Loci on the Argand Plane — Basics
A complex number $z = x + iy$ is a point in the plane. So an equation in $z$ is really an equation in $(x, y)$ — and the locus is the set of points that satisfy it. This lesson teaches the four moves that unlock every loci problem in MEX-N1: read the modulus as distance, read the real part as a vertical line, read the imaginary part as a horizontal line, and substitute $z = x + iy$ when in doubt.
Substitute $z = x + iy$ into $|z| = 5$ and simplify. What does the equation become in $x$ and $y$? What shape does it represent on the $(x, y)$ plane? Before checking, also try $|z - 3| = 5$ and describe the change.
Two habits collapse most loci questions to a one-line answer: read the geometry first ($|z - z_0|$ is distance from $z_0$; $\operatorname{Re}$ and $\operatorname{Im}$ project onto axes), then substitute $z = x + iy$ to convert anything stubborn to Cartesian form.
The read-substitute-simplify flow: (1) read the modulus as a distance; (2) substitute $z = x + iy$ where geometry is not immediate; (3) simplify to a Cartesian relation in $x$ and $y$.
$|z - z_0| = r$ · $\operatorname{Re}(z) = a$ · $\operatorname{Im}(z) = b$
Key facts
- $|z - z_0| = r$ is a circle, centre $z_0$, radius $r$
- $|z - z_0| < r$ is the open disk; $|z - z_0| \leq r$ is the closed disk
- $\operatorname{Re}(z) = a$ is the vertical line $x = a$
- $\operatorname{Im}(z) = b$ is the horizontal line $y = b$
Concepts
- Why modulus equals Euclidean distance on the Argand plane
- Why substituting $z = x + iy$ converts a complex equation into a Cartesian one
- Why strict inequalities exclude the boundary (open) while $\leq$ includes it (closed)
Skills
- Sketch loci of the form $|z - z_0| = r$, $|z - z_0| < r$, $|z - z_0| \leq r$
- Sketch loci $\operatorname{Re}(z) = a$ and $\operatorname{Im}(z) = b$
- Translate any of the above to Cartesian form by substituting $z = x + iy$
If $z_0 = a + bi$ and $z = x + iy$ then $z - z_0 = (x - a) + (y - b)i$, so
Squaring both sides of $|z - z_0| = r$ gives the standard circle
- Equality: $(x - a)^2 + (y - b)^2 = r^2$ — the circle of centre $(a, b)$ and radius $r$.
- Strict inequality: $|z - z_0| < r$ gives $(x - a)^2 + (y - b)^2 < r^2$ — interior, boundary excluded.
- Non-strict: $|z - z_0| \leq r$ gives $(x - a)^2 + (y - b)^2 \leq r^2$ — interior plus boundary.
Worked through the hook: $|z - 2 - i| = 3$ means $|z - (2 + i)| = 3$, a circle centre $(2, 1)$, radius $3$. The inequality $|z - 2 - i| \leq 3$ shades the closed disk: solid boundary, shaded interior.
$|z - z_0|$ = distance from $z$ to $z_0$ · $|z - z_0| = r$ ⇒ circle, centre $z_0$, radius $r$ · $|z - z_0| < r$ ⇒ open disk (dashed boundary) · $|z - z_0| \leq r$ ⇒ closed disk (solid boundary) · Always rewrite as $|z - z_0|$ to read centre correctly
Pause — copy $|z-z_0|=r$ (circle, centre $z_0$, radius $r$), the open-disk/closed-disk boundary convention (dashed vs solid), and the rewrite trick to read the centre into your book.
Quick check: The locus of $z$ satisfying $|z - 1 + 2i| = 4$ is:
We just saw that $|z-z_0| = r$ is a circle and $|z-z_0| < r$ an open disk, identified by rewriting in the form $|z-z_0|$ to read the centre directly. That raises a question: what loci arise from conditions on $\operatorname{Re}(z)$ or $\operatorname{Im}(z)$ alone? This card answers it → $\operatorname{Re}(z) = a$ gives the vertical line $x = a$, and inequalities give half-planes; strict inequalities use dashed boundaries.
If $z = x + iy$ then $\operatorname{Re}(z) = x$ and $\operatorname{Im}(z) = y$. So:
- $\operatorname{Re}(z) = a$ ⇔ $x = a$ — a vertical line through $(a, 0)$.
- $\operatorname{Im}(z) = b$ ⇔ $y = b$ — a horizontal line through $(0, b)$.
Inequalities slice the plane into half-planes:
- $\operatorname{Re}(z) > a$ — the open half-plane to the right of $x = a$.
- $\operatorname{Im}(z) \leq b$ — the closed half-plane on or below $y = b$.
