Complex number summation

Question

For $\alpha=\cos(\frac{2\pi}{7})+i \sin(\frac{2\pi}{7})$, we have $$\left|\sum_{r=0}^{3n-1} (\alpha^{2^{r}})\right|^2=32.$$ Find the value of $n$.

Answer 1

We have $\alpha=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}=e^{2\pi i/7}$, therefore $$\alpha^{2^r}=e^{2^{r+1}\pi i/7}$$ Note that modulo 7 $$2^{r+1}\equiv \begin{cases} 1 & \text{if } r=3k-1 \\ 2 &\text{if } r=3k\\4& \text{if }r=3k+1\end{cases}$$ which implies $$e^{2^{r+1}\pi i/7}\equiv \begin{cases} e^{\pi i/7} & \text{if }r=3k-1 \\ e^{2\pi i/7} & \text{if }r=3k\\e^{4\pi i/7}&\text{if } r=3k+1\end{cases}$$ can you continue?