jmc

Let $w$ be the complex number corresponding to the point $S.$ Since $PQSR$ is a parallelogram, $$w=(1+i)z+2\overline{z}-z,$$so $w=2\overline{z}+iz.$ Then $\overline{w}=2z-i\overline{z},$ so $$\begin{array}{rl}|w{|}^2& =w\overline{w}\\ & =(2\overline{z}+iz)(2z-i\overline{z})\\ & =4z\overline{z}+2i{z}^2-2i{\overline{z}}^2+z\overline{z}\\ & =5|z{|}^2+2i(z^2-{\overline{z}}^2)\\ & =2i(z^2-{\overline{z}}^2)+5.\end{array}$$Let $z=x+yi,$ where $x$ and $y$ are real numbers. Since $|z|=1,$ $x^2+y^2=1.$ Also, $$\begin{array}{rl}2i(z^2-{\overline{z}}^2)& =2i((x+yi{)}^2-(x-yi{)}^2)\\ & =2i(4ixy)\\ & =-8xy,\end{array}$$so $|w{|}^2=5-8xy.$

By the Trivial Inequality, $(x+y{)}^2\ge 0.$ Then $x^2+2xy+y^2\ge 0,$ so $2xy+1\ge 0.$ Hence, $-8xy\le 4,$ so $$|w{|}^2=5-8xy\le 9,$$which implies $|w|\le 3.$

Equality occurs when $z=-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}},$ so the maximum distance between $S$ and the origin is $\boxed{3}.$