Some loci are lines hidden in complex notation. $|z - a| = |z - b|$ is the perpendicular bisector of the segment from $a$ to $b$. $\arg(z - z_0) = \alpha$ is a ray (an open half-line) leaving $z_0$ at angle $\alpha$. Inequalities in $\operatorname{Re}$ and $\operatorname{Im}$ slice the plane into half-planes with definite slopes and intercepts. This lesson decodes each.
+5 XP warm-up $|z - 2|$ is the distance from $z$ to $2$, and $|z - 4i|$ is the distance from $z$ to $4i$.
$|z - 2|$ is the distance from $z$ to $2$, and $|z - 4i|$ is the distance from $z$ to $4i$. Before checking, describe in words the set of points equidistant from $2$ and $4i$. What classical geometric line is this?
+5 XP to read Two readings dominate every line-locus question: $|z - a| = |z - b|$ is the perpendicular bisector of $a$ and $b$ (equidistant set), and $\arg(z - z_0) = \alpha$ is a ray from $z_0$ at angle $\alpha$, the point $z_0$ itself excluded.
Two readings dominate every line-locus question: $|z - a| = |z - b|$ is the perpendicular bisector of $a$ and $b$ (equidistant set), and $\arg(z - z_0) = \alpha$ is a ray from $z_0$ at angle $\alpha$, the point $z_0$ itself excluded.
The identify-substitute-sketch flow: (1) identify the line type (bisector vs ray vs half-plane); (2) if uncertain, substitute $z = x + iy$ and simplify; (3) sketch with correct slope, intercept and endpoint behaviour.
$|z - a| = |z - b|$ · $\arg(z - z_0) = \alpha$ · $\operatorname{Im}(z) \geq mx + c$
The equation says " is the same distance from as from ".
The equation $|z - a| = |z - b|$ says "$z$ is the same distance from $a$ as from $b$". Classical Euclidean geometry: the set of such $z$ is the perpendicular bisector of the segment from $a$ to $b$.
To find its Cartesian equation, substitute $z = x + iy$, square both sides, and simplify, the modulus terms cancel, leaving a linear equation.
Worked through the hook: $|z - 1| = |z - 5i|$. Substitute $z = x + iy$:
So the locus is a straight line of gradient $\tfrac{1}{5}$ and $y$-intercept $\tfrac{12}{5}$. (Geometrically: midpoint of $1$ and $5i$ is $(\tfrac{1}{2}, \tfrac{5}{2})$; perpendicular to direction $(-1, 5)/\sqrt{26}$.)
$|z - a| = |z - b|$ ⇒ perpendicular bisector of segment from $a$ to $b$ · Substitute $z = x + iy$ and square, moduli cancel; result is linear · Bisector passes through midpoint $\tfrac{a+b}{2}$ · Bisector slope is the negative reciprocal of the $ab$ slope
Pause, copy $|z-a|=|z-b|$ as the perpendicular bisector, the substitution-and-square method, and the bisector's midpoint and slope into your book.
Quick check: The locus of $z$ satisfying $|z - 2| = |z + 2|$ is:
We just saw that $|z-a| = |z-b|$ is the perpendicular bisector of the segment from $a$ to $b$, found by substituting $z = x+iy$ and squaring to get a linear equation.
We just saw that $|z-a| = |z-b|$ is the perpendicular bisector of the segment from $a$ to $b$, found by substituting $z = x+iy$ and squaring to get a linear equation. That raises a question: what locus does a condition on the argument give? This card answers it → $\arg(z-z_0) = \alpha$ is a ray from $z_0$ at angle $\alpha$, with $z_0$ excluded (open endpoint).
$\arg(z - z_0)$ is the angle (from the positive real axis, anticlockwise) of the vector from $z_0$ to $z$. Fixing that angle to be $\alpha$ confines $z$ to a ray a half-line, starting at $z_0$ and extending infinitely in direction $\alpha$.
$\arg(z - z_0) = \alpha$ ⇒ ray from $z_0$ at angle $\alpha$ · $z_0$ excluded (open circle at endpoint) · Slope of the ray is $\tan\alpha$ (when defined) · Ray, NOT line, the opposite direction has $\arg = \alpha - \pi$
Pause, copy $\arg(z-z_0) = \alpha$ as a ray from $z_0$ at angle $\alpha$ (with $z_0$ excluded), the slope $\tan\alpha$, and the warning that the opposite direction has argument $\alpha - \pi$ into your book.
Did you get this? True or false: the locus $\arg(z - 1) = \tfrac{\pi}{2}$ is the vertical half-line $x = 1$, $y > 0$ (with $z = 1$ excluded).
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Find the Cartesian equation of the locus $|z - 3| = |z + i|$, and identify the line geometrically.
