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.
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?
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: $\arg(z) \in (\alpha, \beta)$
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
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)
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
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.
- 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.
$|z - z_0| \leq r$: 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:
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 $z_0$.
$\arg(z) = \theta$ 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).
Worked examples · 3 in a row, reveal as you go
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.
Sketch the region $\{z : |z| < 3 \text{ and } \mathrm{Im}(z) \geq 1\}$. Describe the shape.
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.
Fill the gap: The region $|z - z_0| < r$ is an disk — boundary — drawn with a dashed circle.
Misconceptions to fix · the 3 traps that cost marks
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).
Activities · practice with the ideas
Sketch the region $\{z : |z + 1| \leq 3\}$. State centre, radius, and whether the boundary is included.
Sketch the half-plane $\mathrm{Re}(z) < -2$. State the boundary line and indicate solid/dashed.
Sketch the region $\{z : -\pi/4 \leq \arg(z) \leq \pi/4\}$ (excluding the origin).
Sketch the intersection $\{z : 1 < |z| \leq 2\}$ — an annulus. State which boundary circle is dashed and which is solid.
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?
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.
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 region $\{z : |z - 2i| \leq 1\}$ and state whether the boundary is included. (2 marks)
Q2. Sketch the intersection $\{z : |z| \leq 4 \text{ and } \mathrm{Im}(z) > 1\}$. State the corner points of the boundary. (3 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)
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].
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 arenaClimb 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.