If $\langle \cdot, \cdot \rangle$ is a complex inner product and if $\alpha \in \mathbb{C}$ and $\alpha^N = 1$ but $\alpha^2 \not = 1$, then show that
$$\langle x,y \rangle = \frac{1}{N} \sum_{n=0}^{N-1}\lVert x+\alpha^n y \rVert^2\alpha^n$$
where, as usual, $\lVert w \rVert = \langle w, w\rangle^{1/2}$.
Expand the right-hand side \begin{align*} \frac{1}{N} \sum_{n=0}^{N-1}\lVert x+\alpha^n y \rVert^2\alpha^n &= \frac{1}{N} \sum_{n=0}^{N-1}\langle x+\alpha^n y, x+\alpha^n y \rangle \alpha^n \\ &= \frac{1}{N} \sum_{n=0}^{N-1}\langle x+\alpha^n y, x+\alpha^n y \rangle \alpha^n \\ &= \frac{1}{N} \sum_{n=0}^{N-1} \left( \langle x, x \rangle +(\overline{\alpha})^n \langle x, y \rangle + \alpha^n\langle y, x \rangle + \alpha^n(\overline{\alpha})^n\langle y,y \rangle \right) \alpha^n \end{align*} $$= \frac{1}{N} \left( \sum_{n=0}^{N-1} \alpha^n\langle x, x \rangle + \sum_{n=0}^{N-1} \lvert \alpha\rvert^{2n} \langle x, y \rangle + \sum_{n=0}^{N-1} \alpha^{2n} \langle y, x \rangle + \sum_{n=0}^{N-1} \alpha^n \lvert \alpha\rvert^{2n}\langle y,y \rangle \right )$$
We need to use the fact that $1+\alpha + \alpha^2 + \dots + \alpha^{N-1} = \frac{1-\alpha^N}{1-\alpha}$ but I don't know how to proceed. Any help?
Note that $|\alpha|=1$ therefore $$\sum_{n=0}^{N-1} \alpha^n=\dfrac{1-\alpha^{N}}{1-\alpha}=0\\\sum_{n=0}^{N-1} |\alpha|^{2n}=N\\\sum_{n=0}^{N-1} \alpha^{2n}=\dfrac{1-\alpha^{2N}}{1-\alpha^2}=0\\\sum_{n=0}^{N-1} \alpha^n|\alpha|^{2n}=\sum_{n=0}^{N-1} \alpha^n=0$$