Orthogonality with regards to norms in real vs complex fields Consider this problem:
Prove $x$ and $y$ in a complex vector space is orthogonal if and only if $||\alpha x + \beta y|| ^2 = ||\alpha x||^2 + ||\beta y||^2$ for all scalars $\alpha$ and $\beta$.

I have solved a similar problem that asks:
Prove $x$ and $y$ in a real vector space is orthogonal if and only if $|| x + y|| ^2 = ||x||^2 + ||y||^2$.
I see that since $|| x + y|| ^2 = ||x||^2 + 2<x,y> + \space||y||^2$, if $x$ and $y$ are orthogonal, then $2<x,y> = 0$ and hence $|| x + y|| ^2 = ||x||^2 + ||y||^2$.
I also noticed that if the field is complex, then we could have $2<x,y> = 0$ yet $<x,y> \ne 0$, which would invalidate the second statement.

I fail to see how adding a pair of scalars solves the issue. Any help would be appreciated.
 A: First, it is easy to show that if $x$ and $y$ are orthogonal then
$$\|\alpha x + \beta y\|^2 = \|\alpha x\|^2 + \|\beta y\|^2. \tag{1}$$
Now, let's try to prove that if $(1)$ is true for all scalars $\alpha$ and $\beta$ then $x$ and $y$ are orthogonal. To this end, we first observe that
$$\|\alpha x + \beta y\|^2 = \|\alpha x\|^2 + \|\beta y\|^2 + \alpha^* \beta x^H y + \alpha \beta^* y^H x. \tag{2}$$
If $(1)$ holds then
$$\alpha^* \beta x^H y =- \alpha \beta^* y^H x. \tag{3}$$
Since $(3)$ holds for all values of $\alpha$ and $\beta$, put $\alpha =1$ and $\beta=1$ to get
$$x^H y =- y^H x. \tag{4}$$
Now put $\alpha=1$ and $\beta=i$ to get
$$x^H y =+ y^H x. \tag{5}$$
Solving $(4)$ and $(5)$ simultaneously results in $x^H y = y^H x = 0$.
Q.E.D.

Here, $x^H$ denotes the Hermitian transpose of $x$, and $x^H y  = \langle x, y \rangle$.
A: First we show that if $x$ and $y$ are orthogonal then 
$$||\alpha x + \beta y||^2 = ||\alpha x||^2 + || \beta y||^2.$$
Assume $x$ and $y$ are orthogonal. Then by definition $\langle x, y \rangle =0.$ 
Therefore, $ \alpha \beta^{*} \langle x, y \rangle = \alpha^{*} \beta \langle y, x \rangle =0.$
So
$$||\alpha x + \beta y||^2 = \langle \alpha x + \beta y, \alpha x + \beta y \rangle = \alpha \langle x, \alpha x + \beta y \rangle + \beta \langle y, \alpha x + \beta y \rangle$$
$$ = \alpha \alpha^{*}\langle x, x \rangle  + \alpha \beta^{*}\langle x, y \rangle +  \beta \beta^{*}\langle y, y \rangle  + \beta \alpha^{*}\langle y, x \rangle$$
$$= \alpha \alpha^{*}\langle x, x \rangle  + 0 +  \beta \beta^{*}\langle y, y \rangle  + 0 = ||\alpha x||^2 + ||\beta y||^2.$$
Showing the other direction is similar.
Edit:
Note that $\langle x, y, \rangle$ is just some number, so if $\langle x, y, \rangle = 0$ then $\alpha \langle x, y, \rangle = 0$ as well.
For finite dimensional complex vector spaces $\langle x, y \rangle$ is defined:
$$\left [\begin{matrix} x_1 & x_2 & ... & x_{n-1} &x_n \end{matrix} \right] \left [ \begin{matrix} y_1^{*} \\
y_2^{*} \\
\vdots \\
y_{n-1}^{*} \\
y_n^{*}
\end{matrix} \right]
 = x_1 y_1^{*} + ... + x_n y_n^{*}.$$
