| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Given one complex root of cubic or quartic, find all roots |
| Difficulty | Standard +0.3 This is a standard Further Maths FP1 question on complex roots of polynomials. Part (i) requires routine complex arithmetic and substitution. Part (ii) uses the conjugate root theorem and factorization or sum of roots. Part (iii) is straightforward plotting. While it's Further Maths content (making it harder than typical A-level), it follows a very standard template with no novel problem-solving required, placing it slightly above average overall. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02g Conjugate pairs: real coefficient polynomials4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha^2 = (1+\text{j})^2 = 1 + 2\text{j} - 1 = 2\text{j}\) | B1 | |
| \(\alpha^3 = \alpha^2 \cdot \alpha = 2\text{j}(1+\text{j}) = 2\text{j} - 2 = -2 + 2\text{j}\) | B1 | |
| Substituting into \(z^3 + 3z^2 + pz + q = 0\): | M1 | |
| Real: \(-2 + 3(0) + p(1) + q = 0 \Rightarrow p + q = 2\) ... wait, real part: \(-2 + 3(0) + p + q = 0\) | ||
| Real part: \(-2 + 0 + p + q = 0\); Imaginary part: \(2 + 6 + p = 0 \Rightarrow p = -8\) | A1 | |
| \(q = 2 - p = 10\) | A1 | both values shown clearly |
| Answer | Marks |
|---|---|
| Since coefficients are real, \(\alpha^* = 1 - \text{j}\) is also a root | B1 |
| Sum of all roots \(= -3\) (from equation), so third root \(= -3 - (1+\text{j}) - (1-\text{j}) = -5\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Argand diagram showing points \((1,1)\), \((1,-1)\), \((-5, 0)\) correctly plotted | B1 B1 | B1 for conjugate pair symmetric about real axis, B1 for \(-5\) |
# Question 8:
**(i)**
$\alpha^2 = (1+\text{j})^2 = 1 + 2\text{j} - 1 = 2\text{j}$ | B1 |
$\alpha^3 = \alpha^2 \cdot \alpha = 2\text{j}(1+\text{j}) = 2\text{j} - 2 = -2 + 2\text{j}$ | B1 |
Substituting into $z^3 + 3z^2 + pz + q = 0$: | M1 |
Real: $-2 + 3(0) + p(1) + q = 0 \Rightarrow p + q = 2$ ... wait, real part: $-2 + 3(0) + p + q = 0$ |
Real part: $-2 + 0 + p + q = 0$; Imaginary part: $2 + 6 + p = 0 \Rightarrow p = -8$ | A1 |
$q = 2 - p = 10$ | A1 | both values shown clearly
**(ii)**
Since coefficients are real, $\alpha^* = 1 - \text{j}$ is also a root | B1 |
Sum of all roots $= -3$ (from equation), so third root $= -3 - (1+\text{j}) - (1-\text{j}) = -5$ | M1 A1 |
**(iii)**
Argand diagram showing points $(1,1)$, $(1,-1)$, $(-5, 0)$ correctly plotted | B1 B1 | B1 for conjugate pair symmetric about real axis, B1 for $-5$
---8 You are given that the complex number $\alpha = 1 + \mathrm { j }$ satisfies the equation $z ^ { 3 } + 3 z ^ { 2 } + p z + q = 0$, where $p$ and $q$ are real constants.\\
(i) Find $\alpha ^ { 2 }$ and $\alpha ^ { 3 }$ in the form $a + b \mathrm { j }$. Hence show that $p = - 8$ and $q = 10$.\\
(ii) Find the other two roots of the equation.\\
(iii) Represent the three roots on an Argand diagram.
\hfill \mbox{\textit{OCR MEI FP1 2006 Q8 [11]}}