I am trying to show that for a complex inner product space then:
$$ \| x+y \| = \|x-y\| \quad \text{and} \quad \| x+iy \| = \|x-iy\| \implies x\perp y $$
From this answer, I can see that the left hand side of the above implication implies that both Re$\langle x,y\rangle = 0$ and Im$ \langle x,y \rangle =0$ so the result follows from there.
I am confused about the difference between the two conditions though, since $x,y \in \mathbb{C}$, does the first statement $\| x+y \| = \|x-y\|$ imply that $x,y$ are restricted to the reals, and the second the most general case? I don't quite understand why we need both
It implies no restriction whatsoever. It's just that $\|x+y\|=\|x-y\|$ alone, does not imply that $x\perp y$. For instance, in $\mathbb C^2$ let $$ x=(0,1),\ \ \ y=(0,i) $$ Then $$ \|x+y\|=\|(0,1+i)\|=|1+i|=\sqrt2=|1-i|=\|(0,1-i)\|=\|x-y\|,$$ while $\langle x,y\rangle=i\ne0$.
The reason we need both is, as seen above, that one of them is not enough.