Show that $\langle x,y \rangle = 0$.

Question

Suppose that $\left (X, \langle \cdot,\cdot \rangle \right)$ be a complex inner product space. Let $x,y \in X$ be such that $\|\alpha x + \beta y \|^2 = \|\alpha x\|^2 + \|\beta y\|^2$ for all pairs $\alpha, \beta \in \Bbb C$ $($ where $\|\cdot\|$ is the norm induced by the inner product on $X$ $).$ Then $\langle x,y \rangle = 0$ i.e. $x \perp y.$

How do I show that? Please help me in this regard.

Thank you very much.

Answer 1

Taking $\alpha =\beta=1$ and expanding $\langle (x+y),(x+y) \rangle$ we get $\Re \langle x,y \rangle =0$. Take $\alpha =1,\beta =i$ to $Im \langle x,y \rangle =0$.