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My approach is as follow

$\alpha = \cos \frac{\pi }{7} + i\sin \frac{\pi }{7}$

$1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} + {\alpha ^5} + {\alpha ^6} = 0;\alpha = {e^{\frac{{i\pi }}{7}}},{\alpha ^{{2^r}}} = {e^{\frac{{i\pi }}{7} \times {2^r}}}$

${T = \sum\limits_{r = 0}^{3n - 1} {{e^{\frac{{i\pi }}{7} \times {2^r}}}} }$ and $|T|^2=32$

Not able to proceed from here

Samar Imam Zaidi
  • 8,800

1 Answers1

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Hint:

Writing $\alpha$ as $a, a^7=1$

As $2^3\equiv1\pmod8,$ we have a cycle of $3$

$$\sum_{r=0}^{3n-1}a^{2^r}=n(a+a^2+a^4)$$

Now if $P=a+a^2+a^4, Q=a^3+a^5+a^6;$

$$P+Q=\cdots=-1\ \ \ \ (1)$$

$$P\cdot Q=(a+a^2+a^4)(a^3+a^5+a^6)=a^4+a^6+1+a^5+1+a+1+a^2+a^3=3+(a+a^2+\cdots+a^6)=3-1\ \ \ \ (2)$$

Can you find $P?$

lab bhattacharjee
  • 274,582
  • See https://math.stackexchange.com/questions/3611405/simplifying-cos-left-frac2-pi21-right-cos-left-frac8-pi21-righ and https://math.stackexchange.com/questions/1381294/how-to-prove-an-identity-trigonometry-angles-pi-13 https://math.stackexchange.com/questions/1717536/proving-that-cos-frac2-pi13-cos-frac6-pi13-cos-frac8-pi13-frac – lab bhattacharjee Dec 06 '22 at 15:19