Basic subsets of the complex plane

[RuiAce]

Remember to register here for FREE to ask any questions you may come across in your QCE studies!

The syllabus expects that you can identify sets that represent circles and straight lines on the Argand plane. We go through as many basic examples as necessary here.

Many problems involving subsets on the Argand plane can be deduced using only intuition. In general, where possible, going down this road is a time saver. The first few problems can all be handled intuitively.

Circles of a given radius: \( \{ z\in \mathbb{C}: |z-z_0| = r \} \) where \(z_0 = x_0+iy_0\) is a fixed complex number.

Recall that \(|z|\) (i.e. the modulus of \(z\) itself) represents the distance from \(z\) to the origin, which represents the complex number \(0+0i\). In general, \(|z-z_0|\) can be thought of as having two representations:
- Naively, it is the distance from the point representing \(z-z_0\) to the origin. This interpretation is correct, but just not the most useful here.
- The more useful interpretation is that it represents the distance between the points representing \(z\), and \(z_0\). This is because the vector drawn from \(z_0\) to \(z\) can also be used to represent the complex number \(z-z_0\).

Using the latter interpretation, if we fix \( |z-z_0| = r\), we require all points at a distance of \(r\), from the point \(z_0\). The set of all complex numbers satisfying this property will thus be a circle, whose radius is \(r\), and centred at \(z_0\) (represented by the point \( (x_0,y_0)\) on the Argand plane).

Perpendicular bisectors: \( \{z\in\mathbb{C}: |z-z_1| = |z-z_2| \} \) where \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\) are fixed complex numbers, and \(z_1\neq z_2\).

Similar to the above, we can treat \(|z-z_1|\) as the distance from \(z\) to \(z_1\), and \(|z-z_2|\) as the distance from \(z\) to \(z_2\) in the complex plane. But this time, instead of setting things equal to a constant, we require the distance from \(z\) to \(z_1\) AND \(z_2\) be the same.

This means we require all the points equidistant to \(z_1\) and \(z_2\). The set of points satisfying this will be the perpendicular bisector of the line segment joining \( (x_1, y_1) \) and \((x_2, y_2)\). To construct this:
- Indicate the line segment joining \( (x_1, y_1)\) and \((x_2,y_2)\). (I typically do this with dashed lines.)
- Plot the midpoint of this line segment.
- Draw the line perpendicular to the line segment, through the midpoint.

Rays (a.k.a. half-lines): \( \{ z\in \mathbb{C}: \operatorname{Arg}(z-z_0) = \theta \} \) where \(z_0 = x_0+iy_0\) is a fixed complex number.

Recall that \( \arg(z) \) (i.e. the argument of \(z\)) represents the (directed) angle formed between the line segment joining \(z\) and \(0+0i\), and the positive \(x\)-axis. The principal argument \( \operatorname{Arg}(z)\) represents the unique angle in the interval \( (-\pi, \pi]\).

\( \operatorname{Arg}(z-z_0)\) has a similar interpretation as for the modulus \( |z-z_0|\). Again, a more useful interpretation is derived from considering that \(z-z_0\) is represented by the vector from \(z_0\) to \(z\). Using this interpretation, \( \operatorname{Arg}(z-z_0)\) reflects the (directed) angle formed between
- The line segment between \(z\) and \(z_0\), as above, and
- A horizontal starting at \(z_0\) in the right-direction, like the positive \(x\)-axis.

Hence if we fix \( \operatorname{Arg}(z-z_0) = \theta\), we require all complex numbers \(z\) that are inclined at this angle \(\theta\), starting at \(z_0\). These points together form a half-line, since they can only go in one direction from the starting point \(z_0\). However, they can't go backwards, as that would give a different value of \(\theta\) instead!

Note, however, that the point \(z_0\) is excluded. This is because the argument of \(0+0i\) is undefined. Which is a problem, since if we sub \(z=z_0\), we obtain \(\operatorname{Arg}(z_0 - z_0)\). Hence a discontinuity (hole in graph) is placed at \(z_0\).

Now, in terms of algebraic approaches, they will typically work whenever the argument is not involved. This is more or less a consequence of that the argument isn't easy to work with algebraically. The first two examples above can be dealt with algebraically if you prefer - give it a shot! (But you really don't need to.)

Here are some common cases where algebraic methods are superior. In general, we always substitute in \(z=x+iy\) for algebraic approaches.

Classic straight lines: \( \{ z\in \mathbb{C}: a\operatorname{Re}(z)+ b\operatorname{Im}(z) + c = 0 \} \).

Substituting \(z=x+iy\) immediately turns the condition to
\[ ax + by+ c = 0 .\]
This is basically the equation of a line that you'd surely be bored of seeing by now. But at least you got there very easily! (Note: Substituting in \(z=x+iy\) is often a good idea when the real and imaginary parts show up, since they can be replaced with \(x\) and \(y\) immediately.)

Circles by rescaling of distances: \( \{ z\in \mathbb{C}: |z-z_1| = k|z-z_2|\text{ where }k > 0, \, k\neq 1\} \). Again, \(z_1 = x_1+iy_1\) and \(z_2=x_2+iy_2\) are fixed, and we assume \(z_1\neq z_2\).

Usually, identifying that this is a circle is pretty hard intuitively. So instead I present a concrete example to convince you that it works. Here, I take \(k=2\), \(z_1 = 1\) and \(z_2 = i\), which gives the condition
\[ |z-1| = 2|z-i|. \]
Now we manipulate by using the definition of the modulus and squaring. Let \(z=x+iy\).
\begin{align*}
 |z-1|^2 &= 4|z-i|^2\\
 |(x-1)+iy|^2 &= 4| x + i(y-1)|^2\\
 (x-1)^2 + y^2 &= 4\left[ x^2 + (y-1)^2\right]\\
 x^2-2x+1+y^2 &= 4x^2 + 4y^2 - 8y + 4\\
0 &= 3x^2 + 2x + 3y^2 - 8y + 3\\
0 &= x^2 + \frac23x + y^2 - \frac83y + 1
\end{align*}
Then we continue by completing the square. Note that we have to complete the square twice here!
\begin{align*}
\left( x^2 + \frac23 x + \frac19\right) + \left( y^2 - \frac83y + \frac{16}9\right) &= \frac19 + \frac{16}9 - 1\\
\left( x +\frac13\right)^2 + \left( y-\frac43\right)^2 &= \frac89
\end{align*}
So we see that we have a circle of radius \(r=\frac{2\sqrt2}{3}\), with centre \( \left( -\frac13, \frac43\right)\). But this was not something I could've told you immediately (I.e. without all that algebra)!

Note: If \(k=1\), this actually is just a perpendicular bisector. Also, more generally this set takes the form. \( \{ z\in \mathbb{C}: |az+b| = |cz+d| \}\). When in this form, unless \(a = c\), I'd usually recommend algebraic approaches anyhow