3
$\begingroup$

I am trying to understand more about quaternions and I was watching the series of videos from Norman Wildberger, in particular I am a bit stuck on this video: https://www.youtube.com/watch?v=uRKZnFAR7yw

enter image description here

You can see he associates to a complex number a rotation $\varphi$. I forced myself to do some test with numbers, but assuming I have whatever complex number (i.e. $z=3+2i$) and I want to rotate it about the origin of $45^\circ$ what I tried to do is to use the parametrization of the unit circle displayed in the screenshot and enforced the $b/a = 2t/(1-t^2) = 1$ (the slope/trigonometric tangent) to find the complex number that would make such rotation, but that is going nowhere for me.

Probably I misunderstood what he is doing there but I still not see how I can use all of this theory in practice.

Thanks, Daniele

$\endgroup$
7
  • $\begingroup$ I think you're making it harder than is necessary. Remember that when we multiply two complex numbers we multiply their magnitudes and add their arguments. If you want a pure rotation, then, you must multiply by a complex number with magnitude 1. In the case you gave, multiply by $e^{i\frac{\pi}{2}}$ $\endgroup$
    – Kwame Gonzales
    Jun 30, 2018 at 19:42
  • $\begingroup$ yes that is with the usual approch with trigonometry, but if you watch the video he does it in a different way without using any trascendental function and I would like to understand how he does that. $\endgroup$
    – dd95
    Jun 30, 2018 at 19:43
  • 1
    $\begingroup$ You can put lipstick on a pig but it's still a pig. At the end of the day, whatever he is doing must be equivalent to multiplication by that complex number. Also, your question is missing a lot of context; you're supposed to transcribe whatever formulas/equations you're using into your question. Linking a youtube video is not the same thing. $\endgroup$
    – Kwame Gonzales
    Jun 30, 2018 at 19:47
  • $\begingroup$ @TonyS.F. You mean $e^{i\pi/4}$. $\endgroup$
    – J.G.
    Jun 30, 2018 at 19:57
  • 1
    $\begingroup$ As shown in the picture, you can get any complex number of modulus $1$ by dividing a complex number by its conjugate. This is essentially Hilbert 90. That avoids exponential parameterisation.... $\endgroup$
    – Angina Seng
    Jun 30, 2018 at 20:02

1 Answer 1

Reset to default
1
$\begingroup$

What Wildberger says is correct, but the rational parametrization he uses is just a convenience. In the case of $\, 45^\circ, \,$ if you use $\, t=-1+\sqrt2 \,$ you get that $$ r_\theta := \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2}\ i = \frac{\sqrt2}2 (1+i) $$ is the complex number associated with the rotation by $\,45^\circ. \,$ Now, if $\, z = 3 + 2\:i, \,$ then $$ z\, r_\theta = \frac{\sqrt2}2(3 + 2i)(1 + i) = \frac{\sqrt2}2 (1 + 5i). $$

Wildberger is well known for his rational approach to mathematics. In my opinion, it has some merit but it does not solve all problems. This case is an example. I suggest that you continue studying and then you will find that you understand more as you go on. Don't get derailed by temporary confusion.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .