 Solution

The assertion is false. Let $D_{\lambda }(z)=\lambda z$ and $I(z)=1/z$. Their relations areMoving each inversion to the right reduces every word to $D_{\lambda }{I}^{ϵ}$ with $ϵ=0$ or $1$. Thus the subgroup generated by complex scalings and reciprocal inversion consists only of $z\mapsto \lambda z$ and $z\mapsto \lambda /z$. Of these, only the first form fixes zero.

The Möbius transformationfixes zero but is not a scaling: it sends infinity to $1$ instead of infinity. It therefore cannot belong to the subgroup generated by scalings and reciprocal inversion.