2. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant 180^{\circ}\right)$, and the complex numbers $z, (1+i) z, 2 z$ correspond to points $P, Q, R$ on the complex plane, respectively. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $P Q$ and $P R$ is $S$. Then the maximum distance from point $S$ to the origin is $\qquad$.
3
2.【Analysis and Solution】Let the complex number $\omega$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have $\omega+z=2 \bar{z}+(1+i) z$, which simplifies to $\omega=2 \bar{z}+i z$.
Thus, $|\omega|^{2}=(2 \bar{z}+i z)(2 \bar{z}-i z)$
$$
\begin{array}{l}
=4+1+2 i\left(z^{2}-{ }_{-2}^{z}\right) \\
=5-4 \sin 2 \theta \leqslant 5+4=9 .
\end{array}
$$
(When $\theta=135^{\circ}$, the equality holds.)
Therefore, $|\omega|_{\max }=3$.