The Scenario
Imagine you're using GPS navigation. Two cell towers are at fixed locations on a map. You want to find:
- All points that are exactly 5 km from Tower A
- All points that are equidistant from Tower A and Tower B
On an Argand diagram, where complex numbers are represented as points, these sets of points are called loci. Each locus corresponds to a simple condition on $z$.
What Is a Locus?
A locus (plural: loci) is simply the set of all points $z$ in the Argand diagram that satisfy a given condition. Today we'll study two of the most important loci:
- The condition $|z - a| = k$ — all points at distance $k$ from the fixed point $a$ → this is a circle
- The condition $|z - a| = |z - b|$ — all points equidistant from two fixed points $a$ and $b$ → this is the perpendicular bisector of the segment joining $a$ and $b$
💡 Think About It (click to reveal)
Remember: For complex numbers $z = x + y\mathrm{i}$ and $a = p + q\mathrm{i}$, $|z - a| = \sqrt{(x-p)^2 + (y-q)^2}$, which is exactly the Euclidean distance between the points $(x,y)$ and $(p,q)$ in the plane. So complex number modulus conditions directly translate to geometric distance conditions!
A locus is a set of points satisfying a condition. Using the geometric interpretation of $|z - a|$ as the distance between points $z$ and $a$, we can immediately identify two important loci.
If $a$ is a fixed complex number (representing a fixed point in the Argand diagram) and $k > 0$ is a real constant, then the equation:
$$|z - a| = k$$represents all points $z$ at distance $k$ from $a$. This is a circle with:
- Centre at the point representing $a$
- Radius = $k$
Inequality forms:
- $|z - a| < k$ — interior of the circle (all points inside)
- $|z - a| > k$ — exterior of the circle (all points outside)
- $|z - a| \leqslant k$ — circle plus its interior (closed disk)
If $a$ and $b$ are two fixed complex numbers (two fixed points), then the equation:
$$|z - a| = |z - b|$$represents all points $z$ equidistant from $a$ and $b$. This is the perpendicular bisector of the line segment joining the points $a$ and $b$.
Inequality form:
- $|z - a| \leqslant |z - b|$ — the half-plane containing $a$ (all points closer to $a$ than to $b$)
To find the Cartesian equation (in terms of $x$ and $y$ where $z = x + y\mathrm{i}$):
For $|z - a| = k$: Let $a = p + q\mathrm{i}$. Then:
$$\sqrt{(x-p)^2 + (y-q)^2} = k \quad\Longrightarrow\quad (x-p)^2 + (y-q)^2 = k^2$$For $|z - a| = |z - b|$: Let $a = p_1 + q_1\mathrm{i}$, $b = p_2 + q_2\mathrm{i}$. Square both sides:
$$(x-p_1)^2 + (y-q_1)^2 = (x-p_2)^2 + (y-q_2)^2$$Expand and simplify → the $x^2$ and $y^2$ terms cancel, leaving a linear equation in $x$ and $y$ (which confirms it's a straight line).
In your group, you will receive eight clue cards. Each card gives a condition on $z$. For each card, decide: is it a circle or a perpendicular bisector? Then find the center/radius or the Cartesian equation.
Clue A: $|z - 2 - 3\mathrm{i}| = 4$
Clue B: $|z + 2\mathrm{i}| = |z - 4|$
Clue C: $|z - (1 + \mathrm{i})| = |z - (3 - 5\mathrm{i})|$
Clue D: $|z + 4 - 2\mathrm{i}| < 3$
Clue E: $|z - 5| = 5$
Clue F: $|z| = |z - 4 - 6\mathrm{i}|$
Clue G: $|z - 3 - 4\mathrm{i}| \leqslant 5$
Clue H: $|z + 2 + 3\mathrm{i}| \geqslant |z - 1 - \mathrm{i}|$
👁 Hint
For perpendicular bisector: square both sides to remove the modulus signs. The $x^2$ and $y^2$ terms will cancel out, leaving a linear equation.
