Derivation of rotation isometry on the complex plane

[PcumP_Ravenclaw]

Dear all, can you please verify if my derivation of the algebraic formula for the rotation isometry is correct. The handwritten file is attached.

The derivation from the book (Alan F beardon, Algebra and Geometry) which is succinct but rather unclear is given below.

Assume that f (z) = az + b. If a = 1 then f is a translation. If a = 1,
then f (w) = w, where w = b/(1 − a), and f (z) − w = a(z − w). It is now
clear that f is a rotation about w of angle θ, where a = $e^{iθ}$ .

danke...

 

[PcumP_Ravenclaw]

I tried to arrive at my solution beginning with $f(z) − w = a(z − w)$ and they both matched! but the reason for $a !=1$ is so that a will have imaginary part therefore an angle θ to rotate about??

The reason why $|a| =1$ is so that z is not proportionally multiplied/scaled ??

danke..

 

[Fredrik]

It's hard to follow the handwritten stuff, since there are no explanations in it. So I'll just comment on what you said here. You defined f by f(z)=az+b, so it follows immediately that if a=1, then f is a translation. Your definition of w makes sense if and only if a≠1, so I first assumed that the next step is to examine that case. But a few steps later, you wrote $a=e^{i\theta}$, which only makes sense if |a|=1. So you should have stated earlier that the case you're investigating is a≠1, |a|=1. To show that f is a rotation around w when these conditions are met, it's sufficient to show that |f(z)-w|=|z-w|. I would say that your typed statements do that, but you could have included more details. We have
$$f(z)-w=az+b-w=az+w(1-a)-w= a(z-w)+w-w =a(z-w)$$ and therefore
$$|f(z)-w|=|a||z-w|=|z-w|.$$ As you can see, the assumption that |a|=1 is essential.

By the way, you should try to avoid notations like a!=0. The ugliness can be avoided by using the LaTeX code \neq, or just the ≠ symbol that you find when you click on the ∑ symbol in the editor. The notation a!=0 is especially bad since the most obvious interpretation of it is different from what you had in mind.