Complex loci and geometry | Year 12

01 / LEARN

Understand the idea.

Distances define circles and bisectors

Read $∣ z - a ∣$ as the distance from the fixed point $a$ . The locus $∣ z - a ∣ = r$ , for $r > 0$ , is a circle. A strict inequality gives its interior without the boundary; a non-strict inequality includes the boundary.

For distinct fixed points $a$ and $b$ , the equality $∣ z - a ∣ = ∣ z - b ∣$ describes their perpendicular bisector. In Cartesian algebra, squaring both distances cancels the $x^{2}$ and $y^{2}$ terms, leaving a straight-line equation.

Equal-scale Argand plane with the circle |z-(1+i)|=2, a marked centre labelled 1+i and a horizontal radius labelled 2. No intersections are marked.

Arguments define directions, with missing endpoints

An argument condition $Arg (z - a) = θ$ fixes a direction from $a$ , giving a ray with its endpoint excluded. At $z = a$ , the expression is zero and has no argument. Any denominator inside an argument condition must also remain nonzero.

For several locus conditions, translate each into geometry or Cartesian equations, find their intersections, then check the original conditions. Equal non-negative distances can safely be squared; argument conditions still require the correct direction and any excluded points.

Argand diagram showing the real points negative one and three, and their perpendicular-bisector locus, the vertical line with real part one. Real and imaginary scales are equal.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A circle in Cartesian form

Describe $∣ z - (1 + i) ∣ = 2$ and give its Cartesian equation.

Worked solution

1
Express the displacement in coordinates.
$z = x + i y ⟹ z - (1 + i) = (x - 1) + i (y - 1)$
Subtracting the centre gives the vector from that centre to the variable point.
2
Write the distance equation.
$\sqrt{{(x - 1)}^{2} + {(y - 1)}^{2}} = 2$
The modulus is the Euclidean distance from $(1, 1)$ .
3
Square and identify the locus.
${(x - 1)}^{2} + {(y - 1)}^{2} = 4$
Both sides of the distance equation are non-negative, so squaring is equivalent: this is a circle centred at $(1, 1)$ with radius $2$ .

Answer

A circle centred at $(1, 1)$ with radius 2; ${(x - 1)}^{2} + {(y - 1)}^{2} = 4$ .

EXAMPLE 02

Equidistance gives a line

Find the locus $∣ z + 1 ∣ = ∣ z - 3 ∣$ .

Worked solution

1
Translate both distances.
$\sqrt{{(x + 1)}^{2} + y^{2}} = \sqrt{{(x - 3)}^{2} + y^{2}}$
The fixed points are $- 1$ and $3$ on the real axis.
2
Square and cancel equal terms.
$\begin{aligned} {(x + 1)}^{2} + y^{2} & = {(x - 3)}^{2} + y^{2} \\ x^{2} + 2 x + 1 & = x^{2} - 6 x + 9 \end{aligned}$
Squaring preserves equality of non-negative moduli.
3
Solve the remaining linear equation.
$8 x = 8 ⟹ x = 1$
The imaginary coordinate is unrestricted, so the locus is the vertical line $Re z = 1$ .
4
Interpret the line.
$\frac{- 1 + 3}{2} = 1$
It passes through the midpoint of the two fixed points and is perpendicular to the real axis.

Answer

$Re z = 1$ , the line $x = 1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Describe $∣ z - 2 i ∣ < 3$ .

Show answer & reasoning

The open disc centred at $(0, 2)$ with radius 3; boundary excluded.

The quantity $∣ z - 2 i ∣$ measures distance from $(0, 2)$ . Being strictly less than $3$ selects the open disc of radius $3$ and excludes every boundary point.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Give the centre and radius of $∣ z + 2 - i ∣ = 4$ .
2 marks
Write $∣ z - (2 - i) ∣ = 3$ as a Cartesian equation.
2 marks
Find the locus $∣ z - i ∣ = ∣ z + i ∣$ .
2 marks
Describe $Arg (z - 1) = \frac{π}{2}$ .
3 marks
Find the points satisfying $∣ z ∣ = 2$ and $Re z = 1$ .
3 marks
A tracking system reports $∣ z - (3 + 2 i) ∣ \leq 1$ . Interpret its possible positions.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Give the centre and radius of $∣ z + 2 - i ∣ = 4$ .
2 marks
Write the expression as $∣ z - (- 2 + i) ∣$ .
Centre $- 2 + i$ ; radius 4.
Write $∣ z - (2 - i) ∣ = 3$ as a Cartesian equation.
2 marks
The real and imaginary differences are $x - 2$ and $y + 1$ .
${(x - 2)}^{2} + {(y + 1)}^{2} = 9$ .
Find the locus $∣ z - i ∣ = ∣ z + i ∣$ .
2 marks
$x^{2} + {(y - 1)}^{2} = x^{2} + {(y + 1)}^{2}$ simplifies to $- 4 y = 0$ .
The real axis, $y = 0$ .
Describe $Arg (z - 1) = \frac{π}{2}$ .
3 marks
$z - 1$ must be a positive imaginary number. The point $z = 1$ is excluded.
The vertical ray $x = 1$ , $y > 0$ .
Find the points satisfying $∣ z ∣ = 2$ and $Re z = 1$ .
3 marks
$x = 1$ in $x^{2} + y^{2} = 4$ gives $y^{2} = 3$ . Both signs satisfy the circle.
$z = 1 \pm \sqrt{3} i$ .
A tracking system reports $∣ z - (3 + 2 i) ∣ \leq 1$ . Interpret its possible positions.
2 marks
The inequality is a closed disc, rather than only its boundary.
All points at most 1 unit from $(3, 2)$ , including the circle boundary.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Distances define circles and bisectors

Arguments define directions, with missing endpoints

See the method in action.

A circle in Cartesian form

Express the displacement in coordinates.

Write the distance equation.

Square and identify the locus.

Equidistance gives a line

Translate both distances.

Square and cancel equal terms.

Solve the remaining linear equation.

Interpret the line.

Try it yourself.

Check your understanding

Keep practising