= Solution
The assertion is \b[false]. Let $D_\lambda(z)=\lambda z$ and $I(z)=1/z$. Their relations are
$$
I^2=\operatorname{id},\qquad ID_\lambda=D_{\lambda^{-1}}I.
$$
Moving each inversion to the right reduces every word to $D_\lambda I^\epsilon$ with $\epsilon=0$ 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 transformation>
$$
\boxed{M(z)=\frac z{z+1}}
$$
fixes 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.
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