Is the converse of the Pythagorean Theorem false for complex inner products? I was thinking about the converse of the Pythagorean theorem: 
$\lVert x + y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2 \implies x \perp y$
Does this hold if the inner product $\langle \cdot,\cdot\rangle$ is complex-valued and it induces the norm $\lVert \cdot \rVert$ ?
Here's what I've done: 
Assuming an inner product that has linearity in the first argument, i.e. $\langle ax,y\rangle = a\langle x,y\rangle$, and conjugate linearity in the second argument: $\langle x,ay\rangle = a^*\langle x,y\rangle$, we have: 
$\lVert x+y\rVert^2 = \langle x+y,x+y\rangle = \lVert x \rVert^2 + \lVert y \rVert^2 + \langle x,y\rangle + \langle y,x\rangle$ .
Using the assumption that the Pythagorean theorem holds and Hermitian symmetry, i.e. $\langle x,y\rangle^* = \langle y,x\rangle$, then we must have: 
$\langle x,y\rangle + \langle y,x\rangle = 0 \Leftrightarrow \langle x,y\rangle = -\langle x,y\rangle^*$
Since the inner product is complex, let $\langle x,y\rangle = a + bi$.
Then, $\langle x,y\rangle = -\langle x,y\rangle^* \Leftrightarrow a + bi = -a + bi$ .
Equating real and imaginary parts, $a=0$ and $b=b$, i.e $\operatorname{Re}[\langle x,y\rangle] = 0$ .
Thus, we can have two vectors x and y that aren't orthogonal, i.e. $\langle x,y\rangle \ne0$, but do satisfy the Pythagorean Theorem. 
Is this correct? The logic seems right, but I still have some uncertainty.

Pythagorean theorem is valid for non- orthogonal set of vectors as well. There's is good discussion on this at https://arxiv.org/abs/1509.08667
For the Pythagorean theorem, pair wise orthogonality not required, just sum of the cross terms must be zero. For example, there is set of LINOEP vectors for which square of the norm is preserved.
