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Module 12 · L14 of 16 ~40 min ⚡ +90 XP available

Loci — Regions and Inequalities

An equality in the complex plane traces a curve; an inequality fills a region. Replacing $|z - z_0| = r$ with $|z - z_0| \leq r$ turns a circle into a closed disk; $\mathrm{Re}(z) > a$ shades a half-plane; $\arg(z) \in (\alpha, \beta)$ carves out a sector. This lesson teaches you to sketch, shade and combine these regions cleanly.

Today's hook — Sketch the region $\{z : |z| \leq 2 \text{ and } \mathrm{Re}(z) \geq 0\}$. Before drawing, decide: (a) is the circular boundary solid or dashed? (b) which half-plane do you keep? (c) what shape is the final overlap? Compare your answer after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

In a normal Cartesian sketch, $x > 2$ shades the half-plane to the right of the line $x = 2$, with the line drawn dashed (not included). Before checking — write the analogue for the complex plane: which inequality on $z$ shades the same region, and is the boundary line solid or dashed?

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02
The two moves for shading a region
+5 XP to read

Every region problem rewards two habits: first draw the boundary curve as if it were an equality, then test a point (often the origin) to decide which side of the boundary to shade. Strict inequalities ($<$, $>$) use a dashed boundary; non-strict ($\leq$, $\geq$) use a solid one.

The boundary-then-test reading: (1) draw the boundary from the equality, (2) decide solid vs dashed from $\leq$ vs $<$, (3) pick a test point and substitute to find which side satisfies the inequality.

Disk: $|z - z_0| \leq r$  ·  Half-plane: $\mathrm{Re}(z) > a$  ·  Sector: $\arg(z) \in (\alpha, \beta)$

Draw boundary Style solid / dashed Test pt shade side Combine: intersect (AND) or union (OR)
$|z - z_0| \leq r$: closed disk  ·  $|z - z_0| < r$: open disk (dashed boundary)
Disk (interior of a circle)
$|z - z_0| \leq r$ is the closed disk: all points at distance up to $r$ from $z_0$, including the circle itself. Use a solid boundary. $|z - z_0| < r$ excludes the boundary — draw it dashed.
Half-plane
$\mathrm{Re}(z) > a$ shades the half-plane $x > a$; $\mathrm{Im}(z) \geq b$ shades $y \geq b$. The boundary is a vertical or horizontal line; the strictness of the inequality controls solid vs dashed.
Sector by argument
$\alpha < \arg(z) < \beta$ shades the wedge between the rays $\arg(z) = \alpha$ and $\arg(z) = \beta$. The origin is the apex; strict inequality means the rays are dashed and the origin is excluded.
03
What you'll master
Know

Key facts

  • $|z - z_0| \leq r$ is a closed disk; $|z - z_0| < r$ is an open disk
  • $\mathrm{Re}(z) > a$ is the half-plane $x > a$; $\mathrm{Im}(z) > b$ is $y > b$
  • $\alpha < \arg(z) < \beta$ is the sector between rays $\arg(z) = \alpha$ and $\arg(z) = \beta$
  • Strict inequalities give dashed boundaries; non-strict give solid boundaries
Understand

Concepts

  • Why an inequality in a distance fills the interior (or exterior) of a circle
  • Why $\mathrm{Re}(z) = a$ is a vertical line, not a horizontal one
  • Why $\arg(z)$ inequalities require excluding the origin (where $\arg$ is undefined)
Can do

Skills

  • Sketch disks, half-planes, and sectors with correct solid/dashed boundaries
  • Combine two or more inequalities by intersection (AND) and identify the overlap
  • Choose a test point to decide which side of a boundary to shade
04
Key terms
Closed disk$\{z : |z - z_0| \leq r\}$ — the interior of the circle plus the boundary itself. Drawn with a solid boundary.
Open disk$\{z : |z - z_0| < r\}$ — the interior of the circle only; the boundary is excluded. Drawn with a dashed boundary.
Half-planeA region defined by $\mathrm{Re}(z) > a$, $\mathrm{Re}(z) < a$, $\mathrm{Im}(z) > b$ or $\mathrm{Im}(z) < b$. The boundary is a vertical or horizontal line.
SectorA region bounded by two rays from the origin, defined by an inequality on $\arg(z)$. Example: $\arg(z) \in (\alpha, \beta)$.
$\arg(z)$The argument of $z$: the angle from the positive real axis to the vector from $0$ to $z$, measured anticlockwise. Undefined when $z = 0$.
Intersection (AND)The region where two conditions both hold. Shade only points lying in BOTH regions; the result is the overlap.
MEX-N1NESA outcome (Complex Numbers I): represent loci in the complex plane, including regions defined by inequalities on modulus, real or imaginary parts, and argument.
05
Disks, half-planes, and how to shade them
core concept

The recipe is the same in every case: draw the boundary (using the equality), choose solid vs dashed based on strictness, then test a single point to pick the side.

