The shape of the lesson
By the end of the lesson, learners can
- Meet i: No real number squares to \(-1\), so we invent one: \(i = \sqrt{-1}\).
- Add, subtract, multiply: Treat \(i\) like an algebraic letter: collect real and imaginary parts when adding, expand brackets when multiplying, and replace \(i^{2}\) by \(-1\) at the end.
- Dividing with the conjugate: A complex denominator is made real by multiplying top and bottom by its conjugate, since \((c + id)(c - id) = c^{2} + d^{2}\).
- Add/subtract: real parts together, imaginary parts together.
- Multiply: expand the brackets, then replace \(i^{2}\) by \(-1\).
- Divide: multiply top and bottom by the conjugate of the bottom.
- Unknown \(z\): write \(z = x + iy\) (and \(z^{*} = x - iy\)), then equate real and imaginary parts.
- Quadratics: a negative discriminant gives a conjugate pair \(p \pm qi\).
One picture every idea returns to
A complex number is \(z = x + iy\) with \(x, y\) real and \(i^{2} = -1\); \(\text{Re}\,z = x\), \(\text{Im}\,z = y\), and the conjugate is \(z^{*} = x - iy\). Two complex numbers are equal only if their real parts are equal AND their imaginary parts are equal. To divide, multiply top and bottom by the conjugate of the bottom.
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 square root of minus one.. 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: Meet i | No real number squares to \(-1\), so we invent one: \(i = \sqrt{-1}\). Model the worked example Quadratics with complex roots on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Add, subtract, multiply | Treat \(i\) like an algebraic letter: collect real and imaginary parts when adding, expand brackets when multiplying, and replace \(i^{2}\) by \(-1\) at the end. Model the worked example The three operations on the board, one line at a time. Your turn: Expand and simplify \((1 + i)^{4}\). | Pairs |
| 21 to 30 min | Teach and check: Dividing with the conjugate | A complex denominator is made real by multiplying top and bottom by its conjugate, since \((c + id)(c - id) = c^{2} + d^{2}\). Model the worked example Division on the board, one line at a time. Your turn: Express \(\dfrac{1}{2 + i}\) in the form \(x + iy\). | Rally Coach |
| 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, 23 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 Dividing with the conjugate, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| \(\sqrt{-36}\) is \(6i\), not \(-6\) or \(6\). | And \(i^{2} = -1\) must be used every time it appears: leaving \(-9i^{2}\) as \(-9\) is the most common lost mark in the whole topic. |
| \(zz^{*} = x^{2} + y^{2}\), always real and never negative. | If your answer to \((3 + 4i)(3 - 4i)\) is \(-7\), you used \(i^{2} = +1\). |
| You cannot divide parts separately: \(\dfrac{3 + 4i}{1 - 2i}\) is NOT \(3 - 2i\). | Only a REAL denominator can be shared out between the real and imaginary parts. |
Complex numbers: arithmetic in Cartesian form: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
Given \(z = 2 - 5i\) and \(w = -1 + 3i\), find (a) \(z + w\) (b) \(z - w\) (c) \(zw\).
Showing all your working, express \(\dfrac{4 - 3i}{1 + 2i}\) in the form \(x + iy\).
Solve \(z^{2} + 4z + 29 = 0\), and verify that the sum of the roots is \(-4\) and their product is 29.
Find the complex number \(z\) satisfying \(2z - iz^{*} = 4 + i\).
Let \(u = 1 + i\). Find \(u^{2}\) and \(u^{4}\) in the form \(x + iy\), and hence show that \(u^{8}\) is a real number, stating its value.
The complex number \(w\) satisfies \((1 + i)w = 5 - i\).
(a) Find \(w\) in the form \(x + iy\). (b) Find \(ww^{*}\). (c) Find the real number \(k\) for which \(w + \dfrac{k}{w}\) is real.
Answer key and marking notes
- (a) \(1 - 2i\)
- (b) \(3 - 8i\)
- (c) \((2 - 5i)(-1 + 3i) = -2 + 6i + 5i - 15i^{2} = 13 + 11i\).
