The shape of the lesson
By the end of the lesson, learners can
- Circles and discs: Rewrite every modulus in the form \(|z - a|\), so you can read off the centre \(a\).
- 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.
- 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).
- \(|z - a| = k\): 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.
One picture every idea returns to
\(|z - a| = k\) 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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, 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 \(z\). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Circles and discs | Rewrite every modulus in the form \(|z - a|\), so you can read off the centre \(a\). Model the worked example Centre and radius on the board, one line at a time. Your turn: Find the least and greatest values of \(\arg z\) for points on or inside the circle \(|z - 4i| = 2\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: 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. Model the worked example Cartesian equation of a bisector on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: 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). Model the worked example A half-line on the board, one line at a time. Your turn: Is the point \(z = 3 + 3i\) in the region defined by \(|z - 2i| \le |z - 4|\) and \(\arg z \le \dfrac{\pi}{4}\)? | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 24 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Half-lines and regions, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(|z - 3 + 2i|\) has centre \(3 - 2i\), NOT \(-3 + 2i\) or \(3 + 2i\). | Always rewrite as \(|z - (\dots)|\) before reading off the centre. |
| \(|z + 4i|\) is the distance from \(-4i\), not from \(4i\). | Using \((0, 4)\) gives the bisector \(x - 2y + 3 = 0\), a different line. |
| \(\arg(z - a) = \theta\) 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. |
Loci in the Argand diagram: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the loci (a) \(|z + 2 - i| = 3\) (b) \(\arg(z - 3) = -\dfrac{\pi}{4}\).
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.
The complex number \(z\) satisfies \(|z - 3 - 4i| = 2\). Find the greatest and least values of \(|z|\).
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.
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 \(R\).
(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.
Answer key and marking notes
- (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: \(y = 3 - x\), \(x > 3\).
- \((x - 1)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}\).
- \(-2x + 1 - 2y + 1 = -6x + 9 - 10y + 25\), so \(4x + 8y = 32\), i.e. \(x + 2y = 8\).
- Midpoint \((2, 3)\): \(2 + 6 = 8\) ✓.
- Circle centre \(3 + 4i\), radius 2; distance from \(O\) to the centre is \(\sqrt{9 + 16} = 5\).
- Greatest \(|z| = 5 + 2 = 7\); least \(|z| = 5 - 2 = 3\).
- (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: \(|\sqrt{3} + 3i - 4i| = |\sqrt{3} - i| = 2\) ✓.
- \(|z| \le |z - 4|\) 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.).
- (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.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Describe the locus \(|z - 3 + 2i| = 4\) in an Argand diagram.
- AA circle with centre \(3 - 2i\) and radius 4
- BA circle with centre \(-3 + 2i\) and radius 4
- CA circle with centre \(3 - 2i\) and radius 16
- DA circle with centre \(3 + 2i\) and radius 4
Find the Cartesian equation of the locus \(|z - 2| = |z + 4i|\), where \(z = x + iy\).
- A\(y = -\dfrac{1}{2}x - \dfrac{3}{2}\)
- B\(y = \dfrac{1}{2}x + \dfrac{3}{2}\)
- C\(y = 2x - 4\)
- D\(y = -\dfrac{1}{2}x - 2\)
Which describes the locus \(\arg(z - 1 - i) = \dfrac{\pi}{4}\)?
- AA half-line from \(1 + i\) at angle \(\dfrac{\pi}{4}\) to the positive real direction, not including \(1 + i\)
- BThe whole straight line \(y = x\)
- CA half-line from \(-1 - i\) at angle \(\dfrac{\pi}{4}\) to the positive real direction
- DA half-line from the origin through \(1 + i\)
The point \(z\) lies on the circle \(|z - 3 - 4i| = 2\). Find the least possible value of \(|z|\).
- A\(3\)
- B\(7\)
- C\(5\)
- D\(\sqrt{21}\)
The point \(z\) lies on the circle \(|z - 4i| = 2\). Find the greatest possible value of \(\arg z\).
- A\(\dfrac{2\pi}{3}\)
- B\(\dfrac{\pi}{3}\)
- C\(\dfrac{5\pi}{6}\)
- D\(\dfrac{\pi}{2}\)
Which of these points lies in the region defined by \(|z - 2 - 2i| \le 2\) and \(\arg z \ge \dfrac{\pi}{4}\)?
- A\(1 + 2i\)
- B\(3 + i\)
- C\(2 + 4.5i\)
- D\(4 + 4i\)
Find the complex number that satisfies both \(|z - 2| = |z - 6|\) and \(\arg z = \dfrac{\pi}{3}\).
- A\(4 + 4\sqrt{3}\,i\)
- B\(4 + \dfrac{4}{\sqrt{3}}\,i\)
- C\(4\sqrt{3} + 4i\)
- D\(4 + 2\sqrt{3}\,i\)
Find the exact area of the region defined by \(|z| \le 4\) and \(\dfrac{\pi}{6} \le \arg z \le \dfrac{\pi}{2}\).
