# Derivation of rotation isometry on the complex plane

1. Nov 13, 2014

### 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....

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2. Nov 14, 2014

### 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..

3. Nov 14, 2014

### Fredrik

Staff Emeritus
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.