Let $\mathcal{H}$ be a prehilbert space. Two vectors $u,v$ are orthogonal w.r.t. hermitic product if and only if \begin{equation*} \parallel \alpha u + \beta v \parallel^2 = \parallel \alpha u \parallel^2 + \parallel \beta v \parallel^2, \quad \quad \forall \alpha, \beta \in \mathbb{C}. \end{equation*}
Proof.
To the right \begin{equation*} \parallel \alpha u + \beta v \parallel^2 = (\alpha u + \beta v \ | \ \alpha u + \beta v) = \parallel \alpha u \parallel^2 + \parallel \beta v \parallel^2 + \alpha^{\ast}\beta (u \ | \ v) + \alpha\beta^{\ast} (v \ | \ u) \end{equation*} If they are orthogonal then \begin{equation*} \parallel \alpha u + \beta v \parallel^2 = \parallel \alpha u \parallel^2 + \parallel \beta v \parallel^2 \quad \quad \forall \alpha , \beta \in \mathbb{C} \end{equation*} To the left. Similary we obtain the following expresion \begin{equation*} \alpha^{\ast}\beta (u \ | \ v) + \alpha\beta^{\ast} (v \ | \ u) = 0 \quad \quad \forall \alpha , \beta \in \mathbb{C} \end{equation*} If $\alpha, \beta =1$ then $\mathfrak{Re}[(u \ | \ v)] = 0$. How can I prove that $\mathfrak{Im}[(u \ | \ v)] = 0$, so they are orthogonals? (or another method)
Thanks!
Hint: To get $\mathfrak{Re} (u | v) = 0$, you chose the parameters $\alpha$ and $\beta$ in such a way that you can use the equation $z + z^* = 2\mathfrak{Re} z$. The similar equation $z - z^* = 2\mathfrak{Im} z$ will be helpful for the last part. Can you make a different choice for the parameters $\alpha, \beta$ so that this equation can be applied?