The shape of the lesson
By the end of the lesson, learners can
- Conjugate pairs: Take the conjugate of a whole polynomial equation with real coefficients: the coefficients do not change, but every \(z\) becomes \(z^{*}\).
- Quartics and long division: For a quartic, one complex root gives one quadratic factor.
- Square roots of a complex number: Every non-zero complex number has two square roots, negatives of each other.
- Pair: real coefficients mean \(a + bi\) brings \(a - bi\).
- Factor: \(z^{2} - 2az + (a^{2} + b^{2})\).
- Finish: divide or compare coefficients for the remaining factor, then solve.
- Unknown coefficients: substitute the root and equate real and imaginary parts.
- Square roots: \(x^{2} - y^{2} = p\), \(2xy = q\); reject \(x^{2} < 0\); signs of \(x, y\) linked by \(q\).
One picture every idea returns to
For a polynomial with REAL coefficients, if \(a + bi\) is a root then so is \(a - bi\), and \((z - (a + bi))(z - (a - bi)) = z^{2} - 2az + (a^{2} + b^{2})\) is a real quadratic factor. To find \(\sqrt{p + qi}\), write \((x + iy)^{2} = p + qi\) and equate parts: \(x^{2} - y^{2} = p\), \(2xy = q\).
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, Non-real roots come in pairs.. 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: Conjugate pairs | Take the conjugate of a whole polynomial equation with real coefficients: the coefficients do not change, but every \(z\) becomes \(z^{*}\). Model the worked example A cubic from one complex root on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Quartics and long division | For a quartic, one complex root gives one quadratic factor. Model the worked example A quartic on the board, one line at a time. Your turn: Show that \(-1 + i\) is a root of \(z^{3} + 4z^{2} + 6z + 4 = 0\) and find the other roots. | Pairs |
| 21 to 30 min | Teach and check: Square roots of a complex number | Every non-zero complex number has two square roots, negatives of each other. Model the worked example Square roots in exact Cartesian form on the board, one line at a time. Your turn: Find the square roots of \(7 - 24i\). | Round Table |
| 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, 29 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 Square roots of a complex number, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| When you substitute a complex root, ALL the real terms go into one equation and ALL the imaginary terms into the other. | A single combined equation like \(-6 - 3p + q - 2 + 4p + 2q = 0\) throws away information and cannot give two unknowns. |
| The quadratic factor is \(z^{2} - (\text{sum})z + \text{product}\). | For roots \(2 \pm i\) that is \(z^{2} - 4z + 5\); writing \(z^{2} + 4z + 5\) (wrong sign) makes the division leave a remainder. |
| Keep the signs linked: \(xy\) negative means \(x\) and \(y\) have OPPOSITE signs. | Writing the roots as \(\pm 1 \pm 4i\) (four numbers) is wrong; there are exactly two. |
Complex roots and square roots: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
One root of a quadratic equation with real coefficients is \(4 - i\). Find the equation in the form \(z^{2} + bz + c = 0\).
Find the two square roots of \(21 + 20i\), giving your answers in the form \(x + iy\) with \(x\) and \(y\) exact.
Show that \(z = -1 + i\) is a root of \(z^{3} + 4z^{2} + 6z + 4 = 0\), and find the other two roots.
Given that \(2 + i\) is a root of \(z^{3} + pz^{2} + qz - 10 = 0\), where \(p\) and \(q\) are real, find \(p\) and \(q\) and the other two roots.
(a) Find the square roots of \(-3 - 4i\). (b) Hence solve \(z^{2} - 3z + 3 + i = 0\).
(a) Given that \(2i\) is a root of \(z^{4} + 2z^{3} + 6z^{2} + 8z + 8 = 0\), find all four roots. (b) The roots are represented on an Argand diagram. Find the area of the quadrilateral with these four points as vertices.
Answer key and marking notes
- The other root is \(4 + i\). Sum \(= 8\), product \(= 16 + 1 = 17\).
- \(z^{2} - 8z + 17 = 0\).
- \(x^{2} - y^{2} = 21\) and \(2xy = 20\), so \(y = \dfrac{10}{x}\).
- \(x^{2} - \dfrac{100}{x^{2}} = 21\), so \(x^{4} - 21x^{2} - 100 = 0\), \((x^{2} - 25)(x^{2} + 4) = 0\).
- \(x^{2} = 25\): \(x = 5\), \(y = 2\) or \(x = -5\), \(y = -2\).
- Square roots \(5 + 2i\) and \(-5 - 2i\).
- \((-1 + i)^{2} = 1 - 2i - 1 = -2i\); \((-1 + i)^{3} = -2i(-1 + i) = 2i + 2\).