Sketch the locus $\arg(z - 2 - i) = \tfrac{3\pi}{4}$, give the Cartesian equation of the line containing it, and state which half-line is the locus.
Sketch the region $\operatorname{Re}(z) + \operatorname{Im}(z) \leq 4$ on the Argand plane. State the boundary line, its slope and intercept, and which side is shaded.
Fill the gap: The locus $\arg(z - 3) = 0$ is the half-line on the real axis with $x$ , with the point $z = 3$ .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the locus $|z - i| = |z + i|$ is the real axis $y = 0$.
Activities · practice with the ideas
Activities · practice with the ideas
Find the Cartesian equation of the locus $|z - 1| = |z - 3|$ and describe it geometrically.
Sketch the ray $\arg(z + 1) = \tfrac{\pi}{6}$. Give the slope of the containing line and state which half is the locus.
Convert $\operatorname{Im}(z) > 2\operatorname{Re}(z) - 1$ to Cartesian form and sketch the half-plane.
Find the locus of $z$ satisfying $|z - 4 - 2i| = |z|$.
Sketch the set of $z$ with $\arg(z) = -\tfrac{\pi}{4}$ AND $|z| \leq 3$. Describe the locus.
Odd one out: Three of these loci are straight lines. Which is NOT a line?
Earlier you analysed $|z - 1| = |z - 5i|$, the set of points equidistant from $1$ and $5i$.
Substituting $z = x + iy$ and squaring gives $(x - 1)^2 + y^2 = x^2 + (y - 5)^2$, which simplifies to $y = \tfrac{x}{5} + \tfrac{12}{5}$, the perpendicular bisector of the segment from $(1, 0)$ to $(0, 5)$. Note the bisector slope $\tfrac{1}{5}$ is the negative reciprocal of the segment slope $-5$, and it passes through the midpoint $(\tfrac{1}{2}, \tfrac{5}{2})$. The three habits to bank: identify the line type from the form, substitute when uncertain, and sanity-check with midpoint + perpendicular slope.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the Cartesian equation of the locus of $z$ satisfying $|z - 2| = |z - 4i|$. (2 marks)
Q2. Sketch the locus $\arg(z - 1 - i) = \tfrac{\pi}{4}$. State the endpoint, the Cartesian equation of the containing line, and which half-line is the locus. (3 marks)
Q3. Sketch the region defined by $|z - 1| \leq |z + 1|$. Identify the boundary line, decide which half-plane is the region, and state whether the boundary is included. (3 marks)
Activity answers:
1. $(x - 1)^2 + y^2 = (x - 3)^2 + y^2 \Rightarrow -2x + 1 = -6x + 9 \Rightarrow x = 2$. The vertical line $x = 2$, perpendicular bisector of $(1, 0)$ and $(3, 0)$.
2. Endpoint at $(-1, 0)$ (open circle). Slope $\tan(\tfrac{\pi}{6}) = \tfrac{1}{\sqrt{3}}$. The locus is the half-line $y = \tfrac{x + 1}{\sqrt{3}}$ with $x > -1$, heading up-and-right.
3. $y > 2x - 1$. Dashed boundary $y = 2x - 1$. Origin satisfies $0 > -1$, so shade the half-plane above the line containing the origin.
4. $(x - 4)^2 + (y - 2)^2 = x^2 + y^2 \Rightarrow -8x + 16 - 4y + 4 = 0 \Rightarrow 8x + 4y = 20 \Rightarrow y = -2x + 5$. The perpendicular bisector of $0$ and $4 + 2i$.
5. Ray from $0$ (open) at angle $-\tfrac{\pi}{4}$, intersected with $|z| \leq 3$. The locus is the line segment from $(0, 0)$ (excluded) to $\left(\tfrac{3}{\sqrt{2}}, -\tfrac{3}{\sqrt{2}}\right)$ (included).
Q1 (2 marks): $z = x + iy$. $(x - 2)^2 + y^2 = x^2 + (y - 4)^2$ [1]. Expand and simplify: $-4x + 4 = -8y + 16 \Rightarrow y = \tfrac{x + 3}{2}$ [1].
Q2 (3 marks): Endpoint $z_0 = 1 + i$ at $(1, 1)$, open circle [1]. Containing line slope $\tan(\tfrac{\pi}{4}) = 1$, through $(1, 1)$: $y = x$ [1]. Locus is the half-line $y = x$ with $x > 1$ (up-and-right direction matches $\alpha = \tfrac{\pi}{4}$) [1].
Q3 (3 marks): Substitute $z = x + iy$, square: $(x - 1)^2 + y^2 \leq (x + 1)^2 + y^2$ [1]. Expand: $-2x + 1 \leq 2x + 1 \Rightarrow x \geq 0$ [1]. Boundary $x = 0$ (imaginary axis), solid (non-strict); region is the closed right half-plane $x \geq 0$ [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →