}+\omega_{2}+\omega_{4})^{2}&=(\omega_{1}+\omega_{2}+\omega_{4})+2(-1-(\omega_{1}+\omega_{2}+\omega_{4})) \end{aligned} $$

So the positive imaginary root of $m^{2}=-m+2$ which is $\frac{-1+i\sqrt{7}}{2}$

acat3
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  • I think this is something I don’t know. Why did $2(\omega_1\omega_2+2\omega_2\omega_4+\omega_4\omega_1)$ turn into $2(\omega_3+\omega_5+\omega_6)$ – Aditya Mar 26 '20 at 02:30
  • @Aditya because if $\omega$ is roots of unity then $\omega_{n}=\omega^{n}$ – acat3 Mar 26 '20 at 02:51
  • Wait, isn’t $\omega_3=\omega^2$, since the roots are $1,\omega,\omega^2...$? – Aditya Mar 26 '20 at 03:26
  • typically, the first root is $\omega$ since $1=\omega^{7}$ – acat3 Mar 26 '20 at 03:28
  • So the roots are $\omega, \omega^2,\omega^3....\omega^6, 1$? This doesn’t feel right, because I have always known that it starts with 1. – Aditya Mar 26 '20 at 04:09
  • @Aditya that is ok too, since from Your title $\omega_{1}=\omega$ thus we still have $\omega_{n}=\omega^{n}$ – acat3 Mar 26 '20 at 04:22