The shape of the lesson
By the end of the lesson, learners can
- The Argand diagram, modulus and argument: On an Argand diagram the real part goes across and the imaginary part goes up.
- Multiplying and dividing in polar form: Write \(z_{1} = r_{1}e^{i\theta_{1}}\) and \(z_{2} = r_{2}e^{i\theta_{2}}\); the laws of indices do the rest. \(z_{1}z_{2} = r_{1}r_{2}e^{i(\theta_{1} + \theta_{2})}\) and \(\dfrac{z_{1}}{z_{2}} = \dfrac{r_{1}}{r_{2}}e^{i(\theta_{1} - \theta_{2})}\).
- What the operations look like: Every operation has a picture on the Argand diagram (Fig. 1).
- Sketch the point to find its quadrant before using \(\tan^{-1}\).
- Modulus: \(r = \sqrt{x^{2} + y^{2}}\). Argument: adjust the reference angle for the quadrant; \(-\pi < \theta \le \pi\).
- Forms: \(x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}\).
- Multiply: multiply moduli, add arguments. Divide: divide moduli, subtract arguments.
- Geometry: \(z^{*}\) reflects; \(+w\) translates; \(\times re^{i\theta}\) enlarges and rotates.
One picture every idea returns to
\(z = x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}\), where \(r = |z| = \sqrt{x^{2} + y^{2}}\) and \(\theta = \arg z\) with \(-\pi < \theta \le \pi\). Then \(|z_{1}z_{2}| = |z_{1}|\,|z_{2}|\) and \(\arg(z_{1}z_{2}) = \arg z_{1} + \arg z_{2}\); for division, divide the moduli and subtract the arguments.
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, A number with a length and a direction.. Paper 3 only. 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: The Argand diagram, modulus and argument | On an Argand diagram the real part goes across and the imaginary part goes up. Model the worked example Modulus, argument and three forms on the board, one line at a time. Your turn: Find the modulus and argument of \(1 - i\), and write it in exponential form. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Multiplying and dividing in polar form | Write \(z_{1} = r_{1}e^{i\theta_{1}}\) and \(z_{2} = r_{2}e^{i\theta_{2}}\); the laws of indices do the rest. \(z_{1}z_{2} = r_{1}r_{2}e^{i(\theta_{1} + \theta_{2})}\) and \(\dfrac{z_{1}}{z_{2}} = \dfrac{r_{1}}{r_{2}}e^{i(\theta_{1} - \theta_{2})}\). Model the worked example Product and quotient on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: What the operations look like | Every operation has a picture on the Argand diagram (Fig. 1). Model the worked example Rotate and enlarge on the board, one line at a time. Your turn: Without expanding, find \(|(3 + 4i)(5 - 12i)|\). | Numbered Heads Together |
| 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, 26 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 What the operations look like, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Your calculator's \(\tan^{-1}\left(\tfrac{y}{x}\right)\) gives \(\tan^{-1}\left(\tfrac{-4}{-3}\right) = 0.927\) for \(-3 - 4i\), which points into the FIRST quadrant. | The sketch is not optional: it is how you know to adjust. |
| Moduli MULTIPLY but arguments ADD. | Multiplying the arguments, or adding the moduli, are both common slips, and \(\dfrac{4\pi}{3}\) left unadjusted loses the final mark when the range \(-\pi < \theta \le \pi\) is required. |
| Multiplying by \(-i\) (or dividing by \(i\)) rotates CLOCKWISE. | Check the direction with a simple case: \(i \times 1 = i\) moves from the real axis up to the imaginary axis, which is anticlockwise. |
The Argand diagram, modulus-argument and exponential form: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
Find the modulus and argument of (a) \(-5i\) (b) \(-2 + 2i\), giving exact answers.
Express \(z = -3 + 4i\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\), giving \(\theta\) to 3 significant figures.
\(z_{1} = 4e^{2\pi i/3}\) and \(z_{2} = 2e^{-\pi i/6}\). Find \(z_{1}z_{2}\) and \(\dfrac{z_{1}}{z_{2}}\), giving each in the form \(x + iy\) with exact values.
Let \(u = \sqrt{3} + i\). Find \(|u|\) and \(\arg u\), and hence show that \(u^{6} = -64\).
On an Argand diagram, \(U\) and \(V\) represent \(u = 2 + i\) and \(v = 1 + 3i\).
(a) Show that \(v = (1 + i)u\). (b) Hence describe the transformation that maps \(U\) to \(V\), and find angle \(UOV\). (c) Find the area of triangle \(UOV\).
Let \(z = 1 + \sqrt{3}\,i\) and \(w = 1 - i\).
(a) Find \(zw\) in the form \(x + iy\). (b) Find the modulus and argument of \(z\) and of \(w\), and hence of \(zw\). (c) Deduce the exact value of \(\sin\dfrac{\pi}{12}\).
Answer key and marking notes
- (a) \(|-5i| = 5\), \(\arg(-5i) = -\tfrac{\pi}{2}\).