- Multiply by \(\dfrac{1 - 2i}{1 - 2i}\).
- Top: \(4 - 8i - 3i + 6i^{2} = -2 - 11i\). Bottom: \(1 + 4 = 5\).
- \(-\dfrac{2}{5} - \dfrac{11}{5}i\), i.e. \(-0.4 - 2.2i\).
- \(z = \dfrac{-4 \pm \sqrt{16 - 116}}{2} = \dfrac{-4 \pm 10i}{2} = -2 \pm 5i\).
- Sum: \((-2 + 5i) + (-2 - 5i) = -4\).
- Product: \((-2)^{2} + 5^{2} = 4 + 25 = 29\).
- \(z = x + iy\): \(2x + 2iy - i(x - iy) = (2x - y) + (2y - x)i\).
- Equate: \(2x - y = 4\) and \(2y - x = 1\).
- \(y = 2x - 4\), so \(4x - 8 - x = 1\), \(x = 3\), \(y = 2\).
- \(z = 3 + 2i\).
- \(u^{2} = 1 + 2i + i^{2} = 2i\).
- \(u^{4} = (2i)^{2} = 4i^{2} = -4\).
- \(u^{8} = (u^{4})^{2} = (-4)^{2} = 16\), which is real.
- (a) \(w = \dfrac{5 - i}{1 + i} \times \dfrac{1 - i}{1 - i} = \dfrac{5 - 5i - i + i^{2}}{2} = \dfrac{4 - 6i}{2} = 2 - 3i\).
- (b) \(ww^{*} = 2^{2} + 3^{2} = 13\).
- (c) \(\dfrac{1}{w} = \dfrac{w^{*}}{ww^{*}} = \dfrac{2 + 3i}{13}\), so \(w + \dfrac{k}{w} = \left(2 + \dfrac{2k}{13}\right) + \left(-3 + \dfrac{3k}{13}\right)i\).
- Real when \(-3 + \dfrac{3k}{13} = 0\), so \(k = 13\).
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.
Express \((3 + 2i)(1 - 4i)\) in the form \(x + iy\).
- A\(11 - 10i\)
- B\(-5 - 10i\)
- C\(3 - 8i\)
- D\(11 - 14i\)
Express \(\dfrac{5 + i}{2 - 3i}\) in the form \(x + iy\).
- A\(\dfrac{7}{13} + \dfrac{17}{13}i\)
- B\(1 + \dfrac{17}{13}i\)
- C\(-\dfrac{7}{5} - \dfrac{17}{5}i\)
- D\(\dfrac{5}{2} - \dfrac{1}{3}i\)
Given \(z = 2 - 5i\), find \(zz^{*}\).
- A\(29\)
- B\(-21\)
- C\(\sqrt{29}\)
- D\(4 - 25i\)
Find the complex number \(z\) such that \((1 + 2i)z = 7 + 4i\).
- A\(3 - 2i\)
- B\(7 + 2i\)
- C\(-\dfrac{1}{5} + \dfrac{18}{5}i\)
- D\(3 + 2i\)
Solve the equation \(2z + iz^{*} = 4 + 5i\), giving \(z\) in the form \(x + iy\).
- A\(1 + 2i\)
- B\(\dfrac{13}{5} + \dfrac{6}{5}i\)
- C\(2 + i\)
- D\(1 - 2i\)
The number \(\dfrac{a + 3i}{2 - i}\), where \(a\) is real, is real. Find \(a\).
- A\(a = -6\)
- B\(a = 6\)
- C\(a = \dfrac{3}{2}\)
- D\(a = -\dfrac{3}{2}\)
Given \(z = 1 + i\), find \(z^{3}\) in the form \(x + iy\).
- A\(-2 + 2i\)
- B\(1 - i\)
- C\(-2 - 2i\)
- D\(2 + 2i\)
Given \(w = \dfrac{2 + i}{1 - i}\), find \(\text{Re}(w^{2})\).