$\operatorname{Re}(z) = a$ ⇒ vertical line $x = a$ · $\operatorname{Im}(z) = b$ ⇒ horizontal line $y = b$ · Inequality ⇒ half-plane; strict ⇒ dashed boundary, non-strict ⇒ solid · Re fixes the horizontal coordinate (do not swap)
Pause — copy $\operatorname{Re}(z)=a \Rightarrow x=a$ (vertical line), $\operatorname{Im}(z)=b \Rightarrow y=b$ (horizontal line), the half-plane inequality rule, and the dashed/solid boundary convention into your book.
Did you get this? True or false: the locus of $z$ with $\operatorname{Im}(z) = -3$ is the horizontal line $y = -3$.
Worked examples · 3 in a row, reveal as you go
Sketch the locus of $z$ in the Argand plane satisfying $|z - 3 + 4i| = 5$. State the centre and radius.
Sketch $\{z : |z + 2 - i| < 3\}$ on the Argand plane. Indicate clearly which boundary is included.
Sketch the set of $z$ satisfying both $\operatorname{Re}(z) = 2$ and $|z| \leq 3$. Describe the locus.
Fill the gap: The locus $|z - 4| = 2$ is a circle of centre $($, $)$ and radius .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the locus $|z + 1 + i| \leq 2$ is the closed disk of centre $(-1, -1)$ and radius $2$, drawn with a solid boundary.
Activities · practice with the ideas
Sketch $|z - 2 - 2i| = 2$. State the centre and radius, and write the Cartesian equation.
Sketch the region $|z| < 4$ and shade it. Indicate whether the boundary is included.
Sketch the set of $z$ with $\operatorname{Re}(z) \geq -1$. Identify and shade the half-plane.
Sketch the intersection $\operatorname{Im}(z) > 1$ and $|z - i| \leq 2$.
Substitute $z = x + iy$ into $|z - 1| = |z|$ — and identify the locus by simplifying to a Cartesian equation.
Odd one out: Three of these describe a circle (a curve, not a region). Which one is NOT a circle?
Earlier you sketched $|z - 2 - i| = 3$ and $|z - 2 - i| \leq 3$ from the hook.
Pulling the minus sign out gives $|z - (2 + i)| = 3$ — a circle of centre $(2, 1)$ and radius $3$. The non-strict inequality $\leq 3$ shades the entire closed disk: solid boundary, full interior. The key habit is always rewriting as $|z - z_0|$ before reading coordinates, and matching the boundary style (dashed vs solid) to the inequality. These two checks alone fix the most common mistakes in MEX-N1 sketches.
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. Sketch the locus of $z$ satisfying $|z + 1 - 3i| = 2$ on the Argand plane. State the centre and radius. (2 marks)
Q2. Sketch the region in the Argand plane defined by $\operatorname{Re}(z) > 1$ and $|z| \leq 3$. Be explicit about which boundaries are included. (3 marks)
Q3. By substituting $z = x + iy$, show that the locus $|z - 2i| = |z|$ is the horizontal line $y = 1$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Centre $(2, 2)$, radius $2$. Cartesian form $(x - 2)^2 + (y - 2)^2 = 4$.
2. Open disk centred at origin, radius $4$: $x^2 + y^2 < 16$. Dashed boundary, interior shaded.
3. Half-plane on or to the right of the solid vertical line $x = -1$ — boundary included.
4. Solid circle of centre $(0, 1)$ radius $2$; dashed horizontal line $y = 1$. Shade the part of the closed disk strictly above $y = 1$.
5. $(x - 1)^2 + y^2 = x^2 + y^2 \Rightarrow -2x + 1 = 0 \Rightarrow x = \tfrac{1}{2}$. The locus is the vertical line $x = \tfrac{1}{2}$ — the perpendicular bisector of $0$ and $1$.
Q1 (2 marks): $|z + 1 - 3i| = |z - (-1 + 3i)|$ so centre is $(-1, 3)$, radius $2$ [1]. Sketch a solid circle of radius $2$ around $(-1, 3)$ [1].
Q2 (3 marks): $\operatorname{Re}(z) > 1$ is the open half-plane right of $x = 1$ — dashed boundary [1]. $|z| \leq 3$ is the closed disk $x^2 + y^2 \leq 9$ — solid boundary [1]. The required region is the intersection: portion of the closed disk strictly right of $x = 1$ [1].
Q3 (3 marks): Let $z = x + iy$. $|z - 2i|^2 = x^2 + (y - 2)^2$ and $|z|^2 = x^2 + y^2$ [1]. Equating: $x^2 + (y - 2)^2 = x^2 + y^2 \Rightarrow (y - 2)^2 = y^2 \Rightarrow -4y + 4 = 0$ [1]. So $y = 1$, the horizontal line $y = 1$ (which is the perpendicular bisector of $0$ and $2i$) [1].
Five timed questions on circles, disks and lines in the Argand plane. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick locus questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.