AS June 2023 Paper 1 Q12
12
(a) Show that \((1 + \mathrm{i})^4 = -4\) [3 marks]
(b) The function f is defined by\[\mathrm{f}(z) = z^4 + 3z^2 - 6z + 10 \qquad z \in \mathbb{C}\]
(i) Show that \((1 + \mathrm{i})\) is a root of \(\mathrm{f}(z) = 0\) [2 marks]
(ii) Hence write down another root of \(\mathrm{f}(z) = 0\) [1 mark]
(iii) One of the linear factors of \(\mathrm{f}(z)\) is\[\big(z - (1 + \mathrm{i})\big)\]
Write down another linear factor and hence, or otherwise, find a quadratic factor of \(\mathrm{f}(z)\) with real coefficients. [3 marks]
(iv) Find another quadratic factor of \(\mathrm{f}(z)\) with real coefficients. [2 marks]
(v) Hence explain why the graph of \(y = \mathrm{f}(x)\) does not intersect the \(x\)-axis. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Applies the binomial expansion to \((1 + \mathrm{i})^4\) or \((1 + \mathrm{i})^3\) Allow one incorrect term. Or \((1 + \mathrm{i})^2 = 1 + 2\mathrm{i} + \mathrm{i}^2\) (or just \(2\mathrm{i}\)) | M1 | 1.1a |
| Replaces \(\mathrm{i}^2\) with \(-1\), or \(\mathrm{i}^3\) with \(-\mathrm{i}\), or \(\mathrm{i}^4\) with 1 | B1 | 1.2 |
| Completes a reasoned argument to reach the required result. Must include the LHS, at least two intermediate steps, and the RHS. Accept \((1 + \mathrm{i})^2\) replaced with \(2\mathrm{i}\) without explanation. | R1 | 2.1 |
| (3) |
Typical solution
\[\begin{aligned} &(1 + \mathrm{i})^4 \\ &= 1^4 + 4 \cdot 1^3 \cdot \mathrm{i} + 6 \cdot 1^2 \cdot \mathrm{i}^2 + 4 \cdot 1 \cdot \mathrm{i}^3 + \mathrm{i}^4 \\ &= 1 + 4\mathrm{i} - 6 - 4\mathrm{i} + 1 \\ &= -4\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes \((1 + \mathrm{i})\) into f | M1 | 1.1a |
| Equates to 0 and concludes that \((1 + \mathrm{i})\) is a root. | R1 | 2.1 |
| (2) | ||
| (ii) Identifies \(1 - \mathrm{i}\) as a root. Accept \(-1 + 2\mathrm{i}\) or \(-1 - 2\mathrm{i}\) | B1 | 1.2 |
| (1) | ||
| (iii) Identifies a correct linear factor. Accept \(\big(z - (-1 + 2\mathrm{i})\big)\) or \(\big(z - (-1 - 2\mathrm{i})\big)\) Follow through their part (b)(ii). | B1F | 1.1b |
| Forms the product \((z - w)(z - w^*)\) for any non-real \(w\) | M1 | 3.1a |
| Obtains \(z^2 - 2z + 2\) Accept \(z^2 + 2z + 5\) | A1 | 1.1b |
| (3) | ||
| (iv) Obtains a second quadratic factor of \(\mathrm{f}(z)\) with at least two correct terms. | M1 | 3.1a |
| Obtains a correct second quadratic factor. Accept \(z^2 - 2z + 2\) if \(z^2 + 2z + 5\) is the answer to their part (b)(iii) | A1 | 1.1b |
| (2) | ||
| (v) Explains that \(\mathrm{f}(z) = 0\) has no real roots. Condone “no real roots” with an incorrect or no other statement. | M1 | 2.4 |
| Completes a reasoned argument to conclude that \(y = \mathrm{f}(x)\) does not intersect the \(x\)-axis. | R1 | 2.1 |
| (2) | ||
| (13 marks) |
Typical solution
(i)
\[\mathrm{f}(1 + \mathrm{i}) = (1 + \mathrm{i})^4 + 3(1 + \mathrm{i})^2 - 6(1 + \mathrm{i}) + 10 = 0\]\(\therefore\) \((1 + \mathrm{i})\) is a root of \(\mathrm{f}(z) = 0\)
(ii)
\[1 - \mathrm{i}\](iii)
2nd linear factor is \(\big(z - (1 - \mathrm{i})\big)\)
Quadratic factor is
\[\begin{aligned} &\big(z - (1 + \mathrm{i})\big)\big(z - (1 - \mathrm{i})\big) \\ &= z^2 - (1 - \mathrm{i})z - (1 + \mathrm{i})z + (1 + \mathrm{i})(1 - \mathrm{i}) \\ &= z^2 - (1 - \mathrm{i} + 1 + \mathrm{i})z + 1 - \mathrm{i}^2 \\ &= z^2 - 2z + 2\end{aligned}\](iv)
\[\begin{aligned} &z^4 + 3z^2 - 6z + 10 \\ &= (z^2 - 2z + 2)(z^2 + 2z + 5)\end{aligned}\]2nd quadratic factor is \(z^2 + 2z + 5\)
(v)
\(\mathrm{f}(z) = 0\) has no real roots.
Hence \(y = \mathrm{f}(x)\) does not intersect the \(x\)-axis.