Complex loci and geometry | Year 12

01 / LEARN

The idea, step by step

Distances define circles and bisectors

$∣ z - a ∣ = r$ , with $r > 0$ , is a circle centred at the point a with radius r. Replacing equality by $\leq$ gives the closed disc; $<$ gives the open disc. For $z = x + i y$ , square the modulus to obtain a Cartesian equation.

$∣ z - a ∣ = ∣ z - b ∣$ , with distinct a and b, is the perpendicular bisector of the segment joining them: its points are exactly the points equally distant from a and b. Squaring both sides cancels the quadratic terms and gives a line.

Arguments define directions, with missing endpoints

$Arg (z - a) = θ$ describes the ray starting from a in direction $θ$ , but excludes a because zero has no argument. For a quotient, the denominator must also be nonzero.

To solve intersecting locus conditions, first translate each condition, then find and check every intersection in the original conditions. Squaring distances is safe because moduli are non-negative; angle conditions still require direction and exclusions.

A common trap

Including the starting point of an argument ray.

Instead: At the starting point the complex expression is zero, so its argument is undefined.

02 / WORKED EXAMPLES

See the thinking

Example 1

A circle in Cartesian form

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

Write $z = x + i y$ , so $z - (1 + i) = (x - 1) + i (y - 1)$ .
Square the modulus equation to get the Cartesian form.

A circle centred at

(1, 1)

with radius 2;

{(x - 1)}^{2} + {(y - 1)}^{2} = 4

Example 2

Equidistance gives a line

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

Put $z = x + i y$ and square: ${(x + 1)}^{2} + y^{2} = {(x - 3)}^{2} + y^{2}$ .
Expand and cancel to obtain $8 x = 8$ .
The line bisects the real segment from -1 to 3 at right angles.

Re z = 1

, the line

x = 1

Check your understanding

Describe $∣ z - 2 i ∣ < 3$ .

Show answer

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

The modulus is the distance from $2 i$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 $Arg (z - 1) = π / 2$ .
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 $Arg (z - 1) = π / 2$ .
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.

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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 ↗

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