- A\(\dfrac{8\pi}{3}\)
- B\(\dfrac{16\pi}{3}\)
- C\(\dfrac{2\pi}{3}\)
- D\(\dfrac{4\pi}{3}\)
Quiz key
\(|z - 3 + 2i| = |z - (3 - 2i)|\), the distance from \(z\) to \(3 - 2i\), so the locus is the circle with centre \(3 - 2i\) and radius 4. The tempting centre \(-3 + 2i\) reads the signs straight off the expression instead of writing it as \(z - a\). Radius 16 squares the 4 (that would be for \(x^{2} + y^{2}\) form), and \(3 + 2i\) changes only one sign.
\((x - 2)^{2} + y^{2} = x^{2} + (y + 4)^{2}\) gives \(-4x + 4 = 8y + 16\), so \(y = -\dfrac{1}{2}x - \dfrac{3}{2}\). Check: the midpoint \((1, -2)\) satisfies it, and the gradient \(-\dfrac{1}{2}\) is perpendicular to the gradient 2 of the segment from \((2, 0)\) to \((0, -4)\). The tempting \(y = \dfrac{1}{2}x + \dfrac{3}{2}\) takes the second point as \(+4i\) instead of \(-4i\). \(y = 2x - 4\) is the line through the two points, not the bisector, and \(y = -\dfrac{1}{2}x - 2\) has the right gradient but misses the midpoint.
\(\arg(z - (1 + i)) = \dfrac{\pi}{4}\) means the direction from \(1 + i\) to \(z\) is \(\dfrac{\pi}{4}\): a half-line starting at \(1 + i\), going up and to the right, with \(1 + i\) excluded (its argument is undefined). The tempting "whole line \(y = x\)" includes points below \(1 + i\), where the direction is \(-\dfrac{3\pi}{4}\). The half-line from \(-1 - i\) misreads the sign of \(a\), and a half-line from the origin is \(\arg z = \dfrac{\pi}{4}\).
The centre \(3 + 4i\) is at distance \(\sqrt{9 + 16} = 5\) from \(O\), so the nearest point of the circle is \(5 - 2 = 3\) from \(O\). The tempting \(7\) is the greatest value, \(5 + 2\). \(5\) is the distance to the centre, and \(\sqrt{21} = \sqrt{25 - 4}\) is the length of a tangent from \(O\), not the least distance.
The centre is \(4i\), at distance 4 from \(O\), with radius 2. At the tangent point, \(\sin\alpha = \dfrac{2}{4}\), so the tangent makes \(\alpha = \dfrac{\pi}{6}\) with the imaginary axis. The greatest argument is \(\dfrac{\pi}{2} + \dfrac{\pi}{6} = \dfrac{2\pi}{3}\). The tempting \(\dfrac{\pi}{3}\) is the LEAST argument, \(\dfrac{\pi}{2} - \dfrac{\pi}{6}\). \(\dfrac{5\pi}{6}\) uses \(\cos\alpha = \dfrac{2}{4}\) (adding \(\dfrac{\pi}{3}\)), and \(\dfrac{\pi}{2}\) is the argument of the centre.
\(1 + 2i\): distance \(|-1| = 1 \le 2\), and \(\arg = \tan^{-1} 2 = 1.11 \ge 0.785\), so it satisfies both. \(3 + i\) is inside the circle (distance \(\sqrt{2}\)) but its argument \(0.322\) is less than \(\dfrac{\pi}{4}\). \(2 + 4.5i\) has a large enough argument but is 2.5 from the centre. The tempting \(4 + 4i\) has argument exactly \(\dfrac{\pi}{4}\), on the boundary half-line, but is \(2\sqrt{2} \approx 2.83\) from the centre, outside the circle.
\(|z - 2| = |z - 6|\) is the line \(x = 4\). On \(\arg z = \dfrac{\pi}{3}\), \(y = x\tan\dfrac{\pi}{3} = \sqrt{3}\,x\), so \(y = 4\sqrt{3}\) and \(z = 4 + 4\sqrt{3}\,i\). The tempting \(4 + \dfrac{4}{\sqrt{3}}\,i\) uses \(\tan\dfrac{\pi}{6}\), the angle with the imaginary axis. \(4\sqrt{3} + 4i\) swaps \(x\) and \(y\) (argument \(\dfrac{\pi}{6}\)), and \(4 + 2\sqrt{3}\,i\) uses \(\sin\dfrac{\pi}{3}\) times 4.
The sector angle is \(\dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}\), so the area is \(\dfrac{1}{2} \times 16 \times \dfrac{\pi}{3} = \dfrac{8\pi}{3}\). The tempting \(\dfrac{16\pi}{3}\) leaves out the \(\dfrac{1}{2}\) in the sector formula. \(\dfrac{2\pi}{3}\) uses \(r\) instead of \(r^{2}\), and \(\dfrac{4\pi}{3}\) uses only the angle \(\dfrac{\pi}{6}\) instead of the difference of the arguments.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.