A calculus of the absurd

14.2.2 Loci on the Argand diagram

These are the kind of questions which certain examiners love to pose. There are two ways to go about this; geometrically or algebraically. Ordinarily I would always pick the option with lots of algebra over the option with lots of geometry. Unfortunately, in this case the option with lots of algebra also includes pretty much the same amount of geometry.

In essence, problems about loci on the Argand diagram are about finding some complex numbers which satisfy a set of constraints. There are a few very important facts to know, with which most of these questions can be tackled.

The key fact to remember is that complex numbers can be treated as vectors. For example, for any \(z \in \mathbb {C}\) we know (or at least our intuitive notion of length lines up with) that the “length” of \(z\) is equal to \(\abs {z}\). We also know that for two vectors \(a\) and \(b\), the vector between \(a\) and \(b\) is given by \(-a + b\), and thus the distance between \(a\) and \(b\) can be written as \(\abs {a-b}\).

  • Example 14.2.1 Sketch the complete set of points which satisfy

    \begin{equation} \abs {z - 9i} = \abs {z - 12} \end{equation}

We can apply the trick of reading this aloud. We know that \(\abs {z - 9i}\) gives us the distance between \(z\) and the complex number \(9i\). Furthermore, \(2\abs {z - 12}\) means the distance between \(z\) and \(12\), so overall what this means is “we want \(z\) such that the distance between \(z\) and \(-9i\) is the same as the distance between \(z\) and \(12\)”. If we sketch this, this will be the perpendicular bisector of the line between \(12\) and \(9i\) (through the middle of the line).

more examples