OCR MEI FP1 2015 June — Question 8 12 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeMultiplication and powers of complex numbers
DifficultyStandard +0.3 This is a straightforward Further Pure 1 question on complex number arithmetic and polynomial equations. Part (i) requires routine multiplication of complex numbers, part (ii) involves simple substitution and equating real/imaginary parts, part (iii) uses the conjugate root theorem (standard FP1 technique), and part (iv) is algebraic manipulation. While it's a multi-part question requiring several steps, each component uses standard textbook methods with no novel insight required. Being FP1 material makes it slightly above average A-level difficulty, but it remains a routine exercise.
Spec4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02j Cubic/quartic equations: conjugate pairs and factor theorem
8 The complex number \(5 + 4 \mathrm { j }\) is denoted by \(\alpha\).
  1. Find \(\alpha ^ { 2 }\) and \(\alpha ^ { 3 }\), showing your working.
  2. The real numbers \(q\) and \(r\) are such that \(\alpha ^ { 3 } + \mathrm { q } \alpha ^ { 2 } + 11 \alpha + \mathrm { r } = 0\). Find \(q\) and \(r\). Let \(\mathrm { f } ( \mathrm { z } ) = \mathrm { z } ^ { 3 } + \mathrm { qz } ^ { 2 } + 11 \mathrm { z } + \mathrm { r }\), where \(q\) and \(r\) are as in part (ii).
  3. Solve the equation \(\mathrm { f } ( z ) = 0\).
  4. Solve the equation \(z ^ { 4 } + q z ^ { 3 } + 11 z ^ { 2 } + r z = z ^ { 3 } + q z ^ { 2 } + 11 z + r\).
Question 8:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
\((5+4j)^2 = (5+4j)(5+4j) = 25 + 40j - 16 = 9 + 40j\)M1 Use of \(j^2 = -1\) at least once
\((5+4j)^3 = -115 + 236j\)A1, A1
[3]
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha^3 + q\alpha^2 + 11\alpha + r = 0\)
\(\Rightarrow -115 + 236j + 9q + 40qj + 55 + 44j + r = 0\)M1 Substitute for \(\alpha\)
\(\Rightarrow (236 + 40q + 44)j = 0\), \(-115 + 9q + 55 + r = 0\)M1 Compare either real or imaginary parts
\(\Rightarrow q = -7\)A1ft \(q = -7\) ft their \(\alpha^2\) and \(\alpha^3\)
\(\Rightarrow r = 123\)A1ft \(r = 123\) ft their \(\alpha^2\) and \(\alpha^3\)
[4]
Part (iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(f(z) = z^3 - 7z^2 + 11z + 123\)
Sum of roots \(= 7\)M1 Valid method for the third root (division, factor theorem, attempt at linear \(\times\) quadratic with complex roots correctly used)
\(\Rightarrow (5+4j) + (5-4j) + w = 7\)
\(\Rightarrow w = -3\)
Roots are \(5+4j\) and \(5-4j\)B1 quoted
and \(-3\)A1 cao real root identified, A0 if extra roots found
[3]
Part (iv):
AnswerMarks Guidance
AnswerMarks Guidance
\(zf(z) = f(z) \Rightarrow (z-1)f(z) = 0\)
\(\Rightarrow z = 1\) or \(f(z) = 0\)M1 Solving \(z - 1 = 0\) and \(f(z) = 0\) (may be implied)
\(\Rightarrow z = 1, z = -3, z = 5+4j, z = 5-4j\)A1ft For all four solutions [ft (iii)]. NB incomplete method giving \(z=1\) only is M0 A0
[2] 
## Question 8:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(5+4j)^2 = (5+4j)(5+4j) = 25 + 40j - 16 = 9 + 40j$ | M1 | Use of $j^2 = -1$ at least once |
| $(5+4j)^3 = -115 + 236j$ | A1, A1 | |
| | **[3]** | |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^3 + q\alpha^2 + 11\alpha + r = 0$ | | |
| $\Rightarrow -115 + 236j + 9q + 40qj + 55 + 44j + r = 0$ | M1 | Substitute for $\alpha$ |
| $\Rightarrow (236 + 40q + 44)j = 0$, $-115 + 9q + 55 + r = 0$ | M1 | Compare either real or imaginary parts |
| $\Rightarrow q = -7$ | A1ft | $q = -7$ ft their $\alpha^2$ and $\alpha^3$ |
| $\Rightarrow r = 123$ | A1ft | $r = 123$ ft their $\alpha^2$ and $\alpha^3$ |
| | **[4]** | |

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(z) = z^3 - 7z^2 + 11z + 123$ | | |
| Sum of roots $= 7$ | M1 | Valid method for the third root (division, factor theorem, attempt at linear $\times$ quadratic with complex roots correctly used) |
| $\Rightarrow (5+4j) + (5-4j) + w = 7$ | | |
| $\Rightarrow w = -3$ | | |
| Roots are $5+4j$ and $5-4j$ | B1 | quoted |
| and $-3$ | A1 | cao real root identified, A0 if extra roots found |
| | **[3]** | |

### Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $zf(z) = f(z) \Rightarrow (z-1)f(z) = 0$ | | |
| $\Rightarrow z = 1$ or $f(z) = 0$ | M1 | Solving $z - 1 = 0$ and $f(z) = 0$ (may be implied) |
| $\Rightarrow z = 1, z = -3, z = 5+4j, z = 5-4j$ | A1ft | For all four solutions [ft (iii)]. NB incomplete method giving $z=1$ only is M0 A0 |
| | **[2]** | |
8 The complex number $5 + 4 \mathrm { j }$ is denoted by $\alpha$.\\
(i) Find $\alpha ^ { 2 }$ and $\alpha ^ { 3 }$, showing your working.\\
(ii) The real numbers $q$ and $r$ are such that $\alpha ^ { 3 } + \mathrm { q } \alpha ^ { 2 } + 11 \alpha + \mathrm { r } = 0$. Find $q$ and $r$.

Let $\mathrm { f } ( \mathrm { z } ) = \mathrm { z } ^ { 3 } + \mathrm { qz } ^ { 2 } + 11 \mathrm { z } + \mathrm { r }$, where $q$ and $r$ are as in part (ii).\\
(iii) Solve the equation $\mathrm { f } ( z ) = 0$.\\
(iv) Solve the equation $z ^ { 4 } + q z ^ { 3 } + 11 z ^ { 2 } + r z = z ^ { 3 } + q z ^ { 2 } + 11 z + r$.

\hfill \mbox{\textit{OCR MEI FP1 2015 Q8 [12]}}
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