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Cambridge International AS & A Level · Topic 3.9
A-Level Maths 9709 / Topic 3.9 / AS & A Level

Read the modulus as a distance.

Paper 3 only. \(|z - a|\) is the distance from \(z\) to the point \(a\), and \(\arg(z - a)\) is the direction from \(a\) to \(z\). With those two readings, equations and inequalities in \(z\) turn into circles, straight lines, half-lines and shaded regions.

Topic 3.9All complex numbers lessons →
The key idea

\(|z - a| = k\) is the circle with centre \(a\) and radius \(k\); \(|z - a| = |z - b|\) is the perpendicular bisector of the points \(a\) and \(b\); \(\arg(z - a) = \theta\) is the half-line from \(a\) (not including \(a\)) at angle \(\theta\) to the positive real direction. Inequalities give the regions inside or outside these boundaries.

Re Im 2424 2 + 2i P Q |z − 2 − 2i| ≤ 2 centre 2 + 2i, radius 2 OP = 2√2 − 2 = 0.828 OQ = 2√2 + 2 = 4.83
Fig. 1: the disc |z − 2 − 2i| ≤ 2; the line through O and the centre meets the circle at P (least |z|) and Q (greatest |z|), and the circle touches both axes, so arg z runs from 0 to π/2
Section 01

Circles and discs.

Rewrite every modulus in the form \(|z - a|\), so you can read off the centre \(a\). Then \(|z - a| = k\) is a circle, \(|z - a| < k\) is the inside (boundary dashed), \(|z - a| \le k\) includes the circle, and \(|z - a| > k\) is the outside.

Worked example Centre and radius
Describe the locus \(|z - 3 + 2i| = 4\) and find its Cartesian equation.
  1. Write it as \(|z - (3 - 2i)| = 4\): points at distance 4 from \(3 - 2i\).
  2. Circle with centre \((3, -2)\) and radius 4.
  3. With \(z = x + iy\): \(|(x - 3) + (y + 2)i| = 4\), so \((x - 3)^{2} + (y + 2)^{2} = 16\).
Answer: The circle centre \(3 - 2i\), radius 4: \((x - 3)^{2} + (y + 2)^{2} = 16\).
Worked example Greatest and least values in a disc
The complex number \(z\) satisfies \(|z - 2 - 2i| \le 2\). Find the greatest and least values of \(|z|\) and of \(\arg z\).
  1. Disc with centre \(2 + 2i\), radius 2 (Fig. 1). Distance from \(O\) to the centre: \(\sqrt{4 + 4} = 2\sqrt{2}\).
  2. \(|z|\) is the distance from \(O\). Along the line through \(O\) and the centre: least \(= 2\sqrt{2} - 2\), greatest \(= 2\sqrt{2} + 2\).
  3. The centre is 2 from each axis and the radius is 2, so the circle touches the real axis at \(2\) and the imaginary axis at \(2i\).
  4. So \(\arg z\) ranges from \(0\) (at \(z = 2\)) to \(\tfrac{\pi}{2}\) (at \(z = 2i\)).
Answer: \(0.828 \le |z| \le 4.83\) (exactly \(2\sqrt{2} \mp 2\)); \(0 \le \arg z \le \tfrac{\pi}{2}\).
Your turn
Find the least and greatest values of \(\arg z\) for points on or inside the circle \(|z - 4i| = 2\).
Examiner trap

\(|z - 3 + 2i|\) has centre \(3 - 2i\), NOT \(-3 + 2i\) or \(3 + 2i\). Always rewrite as \(|z - (\dots)|\) before reading off the centre.

Section 02

Perpendicular bisectors.

\(|z - a| = |z - b|\) says \(z\) is the same distance from \(a\) as from \(b\): the perpendicular bisector of the segment joining them. \(|z - a| \le |z - b|\) is the half-plane on \(a\)'s side of that line.

