Intuition for complex dot product I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $\mathbb{C}^n$ are viewed as real vectors in $\mathbb{C}^{2n}$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?
 A: A vector $v \in \mathbb{C}^n$ can be expressed as
$$
v\ =\ \begin{bmatrix} a_1 + ib_1 \\ \vdots \\ a_n + ib_n \end{bmatrix}\ =\ \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}\ +\ i\begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}\ =\ \Re(v)+i\Im(v)
$$
As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
\langle v,v \rangle \geq 0,\ \ \text{(equality iff $v \equiv 0$)}
$$
then it is sensible to define (for any $u,v \in \mathbb{C}^n$)
$$
\langle u,v \rangle := \sum_i \bar{u}_i v_i
$$
or equivalently,
\begin{align}
\langle u,v \rangle &= \big{(}\Re(u)^\top-i\Im(u)^\top\big{)}\big{(}\Re(v)+i\Im(v)\big{)}\\
&= \big{(}\Re(u)^\top\Re(v) + \Im(u)^\top \Im(v)\big{)} + i\big{(}\Re(u)^\top \Im(v) - \Im(u)^\top\Re(v)\big{)}\\
&= \Re\big{(}\langle u,v \rangle\big{)} + i \Im\big{(}\langle u,v \rangle\big{)}
\end{align}
We see that,
$$
\frac{\Re\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \frac{\Re(u)^\top\Re(v) + \Im(u)^\top \Im(v)}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \cos(\theta)
$$
where $\theta$ is, real-geometrically, the angle between $\begin{bmatrix} \Re(u) \\ \Im(u) \end{bmatrix}$ and $\begin{bmatrix} \Re(v) \\ \Im(v) \end{bmatrix}$.
On the other hand,
$$
\frac{\Im\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \frac{\Re(u)^\top \Im(v) - \Im(u)^\top\Re(v)}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \cos(\phi)
$$
where $\phi$ is, real-geometrically, the angle between $\begin{bmatrix} \Re(u) \\ \Im(u) \end{bmatrix}$ and $\begin{bmatrix} \Im(v) \\ -\Re(v) \end{bmatrix}$.
The latter real vector is the same as the one that defined $\theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $\mathbb{C}^1$, we have $\phi = \theta - \frac{\pi}{2}$ and thus,
$$
\frac{\Im\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \sin(\theta)
$$
A: The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2\times 2$ matrix $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}.$$ 
(There is another embedding into $M_2(\mathbb R)$ which is just the conjugate.)
Then a complex vector: $$v=\begin{pmatrix}v_1\\v_2\\ \vdots\\ v_n\end{pmatrix}$$
corresponds to a $2n\times 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $\left(\overline{v_1},\dots,\overline{v_n}\right).$ Then  the $2\times 2$ matrix, $A_v^TA_w,$ corresponds to $\sum_{i=1}^n\overline{v_i}w_i$.
