Parallelogram law using complex inner product not adding up

Question

Does the parallelogram law still hold in the complex case?

Using the following definitions:

$\langle \textbf{x}, \textbf{y} + \textbf{z} \rangle = \langle \textbf{x}, \textbf{y} \rangle + \langle \textbf{x}, \textbf{z} \rangle$

$\langle \textbf{x}, \lambda\textbf{y} \rangle = \lambda\langle \textbf{x}, \textbf{y} \rangle$

$\langle \textbf{x}, \textbf{y} \rangle = \overline{\langle \textbf{y}, \textbf{x} \rangle}$

$\langle \lambda\textbf{x}, \textbf{y} \rangle = \overline{\lambda}\langle \textbf{x}, \textbf{y} \rangle$

$\langle \textbf{x}, \textbf{x} \rangle \geq 0$

$\langle \textbf{x}, \textbf{x} \rangle = 0 \text{, if and only if } \textbf{x} = \textbf{0}$,

the equation $\|\textbf{x} + \textbf{y}\|^2 + \|\textbf{x} - \textbf{y}\|^2 = 2\|\textbf{x}\|^2 + 2\|\textbf{y}\|^2$ does not hold.

Is it the case or am I doing something wrong?

Answer 1

Let's try it: \begin{align} \|\textbf{x}+\textbf{y}\|^2 & = \langle\textbf x+\textbf y,\textbf x+\textbf y \rangle = \langle \textbf x,\textbf x \rangle + \langle \textbf x,\textbf y \rangle + \langle \textbf y,\textbf x \rangle + \langle \textbf y,\textbf y \rangle \\[5pt] & = \|\textbf x\|^2 + \langle \textbf x, \textbf y\rangle + \langle \textbf y,\textbf x\rangle + \|\textbf y\|^2. \tag 1 \\[10pt] \|\textbf{x}-\textbf{y}\|^2 & = \langle\textbf x-\textbf y,\textbf x-\textbf y \rangle = \langle \textbf x,\textbf x \rangle - \langle \textbf x,\textbf y \rangle - \langle \textbf y,\textbf x \rangle + \langle \textbf y,\textbf y \rangle \\[5pt] & = \|\textbf x\|^2 - \langle \textbf x, \textbf y\rangle - \langle \textbf y,\textbf x\rangle + \|\textbf y\|^2. \tag 2 \end{align}

Equate the sum of the the left sides of $(1)$ and $(2)$ with the sum of their right sides.