Vertices of a Square Give $x,y\in \mathbb{C}$  are two vertices of a square. I want to find the other vertices in all cases.
My attempt so far is the following:    
Let $x=a+ib$ , $y=c+id$.
If $x$ and $y$ are on the opposite end of the diagonal, then the other two vertices reduces to  $a+id$ and $c+ib$.
Now, how to I find the other cases.
 A: 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 $\overrightarrow{XY}$).
A: If $x$ and $y$ are neighboring vertices, the side is $y-x$.  One choice is to go from $x$ to $y$ and turn left, which is just multiplying by $i$.  So the next vertex will be $y+i(y-x)$, then turn left again, giving $y+i(y-x)+i^2(y-x)=x+i(y-x)$  For the other choice, the $+i$'s become $-i$'s.
