If $f$ is a function defined on $\mathbb{R}^2$, we denote by $f^*$ the function $f \circ \Phi$, defined on $G$ by $f^*(g) = f(\Phi(g))$ where $\Phi : G \rightarrow \mathbb{R}^2$ is the function introduced in question I.A.5.
Prove that for all $g$ in $G$ and $r$ such that $\Phi(r) = \overrightarrow{0}$ we have $f^*(gr) = f^*(g)$.