- (b) \(|-2 + 2i| = \sqrt{8} = 2\sqrt{2}\); second quadrant with reference angle \(\tfrac{\pi}{4}\), so \(\arg = \tfrac{3\pi}{4}\).
- \(r = \sqrt{9 + 16} = 5\).
- Second quadrant: reference angle \(\tan^{-1}\tfrac{4}{3} = 0.9273\), so \(\theta = \pi - 0.9273 = 2.21\).
- \(z = 5e^{2.21i}\).
- \(z_{1}z_{2} = 8e^{i(2\pi/3 - \pi/6)} = 8e^{i\pi/2} = 8i\).
- \(\dfrac{z_{1}}{z_{2}} = 2e^{i(2\pi/3 + \pi/6)} = 2e^{5\pi i/6}\).
- \(= 2\left(-\tfrac{\sqrt{3}}{2} + \tfrac{1}{2}i\right) = -\sqrt{3} + i\).
- \(|u| = \sqrt{3 + 1} = 2\); first quadrant, \(\arg u = \tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}\).
- Multiplying six copies: \(|u^{6}| = 2^{6} = 64\) and \(\arg u^{6} = 6 \times \tfrac{\pi}{6} = \pi\).
- \(u^{6} = 64(\cos\pi + i\sin\pi) = -64\).
- (a) \((1 + i)(2 + i) = 2 + i + 2i + i^{2} = 1 + 3i = v\).
- (b) \(1 + i = \sqrt{2}\,e^{i\pi/4}\): an enlargement with scale factor \(\sqrt{2}\) about \(O\) combined with a rotation of \(\tfrac{\pi}{4}\) anticlockwise. Angle \(UOV = \tfrac{\pi}{4}\).
- (c) \(OU = \sqrt{5}\), \(OV = \sqrt{2}\sqrt{5} = \sqrt{10}\); area \(= \tfrac{1}{2}\sqrt{5}\sqrt{10}\sin\tfrac{\pi}{4} = \tfrac{1}{2}\sqrt{50} \times \tfrac{\sqrt{2}}{2} = \tfrac{5}{2}\).
- (a) \(zw = 1 - i + \sqrt{3}\,i + \sqrt{3} = (1 + \sqrt{3}) + (\sqrt{3} - 1)i\).
- (b) \(|z| = 2\), \(\arg z = \tfrac{\pi}{3}\); \(|w| = \sqrt{2}\), \(\arg w = -\tfrac{\pi}{4}\). So \(|zw| = 2\sqrt{2}\), \(\arg zw = \tfrac{\pi}{12}\).
- (c) Imaginary parts: \(2\sqrt{2}\sin\tfrac{\pi}{12} = \sqrt{3} - 1\).
- \(\sin\tfrac{\pi}{12} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \dfrac{\sqrt{6} - \sqrt{2}}{4}\).
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.
Find the modulus and argument of \(z = -1 + \sqrt{3}\,i\), with \(-\pi < \arg z \le \pi\).
- A\(|z| = 2,\ \arg z = \dfrac{2\pi}{3}\)
- B\(|z| = 2,\ \arg z = -\dfrac{\pi}{3}\)
- C\(|z| = 4,\ \arg z = \dfrac{2\pi}{3}\)
- D\(|z| = 2,\ \arg z = \dfrac{\pi}{3}\)
Find \(\arg(-3 - 4i)\) in radians, where \(-\pi < \arg z \le \pi\), correct to 3 significant figures.
- A\(-2.21\)
- B\(0.927\)
- C\(-0.927\)
- D\(2.21\)
Express \(z = 4e^{\frac{\pi}{6}i}\) in the form \(x + iy\), giving exact values.
- A\(2\sqrt{3} + 2i\)
- B\(2 + 2\sqrt{3}\,i\)
- C\(2\sqrt{3} - 2i\)
- D\(\sqrt{3} + i\)
\(z_{1} = 3\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)\) and \(z_{2} = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)\). Find the modulus and argument of \(z_{1}z_{2}\).
- A\(6,\ \dfrac{7\pi}{12}\)
- B\(5,\ \dfrac{7\pi}{12}\)
- C\(6,\ \dfrac{\pi^{2}}{12}\)
- D\(\dfrac{3}{2},\ -\dfrac{\pi}{12}\)
Given \(z = 1 + i\) and \(w = \sqrt{3} - i\), find the modulus and argument of \(\dfrac{z}{w}\).
- A\(\dfrac{\sqrt{2}}{2},\ \dfrac{5\pi}{12}\)
- B\(\dfrac{\sqrt{2}}{2},\ \dfrac{\pi}{12}\)
- C\(\sqrt{2},\ \dfrac{5\pi}{12}\)
- D\(2\sqrt{2},\ \dfrac{\pi}{12}\)
Given \(z = 2e^{\frac{\pi}{3}i}\), find \(z^{3}\).
- A\(-8\)
- B\(8\)
- C\(-6\)
- D\(8i\)
Express \(u = -2 + 2i\) in the form \(re^{i\theta}\), where \(r > 0\) and \(-\pi < \theta \le \pi\), giving exact values.
