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.

You are right.
And it has to be that way, because the norm that the Hermitian product induces on $\mathbb C^n$ is the same as the Euclidean norm on $\mathbb R^{2n}$.
Therefore, for a fixed $x\in\mathbb C^n$, the set
$$ \{ y\in\mathbb C^n \mid \|x+y\|^2 = \|x\|^2+\|y\|^2 \} $$
has dimension $2n-1$ over $\mathbb R$ -- and therefore it cannot be a linear subspace over $\mathbb C$. In particular it cannot equal $\{y\in\mathbb C^n \mid \langle x,y\rangle = 0 \}$.
