The points $P,$ $Q,$ and $R$ are represented by the complex numbers $z,$ $(1 + i) z,$ and $2 \overline{z},$ respectively, where $|z| = 1.$  When $P,$ $Q$, and $R$ are not collinear, let $S$ be the fourth vertex of the parallelogram $PQSR.$  What is the maximum distance between $S$ and the origin of the complex plane?
Let's start by finding the complex number that represents $S$. Ok, so we know that $S$ is the fourth vertex of the parallelogram $PQSR$. So, we know that $S-R=Q-P$ and $S=R+(Q-P)$. So, we know that $S=2\bar{z}+((1+i)z-z)$ So, we know that $S=2\bar{z}+(z+zi-z)$ So, we know that $S=2\bar{z}+zi$ Now we need to find the maximum distance between $S$ and the origin. So, let's look at the distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. So, let's plug in the values: $\sqrt{(x-0)^2+(y-0)^2}$. So, $S=(2x-y,-2y+x)$. So, the distance between $S$ and the origin is $\sqrt{(2x-y-0)^2+(-2y+x-0)^2}$. That's correct. That expression is equal to $\sqrt{(2x-y)^2+(-2y+x)^2}$. To find the maximum, we can use the AM-GM inequality. So, we can express the sum as $5x^2+5y^2-8xy$. Right. So we want to find the maximum of $5x^2+5y^2-8xy$ subject to $x^2+y^2=1$. Let's write $5x^2+5y^2-8xy$ as $5(x^2+y^2)-8xy$. So we want to find the maximum of $5(x^2+y^2)-8xy$ subject to $x^2+y^2=1$. But since $x^2+y^2=1$, we can write the maximum of $5(x^2+y^2)-8xy$ subject to $x^2+y^2=1$ as $5-8xy$. Ok, let's use the AM-GM inequality. So we want to find the maximum of $5-8xy$ subject to $x^2+y^2=1$. The AM-GM inequality tells us that the arithmetic mean of $x^2$ and $y^2$ is greater than or equal to the geometric mean. So, $\frac{x^2+y^2}{2} \ge \sqrt{x^2y^2}$. But we know that $x^2+y^2=1$. So, $\frac{1}{2} \ge \sqrt{x^2y^2}$. And since $x^2y^2$ is positive, $\sqrt{x^2y^2}=|xy|$. So, $\frac{1}{2} \ge |xy|$. So, $-\frac{1}{2} \le xy \le \frac{1}{2}$. So, $5-\frac{8}{2} \le 5-8xy \le 5+\frac{8}{2}$. So, $1 \le 5-8xy \le 9$. So, the maximum value of $5-8xy$ subject to $x^2+y^2=1$ is $9$. So, the maximum distance between $S$ and the origin is $\sqrt{9}=3$. That's correct. So the answer is $3$.

