Hermitian Product Orthogonality In Lang’s Linear Algebra, after defining the Hermitian product for vector spaces over complex fields, he mentioned that all previously mentioned notions involving orthogonality with scalar products (including theorems) held true for the Hermitian product. I’m not sure if this theorem is though:
For all $v,w\in V$, where $V$ is a vector space over a complex field and $||a||$ is the norm with respect to a Hermitian product $(s,t)$:
$$||v-w||^2=||v+w||^2 \Leftrightarrow (v,w) =0$$.
I tried to prove this and got to the following:
$$||v-w||^2=||v+w||^2 $$$$\Leftrightarrow (v-w,v-w)=(v+w,v+w)$$$$ \Leftrightarrow (v-w,v)-(v-w,w)=(v+w,v)+(v+w,w)$$$$\Leftrightarrow \overline{(v, v-w) - (w, v-w)} = \overline{(v, v+w)+(w,v+w)}$$$$\Leftrightarrow (v,v) - (v,w) - (w,v) + (w,w) = (v,v) + (v,w) + (w,v) + (w,w)$$$$\Leftrightarrow (v,w) + (w,v)=0 $$$$ \Leftrightarrow (v,w) + \overline{(v,w)} = 0.$$
My trouble is with the fact that for all complex numbers $z$, $z+\overline{z}=0$ does not imply that $z$ must be $0$ - take $z=i$ for instance. Is this just not what Lang meant by “all other properties work the same?” I mean this property wouldn’t really affect the proofs of the others anyways, but I’m still curious.
Of course, any input would be appreciated.
 A: Indeed, this statement is not true. Take $v=1,w=i$ in your above notation as you suggest:
$$
\lVert 1-i\rVert^2=2=\lVert1+i\rVert^2
$$
but $\langle 1,i\rangle=i\ne 0.$ What Lang means really is that the notion of orthogonality we get using this definition of Hermitian inner product allows us to (re)prove the standard results that one proves for inner products over $\Bbb{R}$. For instance, given $W\subseteq V$, $\dim W+\dim W^\perp=\dim V$. We also get a pairing of $V$ with $V^*$ but you should be careful as this is in general only $\Bbb{R}-$linear.
A: The statement is indeed not true.  The true statement is

If $\lVert v+\varepsilon w\rVert$ is a constant (independent of $\varepsilon$) for all $\lvert\varepsilon\rvert=1$ then $(v,w)=0$.

Over $\mathbb{R}$ there is only $\varepsilon=\pm 1$.  Over $\mathbb{C}$ there is a whole circle of them, but it turns out you only have to check 4 of them $\pm 1,\pm i$.  This is because of the polarisation identity
$$
(v,w)=\frac1{2\pi i}\int_{\mathbb{T}}\lVert v+\zeta w\rVert^2\,\mathrm{d}\zeta=\frac1{2\pi}\int_{0}^{2\pi}e^{i\theta}\lVert v+e^{i\theta}w\rVert^2\,\mathrm{d}\theta=\frac14\sum_{n=1}^4 i^n\lVert v+i^nw\rVert^2.
$$