Check: the line must pass through the midpoint of AB. You can use this to verify your equation.
👁 Check Answers
Circles:
A: Centre $(2, 3)$, radius $4$
D: Centre $(-4, 2)$, radius $3$, interior only
E: Centre $(5, 0)$, radius $5$
G: Centre $(3, 4)$, radius $5$, closed disk
Perpendicular bisectors:
B: $y = 2x - 3$
C: $x - 3y = 6$
F: $4x + 6y = 26$ or $2x + 3y = 13$
H: $3x + 4y + 2 \geqslant 0$
Work through these sketching problems independently. For each one, label the centre and radius (for circles) or the key points (for perpendicular bisectors), and shade the region if an inequality is given.
👁 Hint for Problem 3
Let $a = 4 + 0\mathrm{i}$ and $b = 0 + 2\mathrm{i}$. The midpoint is at $(2, 1)$. The gradient of AB is $\frac{2-0}{0-4} = -\frac{1}{2}$, so the perpendicular bisector has gradient equal to the negative reciprocal $m = 2$.
Use point-gradient form: $y - 1 = 2(x - 2)$.
👁 Check Your Sketch
Problem 1-2: Circle with centre at $(3, 2)$, radius 4. The inequality $|z - 3 - 2\mathrm{i}| \leqslant 4$ shades the interior including the boundary.
Problem 3: Equation: $y = 2x - 3$. Crosses x-axis at $(\frac{3}{2}, 0)$, crosses y-axis at $(0, -3)$.
Problem 4: $|z - 4| \geqslant |z - 2\mathrm{i}|$ means points closer to $(0, 2)$ than to $(4, 0)$, which is the half-plane on the side of the line containing $(0, 2)$.
(i) On an Argand diagram, show the region $R$ for which $|z - 5 + 4\mathrm{i}| \leqslant 3$.
(ii) Find the greatest and least values of $|z + 3 - 2\mathrm{i}|$ in the region $R$.
👁 Reveal Solution
$|z - 5 + 4\mathrm{i}| = |z - (5 - 4\mathrm{i})| \leqslant 3$.
Centre $C$ at $(5, -4)$, radius $r = 3$.
The region $R$ is the circle including its interior.
We want $|z + 3 - 2\mathrm{i}| = |z - (-3 + 2\mathrm{i})|$. This is the distance from $z$ to the fixed point $P(-3, 2)$.
$C: (5, -4)$, $P: (-3, 2)$.
$$CP = \sqrt{(5 - (-3))^2 + (-4 - 2)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$Greatest value = $CP + r = 10 + 3 = 13$
Least value = $CP - r = 10 - 3 = 7$
$$\boxed{\text{Greatest} = 13, \quad \text{Least} = 7}$$The maximum distance occurs at the point on the circle in the direction away from P along the line CP.
The minimum distance occurs at the point on the circle towards P along the line CP.
Since $CP = 10 > r = 3$, $P$ is outside the circle, so the minimum distance is positive.
Draw an Argand diagram showing the set of points $z$ for which $|z| = 2$.
👁 Show Solution
$|z| = |z - 0| = 2$.
Circle with centre at the origin $(0, 0)$, radius $2$.
Draw an Argand diagram showing the set of points $z$ for which $|z - 4| \leqslant 3$.
👁 Show Solution
$|z - 4| = |z - (4 + 0\mathrm{i})| \leqslant 3$.
Circle with centre at $(4, 0)$, radius $3$. Shade the interior including the boundary.
Draw an Argand diagram showing the set of points $z$ for which $|z - 5\mathrm{i}| = 6$.
👁 Show Solution
$|z - 5\mathrm{i}| = |z - (0 + 5\mathrm{i})| = 6$.
Circle with centre at $(0, 5)$, radius $6$.
Draw an Argand diagram showing the set of points $z$ for which $|z + 3 - 4\mathrm{i}| < 5$.
👁 Show Solution
$|z + 3 - 4\mathrm{i}| = |z - (-3 + 4\mathrm{i})| < 5$.