- A\(2\sqrt{2}\,e^{\frac{3\pi}{4}i}\)
- B\(2\sqrt{2}\,e^{-\frac{\pi}{4}i}\)
- C\(8e^{\frac{3\pi}{4}i}\)
- D\(2\sqrt{2}\,e^{\frac{\pi}{4}i}\)
The complex number \(z\) has \(|z| = 4\) and \(\arg z = \dfrac{\pi}{3}\). Find \(\arg(iz^{*})\), where \(-\pi < \arg \le \pi\).
- A\(\dfrac{\pi}{6}\)
- B\(\dfrac{5\pi}{6}\)
- C\(-\dfrac{5\pi}{6}\)
- D\(-\dfrac{\pi}{6}\)
Quiz key
\(|z| = \sqrt{1 + 3} = 2\). The point \((-1, \sqrt{3})\) is in the second quadrant, with reference angle \(\dfrac{\pi}{3}\), so \(\arg z = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\). The tempting \(-\dfrac{\pi}{3}\) is what a calculator gives for \(\tan^{-1}\left(\dfrac{\sqrt{3}}{-1}\right)\); it points into the fourth quadrant. \(|z| = 4\) forgets the square root, and \(\dfrac{\pi}{3}\) is the first-quadrant angle.
The reference angle is \(\tan^{-1}\dfrac{4}{3} = 0.927\). In the third quadrant, \(\arg z = -(\pi - 0.927) = -2.21\). The tempting \(0.927\) is the calculator value of \(\tan^{-1}\dfrac{-4}{-3}\), which lands in the first quadrant because the two minus signs cancel. \(-0.927\) points into the fourth quadrant, and \(2.21\) into the second.
\(4\cos\dfrac{\pi}{6} + 4i\sin\dfrac{\pi}{6} = 4 \times \dfrac{\sqrt{3}}{2} + 4 \times \dfrac{1}{2}i = 2\sqrt{3} + 2i\). The tempting \(2 + 2\sqrt{3}\,i\) swaps sine and cosine (that is \(4e^{\frac{\pi}{3}i}\)). \(2\sqrt{3} - 2i\) is the conjugate, \(4e^{-\frac{\pi}{6}i}\), and \(\sqrt{3} + i\) forgets to multiply by the modulus 4 (it has modulus 2).
\(|z_{1}z_{2}| = 3 \times 2 = 6\) and \(\arg(z_{1}z_{2}) = \dfrac{\pi}{4} + \dfrac{\pi}{3} = \dfrac{7\pi}{12}\). The tempting \(5\) adds the moduli as well as the arguments. \(\dfrac{\pi^{2}}{12}\) multiplies the arguments, and \(\dfrac{3}{2},\ -\dfrac{\pi}{12}\) is the rule for the quotient \(\dfrac{z_{1}}{z_{2}}\).
\(|z| = \sqrt{2}\), \(\arg z = \dfrac{\pi}{4}\); \(|w| = 2\), \(\arg w = -\dfrac{\pi}{6}\). So \(\left|\dfrac{z}{w}\right| = \dfrac{\sqrt{2}}{2}\) and \(\arg\dfrac{z}{w} = \dfrac{\pi}{4} - \left(-\dfrac{\pi}{6}\right) = \dfrac{5\pi}{12}\). The tempting \(\dfrac{\pi}{12}\) takes \(\arg w = +\dfrac{\pi}{6}\), ignoring that \(w\) lies below the real axis. \(\sqrt{2}\) divides the moduli the wrong way round, and \(2\sqrt{2},\ \dfrac{\pi}{12}\) is the modulus and argument of the product \(zw\).
\(z^{3} = 2^{3}e^{3 \times \frac{\pi}{3}i} = 8e^{\pi i} = 8(\cos\pi + i\sin\pi) = -8\). The tempting \(8\) gets the modulus right but forgets that an argument of \(\pi\) points along the NEGATIVE real axis. \(-6\) multiplies the modulus by 3 instead of cubing it, and \(8i\) uses an argument of \(\dfrac{\pi}{2}\).
\(r = \sqrt{4 + 4} = 2\sqrt{2}\). The point is in the second quadrant, so \(\theta = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\) and \(u = 2\sqrt{2}\,e^{\frac{3\pi}{4}i}\). The tempting \(e^{-\frac{\pi}{4}i}\) is the calculator value of \(\tan^{-1}(-1)\), pointing into the fourth quadrant (that is \(2 - 2i\)). \(8\) is \(r^{2}\), and \(e^{\frac{\pi}{4}i}\) gives \(2 + 2i\).
\(\arg z^{*} = -\dfrac{\pi}{3}\), and \(\arg i = \dfrac{\pi}{2}\), so \(\arg(iz^{*}) = -\dfrac{\pi}{3} + \dfrac{\pi}{2} = \dfrac{\pi}{6}\). The tempting \(\dfrac{5\pi}{6}\) forgets the conjugate and finds \(\arg(iz)\). \(-\dfrac{5\pi}{6}\) is \(\arg(-iz^{*})\), rotating the wrong way, and \(-\dfrac{\pi}{6}\) is \(\arg(-iz)\), which conjugates the final answer instead of \(z\).
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.