trivial motion identity

Question

I am a bit confused about this trivial identity. We have rotation by $\theta$ and translation with a vector $a$. Can someone provide intuition why this is true?

Answer 1

Rotation of a vector in 2D by an angle $\theta$ can be thought of as multiplication by $e^{i\theta}$, i.e. \begin{align*} \rho_{\theta}(z)&=ze^{i\theta} && \text{(rotation)}\\ \tau_{a}(z)&=z+a && \text{(translation)} \end{align*} Thus $$\rho_{\theta}\left(\tau_{a}(z)\right)=\rho_{\theta}(z+a)=e^{i\theta}(z+a).$$ And, $$\tau_{a'}\left(\rho_{\theta}(z)\right)=\tau_{a'}(ze^{i\theta})=ze^{i\theta}+a'.$$ But we are given that $a'=ae^{i\theta}$. Thus the two expressions are equal.