AS & A Level · Practice questions · Loci in the Argand diagram

Distances and directions.

Six original Cambridge-style questions on loci in the Argand diagram: circles, perpendicular bisectors, half-lines, regions, and greatest and least values of the modulus and argument.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rewrite as \(|z - a|\), then read the geometry.

\(|z - a| = k\): circle, centre \(a\), radius \(k\); \(<\) inside, \(>\) outside.
\(|z - a| = |z - b|\): perpendicular bisector of \(a\) and \(b\).
\(\arg(z - a) = \theta\): half-line from \(a\) (excluded) at angle \(\theta\).
Greatest/least \(|z|\): along the line through \(O\) and the centre: distance \(\pm\) radius.
Greatest/least \(\arg z\): tangents from \(O\); use right-angled triangles.

Recall

[2 marks]

Describe the loci (a) \(|z + 2 - i| = 3\) (b) \(\arg(z - 3) = -\dfrac{\pi}{4}\).

(a) \(|z - (-2 + i)| = 3\): circle with centre \(-2 + i\) and radius 3. ✓
(b) Half-line from the point 3 (excluded) going down and to the right at \(45^\circ\) below the real direction: \(y = 3 - x\), \(x > 3\). ✓

Calculation

[3 marks]

Find the Cartesian equation of the locus \(|z - 1 - i| = |z - 3 - 5i|\), and verify that it passes through the midpoint of the two fixed points.

\((x - 1)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}\). ✓
\(-2x + 1 - 2y + 1 = -6x + 9 - 10y + 25\), so \(4x + 8y = 32\), i.e. \(x + 2y = 8\). ✓
Midpoint \((2, 3)\): \(2 + 6 = 8\) ✓. ✓

Calculation

[3 marks]

The complex number \(z\) satisfies \(|z - 3 - 4i| = 2\). Find the greatest and least values of \(|z|\).

Circle centre \(3 + 4i\), radius 2; distance from \(O\) to the centre is \(\sqrt{9 + 16} = 5\). ✓
Greatest \(|z| = 5 + 2 = 7\); least \(|z| = 5 - 2 = 3\). ✓

Analysis

[5 marks]

The complex number \(z\) satisfies \(|z - 4i| \le 2\). (a) Find the least and greatest values of \(\arg z\). (b) Find, in the form \(x + iy\), the value of \(z\) for which \(\arg z\) is least.

(a) Tangent from \(O\): \(\sin\alpha = \tfrac{2}{4}\), \(\alpha = \tfrac{\pi}{6}\) from the imaginary axis. ✓
Least \(\arg z = \tfrac{\pi}{3}\), greatest \(\arg z = \tfrac{2\pi}{3}\). ✓
(b) Tangent length \(\sqrt{16 - 4} = 2\sqrt{3}\), at argument \(\tfrac{\pi}{3}\): \(z = 2\sqrt{3}\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = \sqrt{3} + 3i\). ✓
Check: \(|\sqrt{3} + 3i - 4i| = |\sqrt{3} - i| = 2\) ✓. ✓

Analysis

[5 marks]

On an Argand diagram, shade the region \(R\) of points satisfying both \(|z - 2 - 2i| \le 2\) and \(|z| \le |z - 4|\). Find the greatest value of \(|z|\) for points in \(R\).

\(|z| \le |z - 4|\) is the half-plane \(x \le 2\) (bisector of 0 and 4). \(R\) is the left half of the disc centre \(2 + 2i\), radius 2. ✓
The farthest point of the whole disc from \(O\) is \((2 + \sqrt{2}, 2 + \sqrt{2})\), which has \(x > 2\), so it is not in \(R\). ✓
In \(R\), the farthest point is the top of the dividing diameter, \(2 + 4i\) (on the circle and on \(x = 2\)). ✓
Greatest \(|z| = \sqrt{4 + 16} = 2\sqrt{5} = 4.47\) (3 s.f.). ✓

Analysis

[6 marks]

(a) Find the Cartesian equation of the locus \(|z - 2i| = |z - 4|\). (b) Find the complex number on this locus with \(\arg z = \dfrac{\pi}{4}\). (c) Find the least value of \(|z|\) on the locus, and the complex number where it occurs.

(a) \(x^{2} + (y - 2)^{2} = (x - 4)^{2} + y^{2}\) gives \(-4y + 4 = -8x + 16\), so \(y = 2x - 3\). ✓
(b) \(\arg z = \tfrac{\pi}{4}\) means \(y = x\), \(x > 0\): \(x = 2x - 3\), so \(z = 3 + 3i\). ✓
(c) Perpendicular from \(O\) to \(2x - y - 3 = 0\): direction \((2, -1)\), so \(z = (2t, -t)\) with \(-t = 4t - 3\), \(t = \tfrac{3}{5}\). ✓
\(z = 1.2 - 0.6i\) and \(|z| = \tfrac{3}{\sqrt{5}} = 1.34\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.