I want to find some nice relation for $\alpha_1$ and $\alpha_2$ making \begin{align} \alpha_1^2 \xi_1^2 + \alpha_2^2 \xi_2^2 + 2 \alpha_1 \alpha_2(\xi_1 \cdot \xi_2)=0 \end{align} where $\xi_1, \xi_2$ are vector and $\alpha_1, \alpha_2$ be scalar (real or complex).
For real case, this is nothing but \begin{align} (\alpha_1 \xi_1 + \alpha_2 \xi_2)^2=0 \end{align}
How about complex valued $\alpha_1, \alpha_2$ case? Following are answer in some papers I found. I want to show whether equation A.14 or A.15 satisfies A.13 explicitly.
I found one answer \begin{align} |\xi_1| \alpha_1 = \left[ - \frac{(\xi_1 \cdot \xi_2)}{|\xi_1 ||\xi_2|} + i \sqrt{\Delta} \right] |\xi_2|\alpha_2, \quad \Delta = 1 - \frac{(\xi_1 \cdot \xi_2)^2}{\xi_1^2 \xi_2^2} \end{align} but having a problem showing this is indeed true.
Followings are my trials: Naively from $|z_1 z_2| = |z_1 | |z_2|$, the above relation reduce $|\xi_1| |\alpha_1| = |\xi_2| |\alpha_2|$, but it seems this does not enough for showing this.