Squares in geometry with complex numbers 
$ABCD$ and $AB'C'D'$ are two squares (orientated in the same way) with a common vertex $A$ and not overlapping. The midpoints of the squares are $P$ and $Q$. The midpoint of $BD'$ is $R$, the midpoint of $B'D$ is $S$.
Prove that $PQRS$ is a square, using that $PS$ rotated $90^\circ$ gives $PR$.


My question is: why isn't $PR=i\cdot PS$? And second, if I let $D'$ be $i \beta$ and $B'$ $\beta$, would there be any difference?

Let $a=0$. Then from $b-p=i(a-p)$, and similarly for $c$ and $d$, it follows that
$$b=(1-i)p,\qquad c=2p,\qquad d=(1+i)p\ .$$
In the same vein,
$$b'=(1-i)q,\qquad c'=2q,\qquad d'=(1+i)q\ .$$
This gives
$$r={1\over2}(b+d')={1-i\over2}p+{1+i\over2}q,\qquad s={1\over2}(d+b')={1+i\over2}p+{1-i\over2}q\ .\tag{1}$$
From
$${1\over2}(r+s)={1\over2}(p+q)=:m\tag{2}$$
we infer that $pqrs$ is a parallelogram with center $m$. Therefore it suffices to verify that
$r-m=i(q-m)$, which simplifies, using $(2)$, to $r-s=i(q-p)$. The latter is an immediate consequence of $(1)$.
