\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
\(ABCD\) is a parallelogram if and only if \(AB\) is parallel and equal to \(DC\). This is true if and only if, \[ \bvect {AB} = \bvect {DC}, \] and using complex representation (which is also equivalent) \[ b - a = c - d. \] This is equivalent to \[ a + c = b + d \] so we are done.
In this case, \(ABCD\) is further a square if and only if it is both a rhombus and a rectangle. It is a rhombus if and only if the two diagonals, \(AC\) and \(BD\), are perpendicular to each other, and a rectangle if and only if the two diagonals, \(AC\) and \(BD\), have equal length.
This is equivalent to \(\bvect {BD}\) being \(\bvect {AC}\) rotated \(90\) degrees anti-clockwise exactly (due to the labelling as defined), and using complex representation (which is equivalent) \[ i(c - a) = (d - b). \]
Flipping the signs on both sides (which is reversible) gives \[ i(a - c) = (b - d) \] as desired.
\(X\) is the centre of the square constructed externally along the edge \(PQ\) if and only if \(\bvect {PX}\) is \(\bvect {PQ}\) rotated clockwise by \(45\) degrees and scaled down by a factor of \(\sqrt {2}\). In complex notation, this is equivalent to \[ x - p = (q - p) \cdot \frac {1}{\sqrt {2}} \cdot e^{-i\frac {\pi }{4}}. \]
But notice that \(e^{-i\frac {\pi }{4}} = \cos \frac {\pi }{4} - i\sin \frac {\pi }{4} = \frac {1}{\sqrt {2}}(1 - i)\), and hence this equation is equivalent to \[ x = \frac {1}{2} (q - p) (1 - i) + p = \frac {(1 + i)p + (1 - i)q}{2}, \] as desired.
Similarly, we have \begin {align*} y & = \frac {(1 + i)q + (1 - i)r}{2}, \\ z & = \frac {(1 + i)r + (1 - i)s}{2}, \\ t & = \frac {(1 + i)s + (1 - i)t}{2}. \end {align*}
\(XYZT\) is a square, if and only if \[ x + z = y + t \] and \[ i(x - z) = y - t. \]
For the first one, this is equivalent to \[ (1 + i)p + (1 - i)q + (1 + i)r + (1 - i)s = (1 - i)p + (1 + i)q + (1 - i)r + (1 + i)s, \] which is equivalent to \[ p + r = q + s, \] which is equivalent to \(PQRS\) being a parallelogram.
For the second one, this is equivalent to \[ i \cdot \left ((1 + i)p + (1 - i)q - (1 + i)r - (1 - i)s\right ) = -(1 - i)p + (1 + i)q + (1 - i)r - (1 + i)s, \] which is equivalent to \[ -(1 + i)p + (1 + i)q + (1 - i) r -(1 + i)s = -(1 - i)p + (1 + i)q + (1 - i)r - (1 + i)s, \] which is trivially true.
This shows that \(XYZT\) being square is equivalent to \(PQRS\) being a parallelogram as desired.