| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| 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 Pure 1 question testing standard complex number operations: addition, conjugate multiplication, division (requiring rationalization), forming quadratics from roots, and extending to quartics. All techniques are routine for FP1 students with no novel problem-solving required. While it's multi-part and from further maths (inherently harder), the individual steps are mechanical applications of learned procedures, placing it slightly above average difficulty overall. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha+\beta=3\) | B1 | |
| \(\alpha\alpha^*=(1+j)(1-j)=2\) | M1 | Attempt to multiply \((1+j)(1-j)\) |
| A1 | ||
| \(\frac{\alpha+\beta}{\alpha}=\frac{3}{1+j}=\frac{3(1-j)}{(1+j)(1-j)}=\frac{3}{2}-\frac{3}{2}j\) | M1 | Multiply top and bottom by \(1-j\) |
| A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \((z-(1+j))(z-(1-j))=z^2-2z+2\) | M1 | Or alternative valid methods (condone no "\(=0\)") |
| A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(1-j\) and \(2+j\) | B1 [for both] | |
| \((z-(2-j))(z-(2+j))=z^2-4z+5\) | M1 | Attempt to obtain equation using product of linear factors involving complex conjugates |
| \((z^2-2z+2)(z^2-4z+5)\) | M1 | Using the correct four factors |
| \(=z^4-6z^3+15z^2-18z+10\) | ||
| So equation is \(z^4-6z^3+15z^2-18z+10=0\) | A2 [5] | All correct, \(-1\) each error (including omission of "\(=0\)") to min of 0 |
| Alternative: Use of \(\sum\alpha=6\), \(\sum\alpha\beta=15\), \(\sum\alpha\beta\gamma=18\), \(\alpha\beta\gamma\delta=10\) | M1 | Use of relationships between roots and coefficients |
| A3 [5] | All correct, \(-1\) each error, to min of 0 |
# Question 9(i):
$\alpha+\beta=3$ | B1 |
$\alpha\alpha^*=(1+j)(1-j)=2$ | M1 | Attempt to multiply $(1+j)(1-j)$
| A1 |
$\frac{\alpha+\beta}{\alpha}=\frac{3}{1+j}=\frac{3(1-j)}{(1+j)(1-j)}=\frac{3}{2}-\frac{3}{2}j$ | M1 | Multiply top and bottom by $1-j$
| A1 [5] |
# Question 9(ii):
$(z-(1+j))(z-(1-j))=z^2-2z+2$ | M1 | Or alternative valid methods (condone no "$=0$")
| A1 [2] |
# Question 9(iii):
$1-j$ and $2+j$ | B1 [for both] |
$(z-(2-j))(z-(2+j))=z^2-4z+5$ | M1 | Attempt to obtain equation using product of linear factors involving complex conjugates
$(z^2-2z+2)(z^2-4z+5)$ | M1 | Using the correct four factors
$=z^4-6z^3+15z^2-18z+10$ | |
So equation is $z^4-6z^3+15z^2-18z+10=0$ | A2 [5] | All correct, $-1$ each error (including omission of "$=0$") to min of 0
**Alternative:** Use of $\sum\alpha=6$, $\sum\alpha\beta=15$, $\sum\alpha\beta\gamma=18$, $\alpha\beta\gamma\delta=10$ | M1 | Use of relationships between roots and coefficients
| A3 [5] | All correct, $-1$ each error, to min of 0
---9 Two complex numbers, $\alpha$ and $\beta$, are given by $\alpha = 1 + \mathrm { j }$ and $\beta = 2 - \mathrm { j }$.\\
(i) Express $\alpha + \beta , \alpha \alpha ^ { * }$ and $\frac { \alpha + \beta } { \alpha }$ in the form $a + b \mathrm { j }$.\\
(ii) Find a quadratic equation with roots $\alpha$ and $\alpha ^ { * }$.\\
(iii) $\alpha$ and $\beta$ are roots of a quartic equation with real coefficients. Write down the two other roots and find this quartic equation in the form $z ^ { 4 } + A z ^ { 3 } + B z ^ { 2 } + C z + D = 0$.
\hfill \mbox{\textit{OCR MEI FP1 2009 Q9 [12]}}