  1. Disk. $|z - z_0| \leq r$ — closed disk centred at $z_0$ with radius $r$. Solid circle, shade the interior. $|z - z_0| < r$ — open disk (dashed circle).
  2. Exterior of a disk. $|z - z_0| > r$ — shade everything outside the circle, with a dashed boundary.
  3. Half-plane. $\mathrm{Re}(z) \geq a$ — vertical line $x = a$ (solid), shade $x \geq a$. $\mathrm{Im}(z) < b$ — horizontal line $y = b$ (dashed), shade $y < b$.

Worked through the hook: $\{z : |z| \leq 2 \text{ and } \mathrm{Re}(z) \geq 0\}$.

  • $|z| \leq 2$: closed disk of radius 2 centred at the origin (solid boundary).
  • $\mathrm{Re}(z) \geq 0$: closed half-plane $x \geq 0$ (solid boundary on the imaginary axis).
  • Intersection: the right half of the closed disk — a closed semicircular region.
Connecting to vector intuition. $|z - z_0|$ is a distance; inequalities in a distance fill the "near" or "far" region from $z_0$. $\mathrm{Re}(z)$ is a coordinate; inequalities in a coordinate fill a strip or half-plane.

$|z - z_0| \leq r$: closed disk (solid boundary) · $|z - z_0| > r$: exterior of circle (dashed boundary) · $\mathrm{Re}(z) > a$: half-plane $x > a$ (dashed vertical line) · Always test the origin (or another easy point) to confirm which side to shade

Pause — copy the disk/exterior boundary conventions (solid for $\leq$, dashed for $>$, strict), the half-plane rules, and the test-a-point strategy into your book.

Quick check: The region $\{z : |z - 2i| < 3\}$ in the Argand plane is:

06
Sectors: inequalities on $\arg(z)$
core concept

We just saw that $|z-z_0| \leq r$ is a closed disk (solid boundary) and $|z-z_0| > r$ the exterior (dashed boundary), with the test-a-point rule confirming which side to shade. That raises a question: how do argument inequalities determine a region? This card answers it → $\alpha < \arg(z) < \beta$ is a sector with apex at the origin; $\arg(z-z_0) = \theta$ translates the apex to $z_0$.

The argument $\arg(z)$ measures the angle from the positive real axis to the ray from $0$ to $z$, anticlockwise positive. An inequality $\alpha < \arg(z) < \beta$ shades the wedge of points whose argument lies between $\alpha$ and $\beta$:

  • The two bounding rays $\arg(z) = \alpha$ and $\arg(z) = \beta$ start at the origin.
  • Strict $<$: rays drawn dashed and the origin is excluded ($\arg(0)$ is undefined).
  • Non-strict $\leq$: rays drawn solid (but the origin still has no argument, so technically still excluded).

Translating $\arg(z - z_0)$. Instead of an angle at the origin, $\arg(z - z_0)$ is the angle at $z_0$. So $\arg(z - z_0) = \theta$ is a ray starting at $z_0$ (not the origin), and an inequality between two such conditions gives a sector with apex at $z_0$.

$$\alpha < \arg(z - z_0) < \beta \;\Rightarrow\; \text{sector with apex at } z_0$$
Common mistake. Treating $\arg(z) = \theta$ as a full straight line is wrong — it is only the ray from the origin in direction $\theta$. The opposite ray has argument $\theta + \pi$ (or $\theta - \pi$), which is a different value of $\arg$.

$\arg(z) = \theta$ is a RAY from the origin, not a full line · $\alpha < \arg(z) < \beta$ is a sector with apex at the origin · $\arg(z - z_0) = \theta$ is a ray from $z_0$ — apex translates with the subtraction · The origin ($z = 0$) is always excluded from $\arg$ inequalities

Pause — copy $\arg(z)=\theta$ as a ray from the origin, $\alpha < \arg(z) < \beta$ as a sector, the translation rule for $\arg(z-z_0)$, and that $z=0$ is always excluded from argument conditions into your book.

Did you get this? True or false: the region $\{z : 0 < \arg(z) < \pi/2\}$ is the open first quadrant of the Argand plane (excluding the axes and the origin).

PROBLEM 1 · CLOSED DISK

Sketch the region $\{z : |z - 1 - i| \leq 2\}$ in the Argand plane. State whether the boundary is included, and identify any axis intercepts of the boundary circle.

1
Rewrite as $|z - (1 + i)| \leq 2$. Centre $z_0 = 1 + i$, radius $r = 2$. Inequality is $\leq$, so the boundary is included (solid circle).
Always pull the modulus into $|z - z_0|$ form and read off centre, radius, and boundary type before sketching.
PROBLEM 2 · INTERSECTION OF TWO REGIONS

Sketch the region $\{z : |z| < 3 \text{ and } \mathrm{Im}(z) \geq 1\}$. Describe the shape.

