= Solution
Let $t=e^{iz}$ and $|u\rangle=U|b\rangle=p|\xi\rangle+q|\phi\rangle$. The first factor in $G$ is
$$
(1-t)|u\rangle\langle u|-I.
$$
Since $R(\xi,z)|u\rangle=pt|\xi\rangle+q|\phi\rangle$, the coefficient of $|\phi\rangle$ in $G|u\rangle$ is
$$
q\left[(1-t)(p^2t+q^2)-1\right].
$$
It vanishes when
$$
(1-t)(p^2t+q^2)=1,
$$
or, using $p^2+q^2=1$,
$$
t+t^{-1}=\frac{p^2-q^2}{p^2}.
$$
A unit-modulus solution $t=e^{iz}$ exists exactly when the right-hand side lies in $[-2,2]$. The upper bound is automatic, while the lower bound is
$$
q^2\leq3p^2.
$$
Thus exact preparation by one application of $G$ is possible precisely when
$$
\boxed{p\geq\frac12}
\qquad\text{or equivalently}\qquad
\boxed{q\leq\sqrt3\,p}.
$$
One may choose $z$ so that $\cos z=(p^2-q^2)/(2p^2)$. Then $G|u\rangle$ has no $|\phi\rangle$ component and, by unitarity, equals $|\xi\rangle$ up to phase. This is a <phase-matched amplitude amplification> step.
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