If $\beta=e^{i\frac{2\pi}7}$, and $\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt2$, find $n$ [duplicate]

Question

If $\beta=e^{i\frac{2\pi}7}$, and $\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt2$, find $n$.

My Attempt:

$\beta+\beta^2+\beta^4+\beta^8+\beta^{16}+...=4\sqrt2$

This is no GP. Can we do something with it?

If $\theta=\frac{2\pi}7$, $\beta=\cos\theta+i\sin\theta$, so,

$(\cos\theta+i\sin\theta)+(\cos2\theta+i\sin2\theta)+(\cos4\theta+i\sin4\theta)+...=4\sqrt2$

$\implies (\cos\theta+\cos2\theta+\cos4\theta+...)+i(\sin\theta+\sin2\theta+\sin4\theta+...)=4\sqrt2$

$\implies$ imaginary part is zero. At the same time, real part = $4\sqrt2$

Not able to finish.

Answer 1

Hint: After writing the expression in cis form,you will realize that the first three terms are being repeated, that is the whole expression

$\sum_{r=0}^{3n-1}\beta^{2^r}= n.\sum_{r=0}^{2}\beta^{2^r}$

Also ,I would like to point out that in your attempt ,you equated the resultant complex number to $4\sqrt2$, whereas in the question asked its modulus is equal to $4\sqrt2$.