0
$\begingroup$

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?)

$\endgroup$
  • $\begingroup$ Have you tried writing your point $p$ as $p=\cos \varphi + \mathrm i \sin \varphi$ and multiply it me the rotation matrix? Then compare the result with $p$ with the help of trig. identities. These special matrices are very import for QR-decompisition. You might google givens rotation. Hope that helps. $\endgroup$ – Niklas Sep 14 '17 at 15:41
  • 1
    $\begingroup$ When you say complex plane, do you mean the $\mathbb{C}$-vector space $\mathbb{C}^2$ ($\mathbb{C}$-bidimensional vector space) or the $\mathbb{C}$-vector space $\mathbb{C}$ (usually called complex plane)? $\endgroup$ – trying Sep 14 '17 at 15:53
  • $\begingroup$ And what does “counterclockwise” mean when you talk about the complex angle $i\pi/2$? Are you sure it's not meant to be just $\pi/2$? $\endgroup$ – Hans Lundmark Sep 14 '17 at 18:20
  • $\begingroup$ Rotation in the complex plane is accomplished by multiplying your complex number, say $z$, by $e^i\theta$ to get an anticlockwise rotation of $\theta$ . If the angle is imaginary then you are just scaling $z$ by a real number. Rotation doesn't have any meaning because there's no angle involved. $\endgroup$ – Cye Waldman Sep 16 '17 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.