Loci in the complex plane — method guide

What is a locus problem?

A locus is the set of all points z = x + iy satisfying a given condition. Every locus problem follows the same pipeline — identify the form, substitute z = x + iy, and simplify to a recognisable Cartesian equation.

Universal method

Read the condition carefully

Identify which standard form it resembles: |z − a| = r |z − a| = |z − b| arg(z − a) = θ — or a combination.

Write z = x + iy, identify fixed points

Let z = x + iy. Write each fixed complex number in the form a + bi. Form z − a explicitly as (x − a) + i(y − b).

Compute the modulus or argument

For |z − a|: compute √((x−a)² + (y−b)²). For equations with two moduli, square both sides before expanding — this avoids square-root algebra entirely.

Expand and collect

Expand brackets, collect x² and y² terms, and complete the square where needed. Aim for a recognisable standard form.

State the geometric object

Name the locus clearly — centre and radius for circles, gradient and intercept for lines, starting point and angle for rays. A sketch earns marks and catches sign errors.

Check any restrictions

Half-lines exclude the starting point. Inequalities give regions, not curves. Verify one or two test points satisfy the original condition.

Form |z − a| = r

Locus

Circle with centre at the point corresponding to a and radius r.

Read off centre and radius immediately

If already in the form |z − (a + bi)| = r, the locus is a circle with centre (a, b), radius r. No algebra needed.

When the form is disguised — substitute z = x + iy

Write as √((x−a)² + (y−b)²) = r, then square both sides:

(x - a)² + (y - b)² = r²

Complete the square if the original was in expanded form.

Special case: |z| = r

|z| = r gives x² + y² = r² — a circle centred at the origin.

Scholarship tip: if the question gives |z − a| = k|z − b| with k ≠ 1, this is an Apollonius circle — see Unusual cases.

Form |z − a| = |z − b|

This is the perpendicular bisector of the segment joining a and b. Handle it algebraically to be safe.

Square both sides

(x-a₁)² + (y-a₂)² = (x-b₁)² + (y-b₂)²

Expand and cancel x² and y²

The squared terms always cancel, leaving a linear equation — the perpendicular bisector.

Write in y = mx + c form

Confirm it is perpendicular to the segment joining the two fixed points (gradients multiply to −1).

Lines in general

Sources

Lines arise from equal-distance conditions, from Re(z) = c or Im(z) = c, or from argument conditions where θ = 0 or π.

Re(z) = c and Im(z) = c

Re(z) = c means x = c — a vertical line. Im(z) = c means y = c — a horizontal line. These are immediate with no substitution needed.

Perpendicular bisector — systematic approach

Given |z − z₁| = |z − z₂| with z₁ = a + bi, z₂ = c + di:

Square both sides
Expand fully
Cancel x² and y²
Collect to get 2(c−a)x + 2(d−b)y = c²+d²−a²−b²

Lines from argument conditions

arg(z − a) = 0 gives the ray in the positive real direction from a. arg(z − a) = π gives the opposite ray. These are half-lines — see the Half-lines tab.

Watch signs: |z − (3 + 2i)| = |z + 1 − i| — rewrite the right side as |z − (−1 + i)| before identifying the two fixed points.

Form arg(z − a) = θ

Locus

A half-line (ray) starting from the point corresponding to a, directed at angle θ anticlockwise from the positive real axis. The starting point is excluded.

Identify the starting point and angle

Rewrite a = p + qi. The ray starts at (p, q) and goes in direction θ.

Write the Cartesian equation

Let X = x − p, Y = y − q. Then arg(X + iY) = θ means:

Y/X = tan θ \to y - q = tan θ \cdot (x - p)

A straight line through (p, q) with gradient tan θ.

Apply the half-line restriction

The direction θ determines which half of the line is the locus:

θ	Direction	Restriction
0	Positive x-direction	x > p
π/4	Upper right, 45°	x > p
π/2	Straight up	y > q
3π/4	Upper left, 135°	x < p
π	Negative x-direction	x < p
−π/4	Lower right, −45°	x > p
−π/2	Straight down	y < q

Mark the open circle at the starting point

The point (p, q) is never included — mark it with an open circle on your sketch. This is required for full marks.