- \((2 + 2i) + 4(-2i) + 6(-1 + i) + 4 = (2 - 6 + 4) + (2 - 8 + 6)i = 0\) ✓.
- \(-1 - i\) is also a root; factor \(z^{2} + 2z + 2\).
- \(z^{3} + 4z^{2} + 6z + 4 = (z^{2} + 2z + 2)(z + 2)\), so the roots are \(-1 - i\) and \(-2\).
- \((2 + i)^{2} = 3 + 4i\), \((2 + i)^{3} = (2 + i)(3 + 4i) = 2 + 11i\).
- \((2 + 11i) + p(3 + 4i) + q(2 + i) - 10 = 0\): real \(3p + 2q - 8 = 0\), imaginary \(4p + q + 11 = 0\).
- \(q = -11 - 4p\), so \(3p - 22 - 8p - 8 = 0\), \(p = -6\), \(q = 13\).
- \(z^{3} - 6z^{2} + 13z - 10 = (z^{2} - 4z + 5)(z - 2)\): the other roots are \(2 - i\) and \(2\).
- (a) \(x^{2} - y^{2} = -3\), \(xy = -2\): \(x^{4} + 3x^{2} - 4 = 0\), \(x^{2} = 1\). Roots \(\pm(1 - 2i)\).
- (b) Discriminant \(9 - 4(3 + i) = -3 - 4i\).
- \(z = \dfrac{3 \pm (1 - 2i)}{2}\): \(z = 2 - i\) or \(z = 1 + i\).
- (a) \(-2i\) is also a root, so \(z^{2} + 4\) is a factor.
- \(z^{4} + 2z^{3} + 6z^{2} + 8z + 8 = (z^{2} + 4)(z^{2} + 2z + 2)\) (constants \(4 \times 2 = 8\) ✓, \(z^{3}\): 2 ✓).
- \(z^{2} + 2z + 2 = 0\) gives \(z = -1 \pm i\). Roots: \(\pm 2i\), \(-1 \pm i\).
- (b) Vertices \((0, 2)\), \((-1, 1)\), \((-1, -1)\), \((0, -2)\): a trapezium with parallel vertical sides 4 and 2, a distance 1 apart.
- Area \(= \tfrac{1}{2}(4 + 2) \times 1 = 3\) square units.
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.
Solve the equation \(z^{2} - 6z + 13 = 0\).
- A\(z = 3 \pm 2i\)
- B\(z = -3 \pm 2i\)
- C\(z = 6 \pm 4i\)
- D\(z = 3 \pm \sqrt{22}\)
A cubic equation with REAL coefficients has a root \(2 - i\). Which of these must also be a root?
- A\(2 + i\)
- B\(-2 + i\)
- C\(1 - 2i\)
- D\(-2 - i\)
Find the quadratic equation with real coefficients whose roots are \(1 + 3i\) and \(1 - 3i\).
- A\(z^{2} - 2z + 10 = 0\)
- B\(z^{2} + 2z + 10 = 0\)
- C\(z^{2} - 2z - 8 = 0\)
- D\(z^{2} - 2z + 4 = 0\)
Find the square roots of \(3 + 4i\).
- A\(\pm(2 + i)\)
- B\(\pm(1 + 2i)\)
- C\(\pm(2 - i)\)
- D\(\pm(\sqrt{3} + 2i)\)
Given that \(1 + 2i\) is a root of \(z^{3} - 4z^{2} + 9z - 10 = 0\), find the other two roots.
- A\(1 - 2i\) and \(2\)
- B\(1 - 2i\) and \(-2\)
- C\(-1 + 2i\) and \(2\)
- D\(1 - 2i\) and \(5\)
The equation \(z^{3} + az^{2} + bz + 15 = 0\), where \(a\) and \(b\) are real, has a root \(1 - 2i\). Find \(a\) and \(b\).
- A\(a = 1,\ b = -1\)
- B\(a = -5,\ b = 11\)
- C\(a = -1,\ b = 1\)
- D\(a = 3,\ b = -6\)
Solve \(z^{2} - (2 + 4i)z - 3 + 2i = 0\). (You may use \(\sqrt{8i} = \pm(2 + 2i)\).)
- A\(z = 2 + 3i\) or \(z = i\)
- B\(z = 1 + \sqrt{2} + 2i\) or \(z = 1 - \sqrt{2} + 2i\)
- C\(z = 2 + 3i\) or \(z = -i\)
- D\(z = -2 - 3i\) or \(z = -i\)
A quartic equation with real coefficients has roots \(2i\) and \(1 + i\). Which is the quartic?
