Norm of vector sum So for real-valued vectors the statement

$$ \lVert u+v\rVert = \lVert u-v\rVert $$

holds if and only if $$ \langle u,v\rangle =0 $$
I want to state this now instead for complex-valued vectors

$$\lVert u+v\rVert =\lVert u-v\rVert \iff \langle u,v\rangle=-\langle u,v\rangle $$

Im not seeing the connection, but is the right hand side equal to the conjugate of $\langle u,v\rangle$ or?
 A: Use the polarization identities (since you're in a real vector space):

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2).$$

For complex:

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2+i\lVert u+iv\rVert^2 -i\lVert u-iv\rVert^2).$$


If all they want is a restatement then just write it out:
$$\begin{cases}\lVert u+v\rVert^2=\langle u+v,u+v\rangle=\lVert u\rVert^2+\lVert v\rVert^2+\langle u,v\rangle +\langle v,u\rangle \\
\lVert u-v\rVert^2=\langle u-v,u-v\rangle=\lVert u\rVert^2+\lVert v\rVert^2-\langle u,v\rangle -\langle v,u\rangle\end{cases}$$
Then equality means

$$\langle u,v\rangle +\overline{\langle u,v\rangle} = -(\langle u,v\rangle+\overline{\langle u,v\rangle})$$

So $\text{Re}(\langle u,v\rangle)=0$ is the condition.
You can even verify this in the basic case of $V=\Bbb C$ where the inner product is given as $\langle u,v\rangle= u\overline{v}$. Let $u=1, v=i$ then $\langle u,v\rangle =(1)(-i)$ and clearly $\lVert u+v\rVert^2 = \lVert u-v\rVert^2=2$.