Special case: arg(z) = θ — starting point is the origin, still excluded.

Form arg((z − a)/(z − b)) = θ

The point z sees the segment from a to b at angle θ. By the inscribed angle theorem this gives an arc of a circle.

Use the tan subtraction formula

Let α = arg(z − a), β = arg(z − b). Then:

tan(α - β) = (tan α - tan β)/(1 + tan α \cdot tan β) = tan θ

Substitute tan α = (y−a₂)/(x−a₁) and tan β = (y−b₂)/(x−b₁), then simplify to circle form.

General approach

Substitute z = x + iy everywhere. Separate real and imaginary parts. Square moduli before expanding. Reduce to a known curve.

Case 1 — inequality instead of equation

★

|z − a| < r, |z − a| > r, or combinations

Solve as if an equation to find the boundary curve, then test a point to determine which region satisfies the inequality. Shade the correct region; use dashed boundary for strict inequalities.

Students often find the circle but fail to indicate the region. The locus is the region, not just the boundary.

Case 2 — |z − a| = k|z − b|, k ≠ 1

★

Apollonius circle

Square both sides:

(x-a₁)² + (y-a₂)² = k²[(x-b₁)² + (y-b₂)²]

Expand, group, complete the square. This is a circle whose centre will not generally be at either fixed point — don't assume it will be.

Case 3 — Re or Im of a transformed expression

★

e.g. Re(1/z) = c, Im(z²) = c

Substitute z = x + iy and simplify first:

1/z = (x - iy)/(x² + y²) Re(1/z) = x/(x²+y²) = c \to x² + y² - (1/c)x = 0

A circle through the origin. Always check whether the origin itself is included.

Case 4 — z̄ (conjugate) appears

★

Replace z̄ = x − iy and proceed

Substitute z = x + iy and z̄ = x − iy. Separate real and imaginary parts. The locus is usually a line or circle — the conjugate encodes reflection symmetry about the real axis.

Case 5 — arg condition giving complementary rays

★

arg(z − a) = θ and arg(z − a) = θ + π are different conditions

Together they form a full line, but separately each is a half-line only. The range of arg is (−π, π], so a single condition never gives both rays. State this explicitly if asked for the full line.

Case 6 — z = a excluded from domain

★

Division by (z − a) or |z − a| in the denominator

Even if z = a appears to lie on the locus after simplification, it must be excluded — the expression is undefined there. Mark it as an open circle (hole) on your sketch and note it in your answer.

Case 7 — degenerate cases

★

No solution, a single point, or the whole plane

No solution: |z − a| = −1 — modulus is never negative. State: "No such z exists."

Single point: |z − a| = 0 gives z = a only.

Whole plane: |z − a|² ≥ 0 — always true. State: "Every z satisfies this."

Scholarship markers look for awareness of degenerate cases. Do not skip them.

Before writing your final answer

Algebra

☐ Squared the modulus before expanding?
☐ Cancelled x² and y² correctly?
☐ Completed the square accurately?
☐ Signs on the centre correct?
☐ Radius is positive?
☐ Checked for degenerate cases?

Geometry

☐ Named the locus type?
☐ Centre and radius stated for circles?
☐ Gradient and intercept for lines?
☐ Open circle at start of any ray?
☐ Correct region shaded for inequalities?
☐ Verified a test point satisfies the original?

Standard forms at a glance

Condition	Locus	Key steps
\|z − a\| = r	Circle, centre a, radius r	Read off directly
\|z − a\| = \|z − b\|	Perpendicular bisector	Square, expand, cancel x² y²
\|z − a\| = k\|z − b\|	Apollonius circle	Square, expand, complete square
arg(z − a) = θ	Half-line from a, angle θ	y − q = tanθ·(x − p), restrict
arg((z−a)/(z−b)) = θ	Arc of a circle	Tan subtraction formula
Re(z) = c	Vertical line x = c	Direct
Im(z) = c	Horizontal line y = c	Direct
\|z − a\| < r	Interior of circle	Find circle, test point, shade
Re(1/z) = c	Circle through origin	Multiply by z̄/\|z\|², simplify

The most common Scholarship error: correct algebra, but no clear geometric statement. The locus description is the answer — the algebra is the route.