# Complex number rotation - Wildberger approach

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

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

• 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 $e^{i\frac{\pi}{2}}$ Jun 30, 2018 at 19:42
• 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.
– dd95
Jun 30, 2018 at 19:43
• 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. Jun 30, 2018 at 19:47
• @TonyS.F. You mean $e^{i\pi/4}$.
– J.G.
Jun 30, 2018 at 19:57
• 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.... Jun 30, 2018 at 20:02

## 1 Answer

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.