- A\(-2\)
- B\(\dfrac{1}{4}\)
- C\(\dfrac{5}{2}\)
- D\(2\)
Quiz key
\(3 - 12i + 2i - 8i^{2} = 3 - 10i + 8 = 11 - 10i\). The tempting \(-5 - 10i\) uses \(i^{2} = +1\), so \(-8i^{2}\) becomes \(-8\). \(3 - 8i\) multiplies real parts and imaginary parts separately, which is not how brackets expand, and \(11 - 14i\) slips the sign of \(+2i\).
\(\dfrac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = \dfrac{10 + 15i + 2i + 3i^{2}}{4 + 9} = \dfrac{7 + 17i}{13}\). The tempting \(1 + \dfrac{17}{13}i\) uses \(3i^{2} = +3\) in the numerator. \(-\dfrac{7}{5} - \dfrac{17}{5}i\) works out the denominator as \(4 - 9\), forgetting that \(-9i^{2} = +9\), and \(\dfrac{5}{2} - \dfrac{1}{3}i\) divides the real parts and the imaginary parts separately.
\((2 - 5i)(2 + 5i) = 4 - 25i^{2} = 4 + 25 = 29\), which equals \(|z|^{2}\). The tempting \(-21\) uses \(i^{2} = +1\), computing \(4 - 25\). \(\sqrt{29}\) is \(|z|\), not \(|z|^{2}\), and \(4 - 25i\) leaves the \(i\) in place after squaring \(5i\).
\(z = \dfrac{(7 + 4i)(1 - 2i)}{1 + 4} = \dfrac{7 - 14i + 4i + 8}{5} = \dfrac{15 - 10i}{5} = 3 - 2i\). Check: \((1 + 2i)(3 - 2i) = 3 - 2i + 6i + 4 = 7 + 4i\). The tempting \(7 + 2i\) divides the real parts and imaginary parts separately. \(-\dfrac{1}{5} + \dfrac{18}{5}i\) multiplies the top by \(1 + 2i\) instead of by the conjugate, and \(3 + 2i\) slips the sign of the imaginary part.
\(2(x + iy) + i(x - iy) = (2x + y) + i(x + 2y)\). Equating: \(2x + y = 4\) and \(x + 2y = 5\), so \(x = 1\), \(y = 2\) and \(z = 1 + 2i\). Check: \(2 + 4i + i(1 - 2i) = 2 + 4i + i + 2 = 4 + 5i\). The tempting \(\dfrac{13}{5} + \dfrac{6}{5}i\) treats \(z^{*}\) as \(z\), solving \((2 + i)z = 4 + 5i\). \(2 + i\) gives \(5 + 4i\), with the parts swapped, and \(1 - 2i\) is the conjugate of the answer.
\(\dfrac{(a + 3i)(2 + i)}{5} = \dfrac{(2a - 3) + (a + 6)i}{5}\). Real means \(a + 6 = 0\), so \(a = -6\) (the number is then \(-3\)). The tempting \(a = 6\) multiplies by \(2 - i\) instead of the conjugate, getting imaginary part \(6 - a\). \(a = \dfrac{3}{2}\) sets the REAL part to 0, which makes the number purely imaginary, and \(-\dfrac{3}{2}\) is a sign slip on that.
\(z^{2} = 1 + 2i + i^{2} = 2i\), so \(z^{3} = 2i(1 + i) = 2i + 2i^{2} = -2 + 2i\). The tempting \(1 - i\) cubes each part separately (\(1^{3} + i^{3}\)), which ignores the cross terms. \(-2 - 2i\) slips the sign of \(2i\), and \(2 + 2i\) uses \(i^{2} = +1\).
\(w = \dfrac{(2 + i)(1 + i)}{2} = \dfrac{1 + 3i}{2}\), so \(w^{2} = \dfrac{1 + 6i + 9i^{2}}{4} = \dfrac{-8 + 6i}{4} = -2 + \dfrac{3}{2}i\) and \(\text{Re}(w^{2}) = -2\). The tempting \(\dfrac{1}{4}\) squares only the real part of \(w\). \(\dfrac{5}{2}\) is \(x^{2} + y^{2} = |w|^{2}\), and \(2\) loses the sign of \(9i^{2} = -9\) when combining \(1 - 9\) (writing \(9 - 1\)).
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.