| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Verifying roots satisfy equations |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through standard complex number techniques: verifying a power of a complex number, substituting to verify a root, using conjugate root theorem, forming quadratic factors from complex conjugate pairs, and interpreting discriminants. While it requires multiple steps, each part is routine for Further Maths students with clear scaffolding throughout. The techniques are all standard textbook exercises with no novel problem-solving required. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+i)^4 = 1^4 + 4\cdot1^3\cdot i + 6\cdot1^2\cdot i^2 + 4\cdot1\cdot i^3 + i^4\) | M1 | Applies binomial expansion to \((1+i)^4\) or \((1+i)^3\); allow one incorrect term; or \((1+i)^2 = 1+2i+i^2\) |
| \(= 1 + 4i - 6 - 4i + 1\) | B1 | Replaces \(i^2\) with \(-1\), or \(i^3\) with \(-i\), or \(i^4\) with \(1\) |
| \(= -4\) | R1 | Completes reasoned argument; must include LHS, at least two intermediate steps, and RHS |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1+i) = (1+i)^4 + 3(1+i)^2 - 6(1+i) + 10 = 0\) | M1 | Substitutes \((1+i)\) into \(f\) |
| \(\therefore (1+i)\) is a root of \(f(z) = 0\) | R1 | Equates to 0 and concludes \((1+i)\) is a root |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - i\) | B1 | Identifies \(1-i\) as a root; accept \(-1+2i\) or \(-1-2i\) |
| Answer | Marks | Guidance |
|---|---|---|
| 2nd linear factor is \((z-(1-i))\) | B1F | Identifies a correct linear factor; accept \((z-(-1+2i))\) or \((z-(-1-2i))\); follow through from (b)(ii) |
| \((z-(1+i))(z-(1-i)) = z^2-(1-i)z-(1+i)z+(1+i)(1-i)\) \(= z^2-(1-i+1+i)z+1-i^2 = z^2-2z+2\) | M1 | Forms product \((z-w)(z-w^*)\) for any non-real \(w\) |
| \(z^2 - 2z + 2\) (accept \(z^2+2z+5\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^4+3z^2-6z+10 = (z^2-2z+2)(z^2+2z+5)\) | M1 | Obtains a second quadratic factor of \(f(z)\) with at least two correct terms |
| 2nd quadratic factor is \(z^2+2z+5\) | A1 | Accept \(z^2-2z+2\) if \(z^2+2z+5\) is the answer to (b)(iii) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(z) = 0\) has no real roots | M1 | Explains that \(f(z)=0\) has no real roots; condone "no real roots" with an incorrect or no other statement |
| Hence \(y = f(x)\) does not intersect the \(x\)-axis | R1 | Completes reasoned argument |
## Question 12:
**Part 12(a):**
| $(1+i)^4 = 1^4 + 4\cdot1^3\cdot i + 6\cdot1^2\cdot i^2 + 4\cdot1\cdot i^3 + i^4$ | M1 | Applies binomial expansion to $(1+i)^4$ or $(1+i)^3$; allow one incorrect term; or $(1+i)^2 = 1+2i+i^2$ |
| $= 1 + 4i - 6 - 4i + 1$ | B1 | Replaces $i^2$ with $-1$, or $i^3$ with $-i$, or $i^4$ with $1$ |
| $= -4$ | R1 | Completes reasoned argument; must include LHS, at least two intermediate steps, and RHS |
**Part 12(b)(i):**
| $f(1+i) = (1+i)^4 + 3(1+i)^2 - 6(1+i) + 10 = 0$ | M1 | Substitutes $(1+i)$ into $f$ |
| $\therefore (1+i)$ is a root of $f(z) = 0$ | R1 | Equates to 0 and concludes $(1+i)$ is a root |
**Part 12(b)(ii):**
| $1 - i$ | B1 | Identifies $1-i$ as a root; accept $-1+2i$ or $-1-2i$ |
**Part 12(b)(iii):**
| 2nd linear factor is $(z-(1-i))$ | B1F | Identifies a correct linear factor; accept $(z-(-1+2i))$ or $(z-(-1-2i))$; follow through from (b)(ii) |
| $(z-(1+i))(z-(1-i)) = z^2-(1-i)z-(1+i)z+(1+i)(1-i)$ $= z^2-(1-i+1+i)z+1-i^2 = z^2-2z+2$ | M1 | Forms product $(z-w)(z-w^*)$ for any non-real $w$ |
| $z^2 - 2z + 2$ (accept $z^2+2z+5$) | A1 | |
**Part 12(b)(iv):**
| $z^4+3z^2-6z+10 = (z^2-2z+2)(z^2+2z+5)$ | M1 | Obtains a second quadratic factor of $f(z)$ with at least two correct terms |
| 2nd quadratic factor is $z^2+2z+5$ | A1 | Accept $z^2-2z+2$ if $z^2+2z+5$ is the answer to (b)(iii) |
**Part 12(b)(v):**
| $f(z) = 0$ has no real roots | M1 | Explains that $f(z)=0$ has no real roots; condone "no real roots" with an incorrect or no other statement |
| Hence $y = f(x)$ does not intersect the $x$-axis | R1 | Completes reasoned argument |
---12
\begin{enumerate}[label=(\alph*)]
\item Show that $( 1 + i ) ^ { 4 } = - 4$\\
12
\item The function f is defined by
$$f ( z ) = z ^ { 4 } + 3 z ^ { 2 } - 6 z + 10 \quad z \in \mathbb { C }$$
12 (b) (i) Show that (1+i) is a root of $\mathrm { f } ( \mathrm { z } ) = 0$\\
12 (b) (ii) Hence write down another root of $\mathrm { f } ( \mathrm { z } ) = 0$\\
12 (b) (iii) One of the linear factors of $\mathrm { f } ( \mathrm { z } )$ is
$$( z - ( 1 + i ) )$$
Write down another linear factor and hence, or otherwise, find a quadratic factor of $\mathrm { f } ( \mathrm { z } )$ with real coefficients.\\
12 (b) (iv) Find another quadratic factor of $\mathrm { f } ( \mathrm { z } )$ with real coefficients.\\
12 (b) (v) Hence explain why the graph of $y = \mathrm { f } ( x )$ does not intersect the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q12 [13]}}