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user7530
user7530
00:12
@rschwieb here’s my attempt. Let $a,b,c,d$ be four given real numbers, and consider the curve $\gamma: S^1\to \mathbb{R}^4$ with
$$\gamma(\theta) = \left[\begin{array}{c} a \cos \theta - b \sin\theta \\ a \sin \theta + b \cos \theta \\ c \cos \theta - d \sin \theta \\ c \sin \theta + d \cos \theta \end{array}\right].$$
Does the image of $\gamma$ lie in a two-dimensional subspace of $\mathbb{R}^4$?
 
13 hours later…
rschwieb
rschwieb
13:08
@JoshuaRonis Look: do you want this or not, for two complex numbers $a,b$: $\begin{bmatrix} \cos(\theta) & \ -sin(\theta) \\ \sin(\theta)& \cos(\theta) \end{bmatrix} \begin{bmatrix} a\\ b\end{bmatrix}=\begin{bmatrix} a\cos\theta-b\sin\theta\\a\sin\theta+b\cos\theta\end{bmatrix}$
Joshua Ronis
Joshua Ronis
13:43
@rschwieb yes sorry! but yes, I agree!
And here, I did some of the math out with arbitrary examples: let's say $a=(3+4i)$ and $b=(12+5i)$.
So we've got...
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Now, let's say that $\theta = 30^0$, just like an example...
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Alright! But those numbers don't really mean anything to us, so I rewrote each component as the product of a real number and a unit sized complex number
user image
Okay...the first thing I noticed is that the magnitude of the vector didn't change...
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That's an "approximately equals" sign because of rounding errors
...and I don't know where to go from here. I don't know how to find the plane that both the unrotated vector $\vec{v}$ and rotated vector $R_\theta \vec{v}$ lie on, and I'm not sure if for a different choice of $\theta$ besides $30^0$, the vector would lie on that same plane, or how to even tell...
@rschwieb thanks for helping. Looking forward to ur response! If anything is unclear let me know.
Also, $R_\theta$ is the rotation matrix that rotates by an angle of $\theta$.
If I don't respond in the moment you send it its just that I don't get notifications for the chats for some reason. I did read your last message though!
@user7530 you might want to get in here too?
@user7530 wait...but why would you be rotating the $(a,b,c,d)$ individually, if the complex numbers come in pairs? I now see @rschwieb had the right idea when he first interpreted my question!!!
rschwieb
rschwieb
14:17
@JoshuaRonis Ok, well then that's progress.

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