Four squares are drawn of the edges of a parallelogram ABCD as shown in the diagram. However you change the parallelogram ABCD the quadrilateral formed by joining the centres of the squares is always itself a square. Iain of Madras College used vectors to prove that this quadrilateral is a rhombus and you might like to try to do this yourself. He went on to prove that the angles are all 90 degrees making it a square.

The following proof uses complex numbers (which are, when all is said and done, the same as vectors in two dimensions). The advantage of this method is that it is shorter and there is not so much to write down. You do need to use the result however that the complex number iz represents a vector at right angles to z. The reason for this is that when you multiply any complex number z by the complex number i you get iz which has the same modulus as z but an argument 90 degrees more than the argument of z .

Take the origin O at the centre of the parallelogram and represent the midpoints of the edges of the parallelogram by complex numbers u , - u , v and - v as shown in the diagram.

The midpoint of AB is represented by the complex number - u , and, if we call this point M, then the line segment MB corresponds to the complex number v and the line segment MP corresponds to the complex number iv . Hence P is represented by the complex number - u + iv .

Similarly Q is v + iu , R is u - iv and S is - v - iu . Hence SR is

 z = ( u - iv ) - (- v - iu ) = (1 + i ) u + (1 - i ) v

and PQ is

 z 1 = ( v + iu ) - (- u + iv ) = (1 + i ) u + (1 - i ) v = z .

Hence SR and PQ are equal in length and parallel.

SP is

 w = (- u + iv ) - (- v - iu ) = ( i - 1) u + (1 + i ) v

and RQ is

 w 1 = ( v + iu ) - ( u - iv ) = ( i - 1) u + (1 + i ) v = w .

Hence SP and RQ are equal in length and parallel. It follows that iz = w which shows that SP is perpendicular to, and the same length as, SR and hence that SPQR is a square.