| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Given two complex roots, find all roots |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing understanding that complex conjugate root theorem only applies to polynomials with real coefficients. Part (a) requires recognizing the coefficients are complex (the -2i term), making it conceptually simple. Part (b) involves routine application of factor theorem and polynomial multiplication with complex numbers—standard techniques with no novel insight required. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Can only assume that complex roots of a polynomial equation are in conjugate pairs if the coefficients are all real. | B1 (AO2.3) | Gives a correct explanation. Condone explanation which suggests conjugate pairs only occur if coefficients are real. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1+i)+(-1)+\alpha = 2i\), so \(\alpha = i\) | M1 (AO3.1a) / A1 (AO1.1b) | Writes equation for 3rd root using sum of roots. Condone \(-2i\) or \(2\) or \(-2\) for sum. Obtains correct 3rd root. |
| \(p=(1+i)(-1)+(-1)i+i(1+i) = -1-i-i+i+i^2 = -2-i\) | M1 (AO3.1a) | Writes expression for \(p\) using sum of pairwise products with 3rd root and two given roots. Allow sign errors. |
| \(q=-(1+i)(-1)i = i+i^2 = i-1\) | M1 (AO3.1a) | Writes expression for \(q\) using product of roots with 3rd root and two given roots. Allow sign errors. |
| \(p=-2-i\) and \(q=i-1\) | A1 (AO1.1b) | Both correct. |
## Question 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Can only assume that complex roots of a polynomial equation are in conjugate pairs if the coefficients are all real. | B1 (AO2.3) | Gives a correct explanation. Condone explanation which suggests conjugate pairs only occur if coefficients are real. |
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## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1+i)+(-1)+\alpha = 2i$, so $\alpha = i$ | M1 (AO3.1a) / A1 (AO1.1b) | Writes equation for 3rd root using sum of roots. Condone $-2i$ or $2$ or $-2$ for sum. Obtains correct 3rd root. |
| $p=(1+i)(-1)+(-1)i+i(1+i) = -1-i-i+i+i^2 = -2-i$ | M1 (AO3.1a) | Writes expression for $p$ using sum of pairwise products with 3rd root and two given roots. Allow sign errors. |
| $q=-(1+i)(-1)i = i+i^2 = i-1$ | M1 (AO3.1a) | Writes expression for $q$ using product of roots with 3rd root and two given roots. Allow sign errors. |
| $p=-2-i$ **and** $q=i-1$ | A1 (AO1.1b) | Both correct. |
---8 Stephen is correctly told that $( 1 + \mathrm { i } )$ and - 1 are two roots of the polynomial equation
$$z ^ { 3 } - 2 \mathrm { i } z ^ { 2 } + p z + q = 0$$
where $p$ and $q$ are complex numbers.\\
8
\begin{enumerate}[label=(\alph*)]
\item Stephen states that ( $1 - \mathrm { i }$ ) must also be a root of the equation because roots of polynomial equations occur in conjugate pairs.
Explain why Stephen's reasoning is wrong.
8
\item $\quad$ Find $p$ and $q$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q8 [6]}}