Worked example Cartesian equation of a bisector
Find the Cartesian equation of the locus \(|z - 2| = |z + 4i|\).
  1. The points are \(2\) and \(-4i\), i.e. \((2, 0)\) and \((0, -4)\).
  2. Square both sides with \(z = x + iy\): \((x - 2)^{2} + y^{2} = x^{2} + (y + 4)^{2}\).
  3. Expand: \(x^{2} - 4x + 4 + y^{2} = x^{2} + y^{2} + 8y + 16\), so \(-4x - 8y - 12 = 0\).
  4. Check: the midpoint \((1, -2)\) satisfies \(1 + 2(-2) + 3 = 0\) ✓, and the gradient \(-\tfrac{1}{2}\) is perpendicular to the segment's gradient 2 ✓.
Answer: \(x + 2y + 3 = 0\), i.e. \(y = -\tfrac{1}{2}x - \tfrac{3}{2}\).
Worked example The point on a bisector closest to O
Find the complex number \(z\) on the locus \(|z - 2i| = |z - 4|\) for which \(|z|\) is least.
  1. Bisector of \((0, 2)\) and \((4, 0)\): midpoint \((2, 1)\); segment gradient \(-\tfrac{1}{2}\), so bisector gradient 2: \(y = 2x - 3\).
  2. Least \(|z|\) is at the foot of the perpendicular from \(O\), along the direction \((2, -1)\): \(z = (2t, -t)\).
  3. On the line: \(-t = 4t - 3\), so \(t = \tfrac{3}{5}\) and \(z = \tfrac{6}{5} - \tfrac{3}{5}i\).
  4. \(|z| = \sqrt{\tfrac{36}{25} + \tfrac{9}{25}} = \tfrac{3}{\sqrt{5}}\).
Answer: \(z = 1.2 - 0.6i\), with \(|z| = \tfrac{3}{\sqrt{5}} = 1.34\).
Examiner trap

\(|z + 4i|\) is the distance from \(-4i\), not from \(4i\). Using \((0, 4)\) gives the bisector \(x - 2y + 3 = 0\), a different line.

Stage 1 · Learn

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Describe the locus \(|z - 3i| = 2\).

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What is the locus \(|z - 1| = |z + 1|\)?

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What is the locus \(\arg(z - 2) = \dfrac{\pi}{2}\)?

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Does \(z = 1 + i\) lie in the region \(|z - 2| < 1\)?

Section 03

Half-lines and regions.

\(\arg(z - a) = \theta\) fixes the DIRECTION from \(a\) to \(z\), so the locus is a half-line starting at \(a\); the point \(a\) itself is excluded (its argument is undefined). Draw it with an open circle at \(a\).

Worked example A half-line
Sketch the locus \(\arg(z - 1 - i) = \dfrac{\pi}{4}\) and give its Cartesian description.
  1. Rewrite: \(\arg(z - (1 + i)) = \dfrac{\pi}{4}\): the direction from \((1, 1)\) to \(z\) is \(45^\circ\) above the positive real direction.
  2. So \(z\) lies on the half-line from \((1, 1)\) with gradient 1, going up and to the right.
  3. Cartesian: \(y - 1 = x - 1\), i.e. \(y = x\), but only for \(x > 1\).
Answer: The half-line \(y = x\), \(x > 1\), starting (open) at \(1 + i\).
Worked example Where two loci meet
Find, in exact form, the complex number satisfying both \(|z| = |z - 4i|\) and \(\arg(z - 1) = \dfrac{\pi}{3}\).
  1. \(|z| = |z - 4i|\): perpendicular bisector of \(0\) and \(4i\), the horizontal line \(y = 2\).
  2. \(\arg(z - 1) = \dfrac{\pi}{3}\): half-line from \((1, 0)\) with gradient \(\tan\dfrac{\pi}{3} = \sqrt{3}\): \(y = \sqrt{3}(x - 1)\), \(x > 1\).
  3. Meet: \(2 = \sqrt{3}(x - 1)\), so \(x = 1 + \dfrac{2}{\sqrt{3}} = 1 + \dfrac{2\sqrt{3}}{3}\).
Answer: \(z = \left(1 + \dfrac{2\sqrt{3}}{3}\right) + 2i \approx 2.15 + 2i\)
Your turn
Is the point \(z = 3 + 3i\) in the region defined by \(|z - 2i| \le |z - 4|\) and \(\arg z \le \dfrac{\pi}{4}\)?
Examiner trap

\(\arg(z - a) = \theta\) is a HALF-line, not a full line. The opposite direction from \(a\) has argument \(\theta - \pi\) (or \(\theta + \pi\)), so drawing the whole line through \(a\) loses the mark.

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Find the Cartesian equation of the locus \(|z - 2| = |z + 4i|\).

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The complex number \(z\) satisfies \(|z - 4i| \le 2\). Find the least possible value of \(\arg z\).

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Which complex number lies on the locus \(\arg(z + 1 - i) = \dfrac{3\pi}{4}\)?

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Find the complex number satisfying both \(|z| = |z - 4i|\) and \(\arg(z - 1) = \dfrac{\pi}{3}\).

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Go deeper · practice
Six original Cambridge-style questions
Circles, perpendicular bisectors and half-lines, greatest and least modulus and argument, a shaded region, and the point of a locus nearest the origin.