Electrodynamics : Problem with Notations for fields $\vec{(\vec{r},t)}$ and $\vec{B}(\vec{r},t)$(complex and real notations)

Question

I'm sutyding a course on electrodynamics and am stuck on a few lines I can't make sense of. The professor uses $$\vec{E}(\vec{r},t) = \vec{U_0} cos (\vec{k}\cdot \vec{r} - \omega t + \phi)$$ (so far, so good) for the real part of this complex: $$\vec{U_0}e^{i(\vec{k}\cdot \vec{r} - \omega t + \phi)}\tag{1}$$ which he then equates to: $$\vec{\underline E}(\vec{r})e^{-i\omega t}\tag{2}$$ $\textbf{the underlined letter being a complex vector}$

This would imply that $\vec{U_0}$, a real vector in $\mathbb{R}^2$ can be multiplied by a complex number $e^{i(\vec{k}\cdot \vec{r})}$ and become that thing: $\vec{\underline E}(\vec{r})$. What is this? A vector whose components are all complex numbers, where each real component of $\vec{U_0}$ is multiplied by the complex number $e^{i(\vec{k}\cdot\vec{r} + \phi)}$ ?

Let me be clearer. I expect (1) to be: $$||\vec{U_0}||e^{i(\vec{k}\cdot \vec{r} - \omega t + \phi)},$$ and (2) to be: $$\underline E(\vec{r})e^{-i\omega t}.$$

And I would expect this to be impossible to write: $\vec{A}e^{it}$

But if I were to write it, I would understand it as a vector rotating (phaser notation?)

If you can help out, your help is much appreciated, and I thank you in advance. Note: if you wish to use $j$ for the imaginary unit, be my guest.

Answer 1

Let $\mathbf U$ be triple of real numbers $[U_1,U_2,U_3]$, giving components of a vector $\vec{U}$ in some coordinate system. The expression $$ \mathbf U e^{ir} $$ where $r$ is any real number, is just another triple of numbers, defined as $$ \mathbf U e^{ir} = [U_1e^{ir},U_2e^{ir},U_3e^{ir}]. $$ These three numbers also represent a kind of vector, only in a different vector space. The difference is that the first vector $\mathbf U$ lives in $\mathbb R^3$, but the other, in general (except if $e^{ir}$ is real number), is not, but the second lives in $\mathbb C^3$. Both are vector spaces. The vectors from the complex vector space, however, have no simple geometric interpretation, because they have 6 independent real components.

A tuple vector from $\mathbb C^3$ does not represent an arrow seated in 3D space, but it could be looked upon as a vector in 6D space (most people can't imagine that so they stick with algebra).

The old real tuple vector also belongs to $\mathbb C^3$, and the two vectors are linearly dependent. As you can see, multiplying an ordinary vector by imaginary unit $e^{ir}$ has nothing obvious to do with rotation of the vector in 3D space. It just produces a vector in $\mathbb C^3$ vector space.