suggests use the formula $1+z+\cdots +z^{n-1}=\frac{1-z^n}{1-z}$ with $z\in\mathbb{C}$.
I factorized $1-z^n=(\Pi_{k=1}^{n-1}(\gamma_k-z))(1-z)$ where $\gamma_k$ are the $n$th roots of $1$ $(\gamma_k=\cos(\frac{2k\pi}{n})+i\sin(\frac{2k\pi}{n}))$, so if $1\neq z=|z|(\cos(\theta)+i\sin(\theta))$ then $$1+z+\cdots +z^{n-1}=\frac{1-z^n}{1-z}=\Pi_{k=1}^{n-1}(\gamma_k-z)$$ then if $|z|=1$ we have $|\gamma_k-z|=2\sin\left(\frac{k\pi}{n}-2\theta\right)$, therefore $$|1+z+\cdots +z^{n-1}|=\left|\frac{1-z^n}{1-z}\right|=2^{n-1}\Pi_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}-2\theta\right).$$ And I dont know what to do next or if what I did is a good way.
