Problem source

A quadrilateral drawn in the complex plane has vertices $A$, $B$, $C$ and $D$, labelled anticlockwise.  These vertices are represented, respectively, by the complex numbers $a$, $b$, $c$ and $d$.  Show that $ABCD$ is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if $a+c =b+d\,$.  Show further that, in this case, $ABCD$ is a square if and only if ${\rm i}(a-c)=b-d$.
  Let $PQRS$ be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less  than $180^\circ$.  Squares with centres $X$, $Y$, $Z$ and $T$ are constructed externally to the quadrilateral on the sides $PQ$, $QR$, $RS$ and $SP$, respectively.
  \begin{questionparts}
  \item If $P$ and $Q$ are represented by the complex numbers $p$ and $q$, respectively, show that $X$ can be represented by
    \[
    \tfrac 12 \big( p(1+{\rm i} ) + q (1-{\rm i})\big) \,.
    \]
  \item Show that $XY\!ZT$ is a square if and only if $PQRS$ is a parallelogram.
  \end{questionparts}

Solution source

The vector representing the side $AB$ is $b - a$ and the vector representing the side $DC$ is $c - d$. $ABCD$ is a parallelogram if and only if these opposite sides are parallel and equal in length, which is given by $b - a = c - d$, or equivalently $a + c = b + d$.

Similarly, if $a + c = b + d$, then $c - b = d - a$, so the side $BC$ is parallel and equal in length to the side $AD$. Thus, $a + c = b + d$ is the necessary and sufficient condition for $ABCD$ to be a parallelogram.

In a parallelogram, the shape is a square if and only if the diagonals are equal in length and perpendicular to each other. The diagonals are represented by the vectors $c - a$ and $d - b$. For these to be equal in length and perpendicular, one must be a $90^\circ$ rotation of the other. Since $A, B, C, D$ are labeled anticlockwise, a $90^\circ$ anticlockwise rotation of the vector $\vec{AC}$ (which is $c-a$) would point in the direction of $\vec{DB}$ (which is $b-d$ if we consider the relative orientation). Specifically:
$i(c - a) = d - b \implies -i(a - c) = d - b \implies i(a - c) = b - d$.
Thus, $ABCD$ is a square if and only if $i(a - c) = b - d$.

\begin{questionparts}
\item The midpoint of the side $PQ$ is $\frac{1}{2}(p + q)$. To find the centre $X$ of the square built externally on $PQ$, we start at the midpoint and move a distance equal to half the side length in a direction perpendicular to $PQ$. Since $P, Q, R, S$ are anticlockwise, the outward direction is a $90^\circ$ clockwise rotation of the vector $\vec{PQ}$. A clockwise rotation of $90^\circ$ corresponds to multiplication by $-i$.
\[
x = \frac{p+q}{2} + (-i)\left(\frac{q-p}{2}\right) = \frac{p + q - iq + ip}{2} = \frac{1}{2} \big( p(1+i) + q(1-i) \big)
\]

\item From part (i), we have the representations for the centres:
\begin{align*}
x &= \tfrac{1}{2}(p(1+i) + q(1-i)) \\
y &= \tfrac{1}{2}(q(1+i) + r(1-i)) \\
z &= \tfrac{1}{2}(r(1+i) + s(1-i)) \\
t &= \tfrac{1}{2}(s(1+i) + p(1-i))
\end{align*}

As shown in the first part of the problem, $XYZT$ is a square if and only if:
(1) $x+z = y+t$ (it is a parallelogram)
(2) $i(x-z) = y-t$ (it is a square)

First, examine condition (1):
\begin{align*}
x+z - (y+t) &= \tfrac{1}{2} \big[ (p+r)(1+i) + (q+s)(1-i) - (q+s)(1+i) - (r+p)(1-i) \big] \\
&= \tfrac{1}{2} \big[ (p+r)(1+i - (1-i)) - (q+s)(1+i - (1-i)) \big] \\
&= \tfrac{1}{2} \big[ (p+r)(2i) - (q+s)(2i) \big] \\
&= i(p+r - (q+s))
\end{align*}
Thus, $x+z = y+t$ if and only if $p+r = q+s$, which is the condition that $PQRS$ is a parallelogram.

Next, examine condition (2):
\begin{align*}
i(x-z) &= \tfrac{1}{2} i \big[ p(1+i) + q(1-i) - r(1+i) - s(1-i) \big] \\
&= \tfrac{1}{2} \big[ p(i-1) + q(i+1) - r(i-1) - s(i+1) \big] \\
y-t &= \tfrac{1}{2} \big[ q(1+i) + r(1-i) - s(1+i) - p(1-i) \big] \\
\text{So, } i(x-z) - (y-t) &= \tfrac{1}{2} \big[ p(i-1 + 1-i) + q(i+1 - 1-i) + r(-i+1 - 1+i) + s(-i-1 + 1+i) \big] \\
&= 0
\end{align*}
Since $i(x-z) = y-t$ is an identity (always true for any $PQRS$), $XYZT$ is a square if and only if it is a parallelogram. As established above, this occurs if and only if $PQRS$ is a parallelogram.
\end{questionparts}