What does it mean if the standard Hermitian form of complex two vectors is purely imaginary? If $v,w \in \mathbb{C}^n$, what does it mean geometrically for $\langle v , w \rangle$ to be purely imaginary? 
 A: To understand what geometrically what it means for $\left<v,w\right>$ to be purely imaginary, it is necessary to understand geometrically how a Hermitian inner product $\left<\cdot,\cdot\right>_{\mathbb C}$ on $\mathbb C^n$ relates to the Euclidean inner product $\left<\cdot,\cdot\right>_{\mathbb R}$ on 
$\mathbb R^{2n}$. These are related because $\mathbb C^n$ as a real vector space is simply $\mathbb R^{2n}$.
The key to the relationship is that multiplication by $i$ of $\mathbb C^n$ corresponds to applying a real linear map $J\colon \mathbb R^{2n}\to\mathbb R^{2n}$ such that $J^2=-1$. Concretely, you think of $\mathbb R^{2n}$ as $n$ copies of $\mathbb R^2$, and $J$ as the linear map of rotating all the planes simultaneously by $90^\circ$. Geometrically, this is a choice of axis through the whole space $\mathbb R^{2n}$ around which to do rotations. 
Then the formula expressing the relationship between the Hermitian inner product on $\mathbb C^n$ and the Euclidean inner product is
$$\left<v,w\right>_{\mathbb C}=\left<v,w\right>_{\mathbb R}+i\left<v,Jw\right>_{\mathbb R}$$ 
(this is the formula when the Hermitian inner product is complex-linear in the first variable and conjugate-linear in the second; otherwise the imaginary part $\left<Jv,w\right>$ instead).
From this, the geometric interpretation is easy. The real part of the Hermitian inner product is $0$ if the two vectors are perpendicular in $\mathbb R^{2n}$. The imaginary part of the Hermitian inner product is $0$ if rotating one of the two vectors by $90^\circ$ (around the designated axis!) makes them perpendicular. This is somewhat hard to visualize because the smallest non-trivial example requires thinking about $\mathbb R^4$, which is $1$ dimension higher than comfortable.

You can check the formula with the following easy computation of dot products. Let $(z_1,\dots,z_n),(w_1,\dots,w_n)\in\mathbb C^n$ be given as $(a_1,b_1,\dots,a_n,b_n)\in\mathbb R^{2n}$ and $(c_1,d_1,\dots,c_n,d_n)\in\mathbb R^{2n}$. Then
$$\begin{align*}
(z_1,\dots,z_n)\cdot\overline{(w_1,\dots,w_n)}&=\sum_{j=1}^nz_j\cdot\overline w_j=\sum_{j=1}^n(a_j+ib_j)(c_j-id_j)\\
&=\sum_{j=1}^na_jc_j+b_jd_j+i(a_j(-d_j)+b_jc_j)\\
&=\sum_{j=1}^n(a_j,b_j)\cdot(c_j,d_j)+i((a_j,b_j)\cdot J(c_j,d_j))\\
=(a_1,b_1,\dots,a_n,b_n)\cdot(c_1,d_1,\dots,c_n,d_n)&+(a_1,b_1,\dots,a_n,b_n)\cdot J(c_1,d_1,\dots,c_n,d_n)
\end{align*}$$
A: I'm not sure what it means geometrically, but it means $\| v + w \|^2 = \|v \|^2 + \|w^2\|$. 
$\langle v, w \rangle$ purely imaginary means that $\overline{\langle v, w \rangle } = - \langle v, w \rangle$.  However, we always have $\overline{\langle v, w \rangle } = \langle w, v \rangle$.  This means $$ \langle w, v \rangle = - \langle v, w \rangle \implies \langle w, v \rangle  + \langle v, w \rangle = 0.$$
This implies that \begin{align*} \| v + w \|^2 &= \langle v + w, v + w \rangle \\
&= \langle v, v\rangle + \langle v,w\rangle + \langle w, v \rangle + \langle w, w \rangle \\
&= \| v \|^2 + \| w \|^2.
\end{align*}
