Use complex numbers to prove that the composition of rotations is another rotation

Question

From ACOPS by Paul Zeitz, q4.2.30, a section on complex numbers.

"Let $R_a(\theta)$ denote the transformation of the plane that rotates everything about the center point $a$ by $\theta$ radians counterclockwise. Prove the interesting fact that the composition of $R_a(\theta)$ and $R_b(\phi)$ is another rotation $R_c(\alpha)$. Find $c,\alpha$ in terms of $a,b,\theta,\phi$."

I've tried to argue that $R_a(\theta)$ maps the point $z$ on the complex plane to $z^\prime = e^{i\theta}(z-a)+a,$ and therefore the composition of $R_a(\theta)$ and $R_b(\phi)$ maps $z$ to $z^\prime = e^{i\phi}(e^{i\theta}(z-a)+a-b)+b.$ I have then tried to write this expression in the form $e^{i(\phi+\theta)}(z-c)+c$ but have been unsuccessful.

Answer 1

Hint: Try to find a fixed point $c$, that is $$(R_b\circ R_a) (c) =c$$

that should be a center of a new rotation. Anyway, you are right about rotation angle, which is pretty obvious from formula you derive.

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If my calculation are correct it should be $$ c={{e^{i\phi}(ae^{i\theta}-a+b)-b.}\over e^{i(\phi+\theta)}-1 }$$

Answer 2

You need to solve

$$e^{i\phi}(e^{i\theta}(z-a)+a-b)+b=e^{i(\phi+\theta)}(z-c)+c$$ for $c$.

Rearranging the terms, you get

$$e^{i(\phi+\theta)}z+e^{i\phi}\left(\left(1-e^{i\theta}\right)a-b\right))+b=e^{i(\phi+\theta)}z+c\left(1-e^{i(\phi+\theta)}\right),$$

which validates the claim about composition and yields

$$c=\frac{e^{i\phi}\left(\left(1-e^{i\theta}\right)a-b\right))+b}{1-e^{i(\phi+\theta)}}.$$