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?

I don't know why you have to prove it with complex numbers and not with Euclidean geometry. First, look at quadrilateral $BDB'D'$ and its diagonals $BB'$ and $DD'$. In triangle $BDB'$ the endpoints of the segment $PS$ are the midpoints $P$ and $S$ of the edges $DB$ and $DB'$ of the triangle. Therefore $PS$ is parallel to $BB'$ and $BS = \frac{1}{2} BB'$. The same argument applied to the segment $QR$ in triangle $B'D'B$ yields that $QR$ is parallel to $BB'$ and $QR = \frac{1}{2} BB'$. Thus $PS$ and $QR$ are parallel and equal in length. Apply again the same argument to segment $QS$ in triangle $DB'D'$ and to segment $RP$ in triangle $D'BD$ to conclude that $SQ, \, RP$ and $DD'$ are all parallel to each other and $SQ = RP = \frac{1}{2} DD'$. Consequently, $SQRP$ is a parallelogram with pairs of edges parallel to $BB'$ and $DD'$.
Now, look at triangles $ABB'$ and $AD'D$. Triangle $AD'D$ is a $90^{\circ}$ rotation image of triangle $ABB'$ around the point $A$. Therefore the segment $DD'$ is a $90^{\circ}$ rotation image of segment $BB'$. Therefore segment $BB'$ is orthogonal to $DD'$ and they have equal length. Combining this latter conclusion with our previous findings yield that $SQ = RP = \frac{1}{2} DD' = \frac{1}{2} BB' = QR = PS$, which means that $SQRP$ is a rhombus, and that $SQ$ is orthogonal to $PS$ since they are parallel to the orthogonal segments $BB'$ and $DD'$ respectively. As a rhombus with a $90^{\circ}$ angle, $SQRP$ is a square.  
Edit: Here is another geometric method which could hint you how to prove this by complex numbers the way you are asked. On the line $AS$ draw point $A_1$ such that $S$ is the midpoint of $AA_1$. On the line $AR$ draw point $A_2$ such that $R$ is the midpoint of $AA_2$. Then you obtain two parallelograms $AB'A_1D$ and $AD'A_2B$. They are congruent and in fact if you perform a $90^{\circ}$ degree rotation around $P$, then $AB'A_1D$ is mapped to  $AD'A_2B$. On the other hand, if you perform a $90^{\circ}$ degree rotation around $Q$, then  $AD'A_2B$ is mapped to $AB'A_1D$. Consequently, they are centers (centroids) $R$ and $S$ are mapped to each other during both rotations. Consequently $SQRP$ is a square. 
Put the complex numbers' zero at $P$, i.e. $P = 0$. Then let $A = z \in \mathbb{C}$ and $D' = w \in \mathbb{C}$. Then $B = i z, \,\, D = -iz$ and $R = \frac{1}{2}(iz + w)$. To find $B'$ simply form $AD' = w-z$ which means that $AB' = -i(w-z)$ and thus $B' = z - i(w-z)$. So you have \begin{align}S &= \frac{1}{2}(D + B') = \frac{1}{2}(-iz + z - i(w-z))\\ &= \frac{1}{2}(-iz + z - iw + iz)
= \frac{1}{2}(z - iw)\\ &= (-i) \, \frac{1}{2}(w+iz)\\ &= (-i) R \end{align} Analogous computation holds for $Q$.
Edit: Ro answer your question directly, $PR = i PS$ is correct. As are the formulas written on your piece of paper. I don't know why you got confused.
If you put the zero of the complex numbers at $A$, then in your notations $B = \alpha$ and $D' = \beta$. Hence
$$P = \frac{1}{2}(\alpha + i \alpha), \,\, R = \frac{1}{2}(\alpha + \beta), \,\,  Q = \frac{1}{2}(\beta - i \beta), \,\, S = \frac{1}{2}(i\alpha - i \beta), \,\,  $$ Then
$$PS = S - P = \frac{1}{2}(i\alpha - i\beta) - \frac{1}{2}(\alpha + i \alpha) = \frac{1}{2}(- \alpha -i\beta)$$
 $$PR = R - P = \frac{1}{2}(\alpha + \beta) - \frac{1}{2}(\alpha + i \alpha) = \frac{1}{2}(- i \alpha + \beta) = i \frac{1}{2}(-\alpha - i\beta) = iPS.$$ Indeed $PR = i PS$ as one would expect. And setting $B' = \beta$ and $i\beta$ will lead to the same conclusion. The computations might be slightly different but the result will be the same.   