Circle with centre at $(-3, 4)$, radius $5$. Shade the interior (boundary not included).
Draw an Argand diagram showing the set of points $z$ for which $|6 - \mathrm{i} - z| \geqslant 2$.
👁 Show Solution
$|6 - \mathrm{i} - z| = |z - (6 - \mathrm{i})|$.
Circle with centre at $(6, -1)$, radius $2$. The inequality $\geqslant 2$ means the exterior including the boundary.
Draw an Argand diagram showing the set of points $z$ for which $2 \leqslant |z - 1 + \mathrm{i}| \leqslant 3$.
👁 Show Solution
$|z - 1 + \mathrm{i}| = |z - (1 - \mathrm{i})|$.
Annulus (ring-shaped region) between two concentric circles:
Centre at $(1, -1)$, inner radius $2$, outer radius $3$. Shade the region between them (including both boundaries).
Draw an Argand diagram showing the set of points $z$ for which $\operatorname{Re}(z) = -2$.
👁 Show Solution
If $z = x + y\mathrm{i}$, then $\operatorname{Re}(z) = x = -2$.
This is the vertical straight line $x = -2$, parallel to the imaginary (y) axis, crossing the real (x) axis at $(-2, 0)$.
(Note: This is another simple locus type, though not our main topic today.)
Draw an Argand diagram showing the set of points $z$ for which $|z| = |z - 4|$.
👁 Show Solution
$|z| = |z - 4|$ means points equidistant from $(0, 0)$ and $(4, 0)$.
Perpendicular bisector of the segment from $(0, 0)$ to $(4, 0)$ is the vertical line $x = 2$.
Draw an Argand diagram showing the set of points $z$ for which $|z| \geqslant |z - 2\mathrm{i}|$.
👁 Show Solution
$|z| \geqslant |z - 2\mathrm{i}|$ means points closer or equal distance to $(0, 2)$ compared to $(0, 0)$.
The locus $|z| = |z - 2\mathrm{i}|$ is the horizontal line $y = 1$ (perpendicular bisector of segment from $(0, 0)$ to $(0, 2)$).
The inequality $|z| \geqslant |z - 2\mathrm{i}|$ corresponds to $y \geqslant 1$, the half-plane above the line (containing the point $(0, 2)$).
Draw an Argand diagram showing the set of points $z$ for which $|z + 1 - \mathrm{i}| = |z - 1 + \mathrm{i}|$.
👁 Show Solution
Points equidistant from $a = -1 + \mathrm{i}$ and $b = 1 - \mathrm{i}$.
$A(-1, 1)$, $B(1, -1)$. Midpoint is $(0, 0)$. Gradient of AB is $\frac{-1-1}{1-(-1)} = \frac{-2}{2} = -1$.
Perpendicular gradient is negative reciprocal $m = 1$. Line through origin: $y = x$.
Draw an Argand diagram showing the set of points $z$ for which $|z + 5 + 7\mathrm{i}| \leqslant |z - 2 - 6\mathrm{i}|$.
👁 Show Solution
$A(-5, -7)$, $B(2, 6)$. We want all points closer to A than to B: $|z - a| \leqslant |z - b|$.
Midpoint M: $x = \frac{-5+2}{2} = -\frac{3}{2} = -1.5$, $y = \frac{-7+6}{2} = -\frac{1}{2} = -0.5$.
Gradient AB: $\frac{6 - (-7)}{2 - (-5)} = \frac{13}{7}$.
Perpendicular gradient: $m = -\frac{7}{13}$.
Equation: $y + \frac{1}{2} = -\frac{7}{13}(x + \frac{3}{2})$.
Multiply through by 26: $26y + 13 = -14(x + \frac{3}{2}) = -14x - 21$ → $14x + 26y + 34 = 0$ → $7x + 13y + 17 = 0$.
The region is the half-plane containing A, which is one side of this line.
Exit Ticket
Check the learning objectives you feel confident about: