Complex Equations and Loci in the Complex Plane

An equation or inequality in a complex variable z describes a set of points in the complex plane (a locus): |z − z₀| = r is a circle, |z − p| = |z − q| is a perpendicular bisector and arg(z − z₀) = α is a ray. In brief: |z − 2 + i| = 3 is the circle with center 2 − i and radius 3, and |z − 1| < |z + 1| is the half-plane Re z > 0. To identify a locus, write z = x + iy and simplify, or interpret |z − a| as a distance. Recall that |z − w| is the distance between the points z and w in the complex plane.

Problems about loci translate geometry into complex numbers: the distance between two points becomes the modulus of a difference, and an angle becomes an argument. This guide lists the standard loci (circle, line, ellipse, hyperbola, ray, arc), shows how to find equations of loci by two methods, solves equations in complex numbers and gives a table that connects the equation with the geometric figure.

What are the basic loci?

• |z − z₀| = r: the circle with center z₀ and radius r.
• |z − p| = |z − q|: all points equally far from p and q, the perpendicular bisector of the segment pq.
• Re z = c and Im z = c: a vertical and a horizontal line.
• Re(az) = c for a complex a: a line perpendicular to the direction of ā.

Example 1: a circle

|z − 3 + 2i| = 4 means |z − (3 − 2i)| = 4. The circle has center (3, −2) and radius 4. In coordinates, (x − 3)² + (y + 2)² = 16.

Example 2: a perpendicular bisector

|z − 1| = |z − i|: the points equally far from 1 and i. Let z = x + iy: (x − 1)² + y² = x² + (y − 1)², so −2x + 1 = −2y + 1 and y = x. The line y = x is the perpendicular bisector of the segment from (1, 0) to (0, 1).

How do you identify a locus?

1. Geometric method: interpret each modulus as a distance and recognise a known figure.
2. Algebraic method: put z = x + iy, replace |…|² by sums of squares, and simplify to an equation in x and y.
3. Check with a test point.

Example 3: not obviously a circle

|z − 1| = 2|z + 2|. Square both sides: (x − 1)² + y² = 4[(x + 2)² + y²], so x² − 2x + 1 + y² = 4x² + 16x + 16 + 4y². Collect: 3x² + 18x + 3y² + 15 = 0, i.e., x² + 6x + y² + 5 = 0, so (x + 3)² + y² = 4. This is a circle with center −3 and radius 2, a circle of Apollonius (points whose distances from two fixed points have a fixed ratio ≠ 1). Check: z = −1 is on it: |−2| = 2 and 2|1| = 2 ✓.

Example 4: a line through Re

Re((1 + i)z) = 2. With z = x + iy: (1 + i)(x + iy) = (x − y) + (x + y)i. The real part is x − y = 2, a line with slope 1.

Example 5: z + 1/z real

For which z ≠ 0 is z + 1/z real? z + z̄/|z|² has the imaginary part y − y/|z|² = y(1 − 1/|z|²). This is 0 if y = 0 (the real axis) or |z| = 1 (the unit circle). So the locus is the real axis (without 0) together with the unit circle.

Which curves arise from sums and differences of distances?

• |z − p| + |z − q| = 2a (with 2a > |p − q|): an ellipse with foci p and q.
• ||z − p| − |z − q|| = 2a (with 2a < |p − q|): a hyperbola with foci p and q.
• |z − p| = Re z + c (the distance to a point equals the distance to a line, up to a shift): a parabola.

Example 6

|z − 2| + |z + 2| = 6. The foci are ±2 and the sum of distances is 6, so a = 3, c = 2 and b² = 9 − 4 = 5. The ellipse: x²/9 + y²/5 = 1. The point 3 on the real axis satisfies |1| + |5| = 6 ✓.

Example 7

|z − 3| − |z + 3| = 4 (a branch of a hyperbola): the foci are ±3, 2a = 4, a = 2, b² = 9 − 4 = 5: x²/4 − y²/5 = 1. The sign selects the left branch (the points closer to −3).

What about loci defined by arguments?

• arg(z − z₀) = α: a ray from z₀ (without the point z₀) in the direction α.
• arg((z − p)/(z − q)) = α: an arc of a circle through p and q, from which the segment pq is seen at the angle α. For α = π/2, it is a semicircle on pq as the diameter.
• arg((z − p)/(z − q)) = 0 or π: the line through p and q (outside or inside the segment).

Example 8

arg(z − 1 − i) = π/4: the ray from 1 + i making the angle 45° with the positive real axis, the points 1 + i + t(1 + i) for t > 0. For example 2 + 2i.

Example 9

Re((z − 1)/(z + 1)) = 0 with z ≠ ±1: the quotient is purely imaginary, which happens when (z − 1) ⟂ (z + 1), i.e., the angle at z in the triangle with vertices −1, 1 and z is 90°. So z lies on the circle with diameter from −1 to 1, which is |z| = 1.

How do you describe regions?

• |z − z₀| < r: the open disc. |z − z₀| ≥ r: the outside, including the circle.
• r₁ ≤ |z| ≤ r₂: an annulus (a ring).
• |z − p| < |z − q|: the half-plane containing p, bounded by the perpendicular bisector.
• α < arg z < β: a sector (a wedge).
• Re z > 0: the right half-plane.

Example 10

1 ≤ |z − 2i| ≤ 3 and Im z ≥ 2: the upper half of an annulus centered at 2i with the radii 1 and 3.

Example 11

Maximum and minimum of |z| if |z − 3 − 4i| ≤ 2. The disc has the center 3 + 4i at distance 5 from 0 and radius 2. The nearest point to the origin is at distance 5 − 2 = 3 and the farthest at distance 5 + 2 = 7. So 3 ≤ |z| ≤ 7.

Table of loci

In the last row, z is closer to 1 than to −1, so it lies in the half-plane x > 0. A test point confirms this: z = 2 gives |1| < |3|.

Where does this topic come from?

The use of complex numbers for geometric problems goes back to Wessel, Argand and Gauss, and to the later work of Möbius and Riemann. The description of curves by equations in z, with the modulus and the argument, became part of the standard approach to geometry problems in the complex plane, and the idea of studying a figure by its defining distance property goes back to Apollonius and the classical definition of conics.

Where is it used today?

In control theory, regions of the complex plane (such as the left half-plane Re s < 0) describe stable systems. In electrical engineering, impedance loci are drawn as circles and lines in the complex plane. In geometry competitions, complex numbers give short solutions of problems about triangles and circles, and in fluid dynamics, the complex plane is used to describe potential flow around objects.

Which mistakes are common?

Forgetting the modulus is a distance

|z − a| = r is not “z − a = r”. It describes all points at distance r, a circle, not a single point.

Squaring incorrectly

Square the moduli: |z − 1|² = (x − 1)² + y², which has no cross terms with i.

Missing the excluded points

For loci with arg((z − p)/(z − q)), the points p and q are excluded.

Mixing up the sides of an inequality

|z − p| < |z − q| describes the half-plane on the side of p. Test a point such as p itself.