Suppose that ABCD is a quadrilateral (labelled anticlockwise) with its vertices being represented by complex number a,b,c,d respectively.
The conditions for ABCD to be a square are (a-b)=(d-c) and i(a-c)=(b-d) [i.e. edges are parallel and diagonals intersect at right angle].
Why is the second condition not sufficient? surely it is only met if the two diagonals are perpendicular and have equal length; and ABCD is therefore a square (without showing that the first condition is true)?
As mentioned by the comment, it could be a kite.
An explicit example, let $a=2,b=i,c=-2,d=-3i,$
then we have
$$i(a-c)=4i=b-d$$
but
$$a-b=2-i$$ and
$$d-c=-3i-2 \neq a-b$$