A quadrilateral drawn in the complex plane has vertices $A,B,C,D$, labelled anticlockwise. These vertices are represented, respectively, by the complex numbers a, b, c, and d. Show that ABCD is a square if and only if $i(a-c) = (b-d)$.
Let PQRS be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than 180. Squares with centres $X, Y, Z, T$ are constructed externally to the quadrilateral on the sides $PQ, QR, RS, SP$ respectively.
(1) If $P$ and $Q$ are represented by the complex numbers $p$ and $q$, respectively, show that $X$ can be represented by $$\frac 1 2(p(1+i)+q(1-i))$$ (2) Show that XYZT is a square if and only if $PQRS$ is a parallelogram.
I got stuck from the first part above.
I know for a square, $\left|AB\right| =\left|BC\right|$ and they are 90 angles to each other.
So given they are in complex plane : $c-b = i(b-a)$ or $c-b = i(a-b)$
I am confused with which direction to take.
After this, how do i get to $i(a-c) = (b-d)$ and go the other way? Since the question is 'iff'?
As for Part (1) and (2) I do not even know where to start..
Many thanks in advance.