# Ask Uncle Colin: What shape is this?

Dear Uncle Colin,

The question says that $z_1$, $z_2$, $z_3$ and $z_4$ are distinct complex numbers representing the vertices of a quadrilateral ABCD, in order. Further, $z_1 - z_4 = z_2 - z_3$ and $\arg\left( \frac{z_4 - z_1}{z_2 - z_1}\right) = \piby 4$. The shape is supposed to be one of a rectangle, a square, a rhombus and a trapezium; the answer given is rectangle, but I think it could also be a square. What do you think?

- Quite Unhappy About Definitions

Hi, QUAD, and thanks for your message!

The short answer is, I think a square *is* a rectangle (and a rhombus and a trapezium, for that matter). In fact, all possible scenarios also form a trapezium (UK definition).

### A quick run through

If $z_1 - z_4 = z_2 - z_3$, then we know that $\vec{DA} = \vec{CB}$, so the shape has two disjoint sides that are parallel and the same length. We’re looking at something that’s definitely a parallelogram. All parallelograms are trapeziums.

The argument restriction says that angle $D\hat AB$ is a right angle, so we have a parallelogram that includes a right angle. By virtue of what we know about parallelograms (opposite angles are equal, adjacent angles sum to two right angles), that means that all four of its angles are right angles. All right-angled parallelograms are rectangles.

We don’t know anything at all about $\vec{AB}$ or $\vec{CD}$ (other than that we’ve worked out they’re the same length) - so they *could* be a square (and all squares are rhombuses), but it’s not *necessary*.

### Summary

All of the possible shapes are rectangles (and all rectangles are trapeziums).

It is not impossible for the shape to be a square (and all squares are rhombuses).

The question ought to be a bit more specific, but a rectangle is the most restricted shape that satisfies the conditions and I agree with the given answer (or else you could just write “quadrilateral” and claim that what you’ve written is perfectly true.)

Hope that helps!

- Uncle Colin