What does this assumption imply in case $X$ is a complex inner product space?

Question

If $X$ is a real inner product space and $x,y\in X$ satisfy $\|x\|=\|y\|$, then $(x-y)\perp (x+y)$. What does this assumption imply in case $X$ is a complex inner product space?

My Work:

I proved the first part. Stuck at the second. That is if $X$ is a complex inner product space such that $\|x\|=\|y\|$, it asks to get a result similar to the case where $X$ is real inner product space. We cannot get anything special on $\langle x-y,x+y\rangle$. Can anybody please give me a hint?

Answer 1

For complex inner product space, $(x−y)⊥(x+y)$ means

$(x-y, x+y)=(x,x)-(y,x)+(x,y)-(y,y)=(x,y)-(y,x)$ (since $\|x\|=\|y\|$)

So the exact condition is

For all $x,y \in X$, $(x−y)⊥(x+y)\iff(x,y)-(y,x)=0$ or

$(x−y)⊥(x+y)\iff(x,y)=\overline{(x,y)}$ or $(x,y)$ must be real valued.

(since $(y,x)=\overline{(x,y)}$)