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.
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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).
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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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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...
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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.
$|z - z_0| \leq r$: 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.
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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
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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.
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Disks, half-planes, and how to shade them
The recipe is the same in every case: draw the boundary (using the equality), choose solid vs dashed based on strictness, then...
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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.
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).
Exterior of a disk. $|z - z_0| > r$, shade everything outside the circle, with a dashed boundary.
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:
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Sectors: inequalities on $\arg(z)$
We just saw that is a closed disk (solid boundary) and the exterior (dashed boundary), with the test-a-point rule confirming...
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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).
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Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
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.
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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.
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Cartesian form of the boundary: $(x - 1)^2 + (y - 1)^2 = 4$. Real-axis crossings ($y = 0$): $(x - 1)^2 + 1 = 4 \Rightarrow x = 1 \pm \sqrt{3}$.
Axis crossings give two anchor points to plot, making the sketch accurate. Setting $y = 0$ finds real-axis crossings.
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Imaginary-axis crossings ($x = 0$): $1 + (y - 1)^2 = 4 \Rightarrow y = 1 \pm \sqrt{3}$. Shade the closed interior (including boundary). Test the centre $(1, 1)$: $|0| \leq 2$ ✓, so the centre is in the region, confirming shade interior.
A test point on the inside confirms which side to shade. The centre always satisfies any disk inequality with $\leq$ or $<$.
PROBLEM 2 · INTERSECTION OF TWO REGIONS
Sketch the region $\{z : |z| < 3 \text{ and } \mathrm{Im}(z) \geq 1\}$. Describe the shape.
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$|z| < 3$: 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.
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Where does the line $y = 1$ meet the circle $x^2 + y^2 = 9$? $x^2 + 1 = 9 \Rightarrow x = \pm 2\sqrt{2}$. Intersection corners at $(2\sqrt{2}, 1)$ and $(-2\sqrt{2}, 1)$.
Find the corner points, they bound the overlap. These corner points are on the line $y = 1$ (included) but on the circle boundary (excluded), so they are not in the region.
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Intersection: a circular segment ("cap") above the chord $y = 1$, inside the open disk. The chord $y = 1$ is solid (included). The arc above is dashed (excluded). Result: a closed-on-bottom, open-on-top cap.
When combining a closed and an open region, each boundary keeps its original style. A common mark-loser is drawing both boundaries the same way.
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.
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$0 < \arg(z) < \pi/3$: 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.
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$|z| \leq 2$: closed disk of radius 2 about the origin, solid boundary circle.
A modulus inequality is independent of argument, it just caps the distance from the origin.
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Intersection: a circular sector (pie slice) of radius $2$ and angle $\pi/3$, with apex at the origin (excluded), bounded by two dashed rays and a solid arc. The slice spans from the positive real direction to direction $\pi/3$.
A sector $\cap$ disk gives a pie slice. Mark the boundary style on each edge separately, the curved arc and the two straight rays may differ.
Fill the gap: The region $|z - z_0| < r$ is an disk, boundary drawn with a dashed circle.
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Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
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).
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Revisit your thinking
Activities · practice with the ideas
Activities · practice with the ideas
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Sketch the region $\{z : |z + 1| \leq 3\}$. State centre, radius, and whether the boundary is included.
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Sketch the half-plane $\mathrm{Re}(z) < -2$. State the boundary line and indicate solid/dashed.
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Sketch the region $\{z : -\pi/4 \leq \arg(z) \leq \pi/4\}$ (excluding the origin).
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Sketch the intersection $\{z : 1 < |z| \leq 2\}$, an annulus. State which boundary circle is dashed and which is solid.
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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?
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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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Show what you have learned
Multiple choice, then short answer under exam conditions.
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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.
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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)
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].
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Consolidate and move on
Sit the module quiz, then close the lesson off.
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Take the full module quiz
quiz
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.
