Mastering Mathematics Smartly by Wee Wen Shih

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Argand diagrams & complex loci

It is not difficult to represent any complex number z on an Argand diagram if we understand the approach of finding the modulus and argument of z, as described in this page.

This short summary helps the learner to understand the complex loci, especially the circle and the half-line.

This resource page has sets of notes, exercises and worked solutions as well as animations on Complex Numbers.

  • To learn about the complex loci (circle), click "+" beside "Complex numbers 2: Argand diagram" and select "Notes and examples" for details.

  • To learn about the complex loci (half-line), click "+" beside "Complex numbers 3: Modulus-argument" and select "Notes and examples" for details.

This document contains a short summary on complex loci and features some nice worked examples sourced from 2 books.

Below is a nice question a student asked on SgForums:
Sketch the locus of the point representing complex number z such that |z - 2 - 2i| =1.
Sketch the locus of the point representing complex number w such that |w + 4 - 2i| =2.
Hence find the max and min values of arg (z - w).

The approach to solve is described as follows:
1. Draw the two circles.

2. To find max arg(z - w), we draw a straight line with a positive slope in a way that it is tangent to the two circles. Then use similar triangles and the sine ratio.

3. To find min arg(z - w), we draw a straight line with a negative slope in a way that it is tangent to the two circles. Then use similar triangles and the sine ratio.

Additional comments:
(i) arg(z - w) refers to the angle that a line, joining the points indicated by z and w, makes with respect to the horizontal.

(ii) This diagram shows the situation in step 2.

M and N are centres of respective circles. Triangles ACN and BCM are similar. It is not difficult to see that sin \theta = \frac{1}{2}.

(iii) Use the same method for step 3.

Non-standard equations of loci

What do you do when you face a non-standard equation of a locus? The trick is to perform the appropriate algebraic manipulation to make it back to the standard form.

Let's look at two worked examples.

Always real or imaginary complex numbers

Suppose z = \text{cos}\:\theta + \text{i}\:\text{sin}\:\theta. Then by de Moivre's theorem, z^n = \text{cos}\:n\theta + \text{i}\:\text{sin}\:n\theta.
A typical question (e.g. in the 2008 exam) is to ask when z^n is always real or always imaginary. Let's investigate these further.

For z^n to be always real, \text{sin}\:n\theta = 0.
Therefore n\theta = k\pi for k \in \mathbb{Z}, which is observable from the sine graph (sketch it!).

For z^n to be always imaginary, \text{cos}\:n\theta = 0.
Therefore n\theta = \frac{(2k+1)\pi}{2} for k \in \mathbb{Z}, which is observable from the cosine graph (sketch it!).
Equivalently, we may write n\theta = \frac{(2k-1)\pi}{2} for k \in \mathbb{Z}.
Why the equivalence? Because in either case (2k+1 or 2k-1), a list of odd numbers is produced.