@user7530 The user writes: "$\begin{bmatrix} \cos(90) & \ -sin(90) \\ \sin(90)& \cos(90) \end{bmatrix} \begin{bmatrix} (2+3i)\\ (1+4i) \end{bmatrix}=\begin{bmatrix} -(1+4i)\\ (2+3i) \end{bmatrix}$" which is not what you are conjecturing they mean, (which I believe is "apply the rotation to the two complex entries individually.) The appear to be simply treating the matrix as a $\mathbb C$-linear operator on $\mathbb C^2$.

@user7530 that's exactly the question...if I understand the notation correctly. For a specific vector (with possibly complex components), does its multiplication by any rotation matrix $R_\theta$, regardless of what $\theta$ is, give us a vector that lies on the same plane? The same two-dimensional linear subspace?

@rschwieb you may be correct as well on what I mean...I'm not very good at understanding all the notation :( Also, thanks for your answer, +1, but I'm still confused since I don't know if you understood my question...

@JoshuaRonis What is "complex notation"? It makes no difference if you are using $\mathbb R$ or $\mathbb C$, the matrix algebra is all the same... If you meant "complex" in the literal english sense, then never mind :)

I mean what @user7530 wrote in his comment. Is the way I rewrote it in English in my response to his comment what you meant with your answer? Yep, I meant complex in the English sense! Still a high school student here!

@JoshuaRonis What you wrote in your post suggests you are just using a real $2\times 2$ matrix to multiply elements of $\mathbb C^2$ via matrix multiplication. My understanding (possibly wrong) of what user7530 wrote is that $R_\theta (a,b)$ for complex $a$ and $b$ would be $(R_\theta(a), R_\theta(b))\in \mathbb C^2$. These two operations are completely different, so I suggest you carefully determine which one you want to write about.

@JoshuaRonis Doing it the second way would send $(2+3i, 1+4i)$ to $(-3+2i, -4+i)$

@user7530 Hmm, OK, thanks for clarifying. I will think about it. I don't see anything about separating the parts though...

@user7530 okay so no, I don't mean what rscwieb thought user7530 meant in his comment, but I think I mean what user7530 actually meant in his comment! Regardless, that last way of putting it is the best: does the entire orbit of any complex vector getting rotated by an angle of $\theta$, which by entire orbit I mean we let $\theta$ range from $0$ to $360^0$, lie in the same plane? Thanks to both of you!

@JoshuaRonis Can you rewrite your math so it actually reflects what you want, then? This is very confusing when you say you want one thing but write another.

@rschwieb could you please do it, and I'll accept the edit? I'm not sure I'll do it correctly...I'd really appreciate it!

@JoshuaRonis How can I? I am not very sure what you mean at all. If it's a problem of formatting I'm sure we can tidy it up once you supply the raw material.

@rschwieb wait I'm a little confused...is there a part of my math that's incorrect? Also, should we continue this discussion in chat? Well, later today at least, I gtg right now, but I don't want to keep on cluttering up the comments...

@JoshuaRonis By all means create a chat. I'd be happy to do that. My very first comment in the thread here supports my original viewpoint of what you meant. If I've got your meaning wrong, then you'd have to rewrite that to be whatever you really meant.

Moved it to chat! I can't stay long now, but how do you know they'll remain on a plane? And given a vector with two complex components, how can you identify that plane?

Until you decide I was right when I described how your transformation operates, or wrote it in a way that I can understand it, I can’t make any progress.