Question: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}\newcommand{\i}{\mathrm{i}}\newcommand{\text}[1]{\mathrm{#1}}\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}\newcommand{\i}[0]{\mathrm{i}}\newcommand{\kume}[1]{\mathbb{#1}}\newcommand{\bold}[1]{\textbf{#1}}\newcommand{\italic}[1]{\textit{#1}}\newcommand{\kumedigerBETA}[1]{\rm #1\!#1}$ $$\text{Imagine\s yourself\s rotating \s a \s point\s in \s a\s complex \s plane\s by \s }\frac{\i\pi}{2} \text{radians\s counterclockwise}$$
Here we have the rotation matrix:
$$\begin{bmatrix}\cos x & -\sin x \\\sin x & \cos x \end{bmatrix}$$
Putting in the values
$$\begin{bmatrix}\cos \frac{\i\pi}{2} & -\sin \frac{\i\pi}{2} \\\sin \frac{\i\pi}{2} & \cos \frac{\i\pi}{2} \end{bmatrix}$$
I know that they're the basis vectors, one of their components became complex.
I can imagine each basis vector in imaginary plane, but not all at the same time.
How can I imagine where the point has gone?
I know that I can't imagine 4D, I'm just trying to imagine the complex planes of the new point.
(Note, I know that $\cosh x=\cos \i x$ and $\sinh x=-\sin\i x$. Will that help?)