If $1, \omega_1,\omega_2...\omega_6$ are $7^{th}$ roots of unity, then find the value of $Im(\omega_1+\omega_2+\omega_4)$ $\omega_1$ 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 ?
 A: 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.
A: 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, 
$$Im(\omega_1+\omega_2+\omega_4) = A = \frac{\sqrt7}2$$
A: 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 $$\sin\dfrac{2\pi}7+\sin\dfrac{4\pi}7+\sin\dfrac{8\pi}7=2\sin\dfrac{5\pi}7\cos\dfrac{3\pi}7+\sin\dfrac{4\pi}7>0$$
A: $$
\begin{aligned}
(\omega_{1}+\omega_{2}+\omega_{4})^{2}&=(\omega_{1}+\omega_{2}+\omega_{4})+2(\omega_{3}+\omega_{5}+\omega_{6})\\
(\omega_{1}+\omega_{2}+\omega_{4})^{2}&=(\omega_{1}+\omega_{2}+\omega_{4})+2(-1-(\omega_{1}+\omega_{2}+\omega_{4}))
\end{aligned}
$$
So the positive imaginary root of $m^{2}=-m+2$ which is $\frac{-1+i\sqrt{7}}{2}$
