Edit: Let $X$ and $Y$ be the vertices represented by $x$ and $y$ respectively. Then $y-x$ represents the vector $\overrightarrow{XY}$. On $\mathbb{C}^2$, multiplying a complex number by $i=\sqrt{-1}$ means rotation by 90 degrees anticlockwise. Thus the vector $\vec{v}$ obtained by rotating $\overrightarrow{XY}$ ninety degrees to the left is given by $i(y-x)$.
Now, if $x$ and $y$ are opposite vertices on the diagonal, then $\frac{y+x}{2}$ is the center of the square, $\overrightarrow{XY}$ is the diagonal and $\vec{v}$ is the other diagonal. The other two vertices are obtained by traveling from the center half the diagonal along $\vec{v}$ in both directions. Hence they are given by $\frac{y+x}{2}\pm i\frac{y-x}{2}$.
If $x$ and $y$ are adjacent vertices, then $\overrightarrow{XY}$ represents an edge and $\vec{v}$ represents a perpendicular edge. Hence the other two vertices are obtained by shifting both $x$ and $y$ by $\vec{v}$ or $-\vec{v}$. Hence they are given by $\{x+i(y-x),\ y+i(y-x)\}$ (if the other two vertices lying on the left of $\overrightarrow{XY}$) or $\{x-i(y-x),\ y-i(y-x)\}$ (if the other two vertices lying on the right of $\overrightarrow{XY}$).