1
$|z| < 3$: open disk, centre origin, radius $3$, dashed boundary. $\mathrm{Im}(z) \geq 1$: closed half-plane $y \geq 1$, solid horizontal boundary at $y = 1$.
Sketch the two regions separately first; only afterwards overlay them. Mixing boundary styles is the most common slip.
PROBLEM 3 · SECTOR INSIDE A DISK

Sketch the region $\{z : 0 < \arg(z) < \pi/3 \text{ and } |z| \leq 2\}$. Describe the shape and state which boundaries are solid versus dashed.

1
$0 < \arg(z) < \pi/3$: open sector with apex at the origin, between the positive real axis and the ray at $60$°. Both rays dashed; origin excluded.
The argument inequality is strict, so the bounding rays are dashed. Always exclude the origin from any argument condition.

Fill the gap: The region $|z - z_0| < r$ is an disk — boundary — drawn with a dashed circle.

Trap 01
Solid vs dashed boundary confusion
$\leq$ and $\geq$ give SOLID boundaries (included); $<$ and $>$ give DASHED boundaries (excluded). When combining regions, each boundary keeps its own style — do not unify them. Markers explicitly check this.
Trap 02
Treating $\arg(z) = \theta$ as a line
It is a RAY from the origin in direction $\theta$, not a full line. The opposite direction has argument $\theta + \pi$ — a different argument value. Sectors are wedges, not strips.
Trap 03
Forgetting to exclude the origin in $\arg$ inequalities
$\arg(0)$ is undefined, so any inequality involving $\arg(z)$ implicitly excludes $z = 0$. When sketching, leave a small open circle at the apex even if the bounding rays are solid.

Did you get this? True or false: the region $\{z : \mathrm{Re}(z) \geq 0 \text{ and } \mathrm{Im}(z) \geq 0\}$ is the closed first quadrant (both axes included).

Work mode · how are you completing this lesson?
1

Sketch the region $\{z : |z + 1| \leq 3\}$. State centre, radius, and whether the boundary is included.

2

Sketch the half-plane $\mathrm{Re}(z) < -2$. State the boundary line and indicate solid/dashed.

3

Sketch the region $\{z : -\pi/4 \leq \arg(z) \leq \pi/4\}$ (excluding the origin).

4

Sketch the intersection $\{z : 1 < |z| \leq 2\}$ — an annulus. State which boundary circle is dashed and which is solid.

5

Sketch $\{z : |z - 2| < |z + 2|\}$. (Hint: rewrite as a comparison of distances to two fixed points.)

Odd one out: Three of these regions are half-planes. Which one is NOT?

11
Revisit your thinking

Earlier you predicted what the region $\{z : |z| \leq 2 \text{ and } \mathrm{Re}(z) \geq 0\}$ looks like.

It is the closed right-half of the closed disk of radius 2 centred at the origin — a closed semicircular region. Both boundaries (the semicircular arc and the imaginary-axis diameter) are solid because both inequalities are non-strict. The recipe to internalise: boundary first, strictness second, test point third, intersection last. Apply this template and any combination of disks, half-planes, and sectors becomes routine.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 32 marks

Q1. Sketch the region $\{z : |z - 2i| \leq 1\}$ and state whether the boundary is included. (2 marks)

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ApplyBand 43 marks

Q2. Sketch the intersection $\{z : |z| \leq 4 \text{ and } \mathrm{Im}(z) > 1\}$. State the corner points of the boundary. (3 marks)

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AnalyseBand 53 marks

Q3. Sketch the region $\{z : 1 \leq |z| \leq 2 \text{ and } 0 \leq \arg(z) \leq \pi/2\}$. Describe the shape and indicate solid/dashed for each boundary. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Rewrite $|z + 1| = |z - (-1)| \leq 3$. Centre $(-1, 0)$, radius $3$. Closed disk, solid boundary.

2. Boundary $x = -2$ (vertical line), dashed (strict $<$). Shade everything left of the line.

3. Closed sector symmetric about the positive real axis between $\arg(z) = -\pi/4$ and $\arg(z) = \pi/4$. Both rays solid; origin excluded.

4. Annulus between $|z| = 1$ and $|z| = 2$. Inner circle dashed (strict $1 <$); outer circle solid ($\leq 2$).

5. $|z - 2| < |z + 2|$: $z$ closer to $2$ than to $-2$. Boundary is the perpendicular bisector — the imaginary axis $x = 0$ (dashed). Shade the half-plane $\mathrm{Re}(z) > 0$.

Q1 (2 marks): Centre $(0, 2)$, radius $1$, closed boundary (solid) [1]. Correct sketch showing disk touching $y$-axis between $y = 1$ and $y = 3$ [1].

Q2 (3 marks): Identify both regions [1]. Corner points $(\pm\sqrt{15}, 1)$ [1]. Sketch with solid arc and dashed chord [1].

Q3 (3 marks): Annular sector in first quadrant [1]. All four boundary edges solid because all inequalities non-strict [1]. Two arcs and two ray segments, apex region near origin excluded because $\arg(0)$ undefined [1].

01
Boss battle · The Region Shader
earn bronze · silver · gold

Five timed questions on disks, half-planes, sectors, and combined regions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering quick region-shading questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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