- A\((z^{2} + 4)(z^{2} - 2z + 2) = 0\)
- B\((z^{2} - 4)(z^{2} - 2z + 2) = 0\)
- C\((z^{2} + 4)(z^{2} + 2z + 2) = 0\)
- D\((z - 2i)(z - 1 - i)(z^{2} + 2) = 0\)
Quiz key
\(z = \dfrac{6 \pm \sqrt{36 - 52}}{2} = \dfrac{6 \pm 4i}{2} = 3 \pm 2i\). Check: \((z - 3)^{2} = -4\). The tempting \(-3 \pm 2i\) uses \(+b\) instead of \(-b\) in the formula. \(6 \pm 4i\) forgets to divide by \(2a = 2\), and \(3 \pm \sqrt{22}\) adds \(36 + 52\) instead of subtracting.
The conjugate of \(2 - i\) is \(2 + i\), and it must also be a root, because the coefficients are real. The tempting \(-2 + i\) is \(-(2 - i)\), the negative, not the conjugate; only the sign of the imaginary part changes. \(1 - 2i\) swaps the real and imaginary parts, and \(-2 - i\) changes the wrong sign.
Sum \(= 2\), product \(= (1 + 3i)(1 - 3i) = 1 + 9 = 10\), so \(z^{2} - 2z + 10 = 0\). The tempting \(+2z\) forgets the minus sign in front of the sum. \(-8\) computes the product as \(1 - 9\), and \(+4\) adds 1 and 3 instead of their squares.
\(xy = 2\) and \(x^{2} - y^{2} = 3\): substituting \(y = \dfrac{2}{x}\) gives \(x^{4} - 3x^{2} - 4 = 0\), so \(x^{2} = 4\), \(x = \pm 2\), \(y = \pm 1\). The roots are \(\pm(2 + i)\); check \((2 + i)^{2} = 4 + 4i - 1 = 3 + 4i\). The tempting \(\pm(1 + 2i)\) swaps \(x\) and \(y\): \((1 + 2i)^{2} = -3 + 4i\). \((2 - i)^{2} = 3 - 4i\), the conjugate, and \(\sqrt{3} + 2i\) takes square roots of each part separately.
\(1 - 2i\) is also a root, so \(z^{2} - 2z + 5\) is a factor. Comparing constant terms, \(5 \times (-2) = -10\), so \(z^{3} - 4z^{2} + 9z - 10 = (z^{2} - 2z + 5)(z - 2)\) and the real root is \(2\). The tempting \(-2\) reads the factor \((z - 2)\) as giving root \(-2\). \(-1 + 2i\) is the negative of the conjugate, and \(5\) is the constant term of the quadratic factor, not a root.
\((z - 1 + 2i)(z - 1 - 2i) = z^{2} - 2z + 5\). For constant term 15 the other factor is \((z + 3)\): \((z^{2} - 2z + 5)(z + 3) = z^{3} + z^{2} - z + 15\), so \(a = 1\), \(b = -1\) (the real root is \(-3\)). The tempting \(a = -5,\ b = 11\) uses \((z - 3)\), which gives constant term \(-15\). \(a = -1,\ b = 1\) has both signs reversed, and \(a = 3,\ b = -6\) collects only some of the terms in the expansion.
\(b^{2} - 4ac = (2 + 4i)^{2} - 4(-3 + 2i) = (-12 + 16i) + 12 - 8i = 8i\). So \(z = \dfrac{2 + 4i \pm (2 + 2i)}{2}\), giving \(2 + 3i\) or \(i\). Check: the sum is \(2 + 4i\) and the product \(i(2 + 3i) = -3 + 2i\). The tempting \(1 \pm \sqrt{2} + 2i\) takes \(\sqrt{8i}\) as \(2\sqrt{2}\), dropping the \(i\). \(-i\) mishandles the subtraction \((2 + 4i) - (2 + 2i)\), and \(-2 - 3i\), \(-i\) uses \(+b\) instead of \(-b\).
\((z - 2i)(z + 2i) = z^{2} + 4\) and \((z - 1 - i)(z - 1 + i) = (z - 1)^{2} + 1 = z^{2} - 2z + 2\). The quartic is \((z^{2} + 4)(z^{2} - 2z + 2) = z^{4} - 2z^{3} + 6z^{2} - 8z + 8 = 0\). The tempting \(z^{2} - 4\) treats \((2i)^{2}\) as \(+4\), giving real roots \(\pm 2\). \(z^{2} + 2z + 2\) has roots \(-1 \pm i\), and the last option keeps only one of each conjugate pair, so its coefficients are not real.
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.