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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.

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    @Rahul Thanks for pointing it out. I'll fix it.2011-08-30

2 Answers 2

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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 $\overrightarrow{XY}$).

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    @Joe: Done. Hopes it's clearer now.2011-08-30
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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.

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    Yes-try sketching it. One point can be assigned to corner A of square ABCD. Then the other can be any of the other three.2011-08-30