Let $V$ be a complex inner product space. I need to show the following: $(x\ and \ y\ are\ orthogonal)\ \Rightarrow (\left \| \lambda x+\beta y \right \|^{2}=\left | \lambda \right |^{2}\left \| x \right \|^{2}+\left | \beta \right |^{2}\left \| y \right \|^{2})$ for all $x,y$ in $V$ and for all $\lambda$, and $\beta$ in $\mathbb{C}$
If $x$ and $y$ are orthogonal, then $\left \langle x,y \right \rangle=0$.Then $\left \| \lambda x+\beta y \right \|^{2}=\left \langle\lambda x+\beta y,\lambda x+\beta y \right \rangle=\left | \lambda \right |^{2}\left \| x \right \|^{2}+\left | \beta \right |^{2}\left \| y\right \|^{2}+\lambda \beta ^{*}\left \langle x,y \right \rangle+\lambda ^{*}\beta \left \langle x,y \right \rangle^{*}=\left | \lambda \right |^{2}\left \| x \right \|^{2}+\left | \beta \right |^{2}\left \| y \right \|^{2}$)
By $*$ above $\lambda$, I mean complex conjuguate.
However, to prove the other direction, all what I can say is that Re($\lambda \beta ^{*}\left \langle x,y \right \rangle$)=$0$. Anyone knows how to continue from here to prove $\left \langle x,y \right \rangle=0$
If $\lambda=\beta=1$ then ${\rm Re}(\lambda\bar\beta\langle x,y\rangle)=0$ means ${\rm Re}(\langle x,y\rangle)=0$, and for $\lambda=i,\,\beta=1$ it means ${\rm Im}(\langle x,y\rangle)=0$.
Anyway, for the direction $\Leftarrow$, one usually uses only that the original condition holds with $\lambda=1$ and $\beta\in\{1,i,-1,-i\}$. (See also polarization identity.)