Theorems in complex analysis are stated on regions: "analytic on the disk", "continuous on the closed annulus". The modulus turns most of these sets into geometry, because $|z - a|$ is the distance from $z$ to $a$. This lesson is a short dictionary from inequalities to pictures, plus the words open, closed, bounded and connected.
Key idea
$|z - a| < r$ is the open disk with center $a$ and radius $r$; $|z - a| \le r$ is the closed disk; $|z - a| = r$ is the circle; $r_1 < |z - a| < r_2$ is an annulus; $\operatorname{Re} z > c$ and $\operatorname{Im} z < c$ are half-planes; $|z - a| = |z - b|$ is the perpendicular bisector of the segment from $a$ to $b$.
A set is open if each of its points has a small disk around it inside the set, closed if it contains its boundary, bounded if it fits inside some disk, and connected if any two of its points can be joined by a path inside it. An open connected set is called a region (or domain).
Why it works
With $z = x + iy$ and $a = a_1 + ia_2$, $|z - a| = \sqrt{(x - a_1)^2 + (y - a_2)^2}$: the distance formula. So "$|z - a| < r$" says "$z$ is less than $r$ away from $a$". Strict inequalities between continuous quantities give open sets (wiggling a point a little keeps the inequality true), while $\le$ and $=$ keep the boundary and give closed sets.
Worked examples
Worked example: Reading a disk
Describe $|z + 1 - 2i| \le 3$.
- Rewrite the inside as $z - a$: $z + 1 - 2i = z - (-1 + 2i)$.
- So the set is all $z$ at distance at most $3$ from $-1 + 2i$: the closed disk with center $-1 + 2i$ and radius $3$.
- It is closed (it includes the circle), bounded and connected, but not open.
Answer: closed disk, center $-1 + 2i$, radius $3$
Worked example: Equal distances
Describe $|z - 1| = |z + i|$.
- The points equally far from $1$ and from $-i$ form the perpendicular bisector of the segment joining them.
- Algebra: square both sides, $(x - 1)^2 + y^2 = x^2 + (y + 1)^2$.
- Expand and cancel $x^2$ and $y^2$: $-2x + 1 = 2y + 1$, so $y = -x$.
Answer: the line $y = -x$
Worked example: A circle in disguise
Describe $|z| = 2|z - 3|$.
- Square: $x^2 + y^2 = 4\big((x - 3)^2 + y^2\big)$, so $3x^2 - 24x + 36 + 3y^2 = 0$.
- Divide by $3$ and complete the square: $(x - 4)^2 + y^2 = 4$.
- A circle with center $4$ and radius $2$. Check: $z = 2$ gives $|2| = 2$ and $2|2 - 3| = 2$.
Answer: the circle $|z - 4| = 2$
Common mistake
Reading the sign of the center backwards. $|z - 2i| < 3$ is the disk centered at $2i$, not at $-2i$: the expression is the distance from $z$ to $2i$. And $|z + 2i| < 3$ means $|z - (-2i)| < 3$, centered at $-2i$. To check, plug in your claimed center: it should make the inside $0$.