| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Verifying roots satisfy equations |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question testing standard complex number techniques: expanding a complex cube (routine algebra), substituting to verify a root (direct calculation), and factorising using the conjugate root theorem. While it requires multiple steps and careful arithmetic, it involves no novel insight—just methodical application of well-practiced procedures. Slightly easier than average due to its predictable structure. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2^3+3\times2^2\times3\text{i}+3\times2\times(3\text{i})^2+(3\text{i})^3\) | M1 | Binomial expansion. Must be 4 terms with \(1,3,3,1\) and correct powers. Condone missing brackets. Or by \((2+3\text{i})^2\times(2+3\text{i})\) but marks only awarded once all binomial brackets expanded. |
| \(2^3+3\times2^2\times3\text{i}-3\times2\times3^2-3^3\text{i}\) or better | A1 | All correct and \(\text{i}^2=-1\) twice. Must see evidence of \(\text{i}^2\) becoming \(-1\). |
| \(-46+9\text{i}\) | A1 | SC if only working seen is \((-5+12\text{i})(2+3\text{i})=-46+9\text{i}\) award B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2+3\text{i})^2=-5+12\text{i}\) | B1 | May be seen in (i). If only seen in (i) and not implied by working for this part award B0. |
| \(3(-46+9\text{i})-8(-5+12\text{i})+23(2+3\text{i})+52\) | M1 | Attempt to substitute their \(z^2\) and \(z^3\) into \(3z^3-8z^2+23z+52\). \(3(2+3\text{i})^3-8(2+3\text{i})^2+23(2+3\text{i})+52\) enough for M1. |
| \(=-138+27\text{i}+40-96\text{i}+46+69\text{i}+52\) \(=-138+138+96\text{i}-96\text{i}=0\) | A1 (AG) | Convincingly cancels to 0. Must show some collection/cancellation. Could be gathering real and imaginary terms. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2-3\text{i}\) is also a root | B1 | Seen or implied |
| \((z-(2+3\text{i}))(z-(2-3\text{i}))\) | M1 | ...is the required quadratic factor |
| \(=z^2-4z+13\) | A1 | |
| \((z^2-4z+13)(3z+4)\) | A1 | Can be deduced by inspection. Must be written as product of linear factor and quadratic factor. Condone "\(=0\)" present. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2-3\text{i}\) is also a root | B1 | Seen or implied |
| \(z-2=\pm3\text{i}\), \((z-2)^2=-9\) | M1 | Using the two roots to get an equation in \(z^2\) |
| \(z^2-4z+13\ (=0)\) | A1 |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2^3+3\times2^2\times3\text{i}+3\times2\times(3\text{i})^2+(3\text{i})^3$ | M1 | Binomial expansion. Must be 4 terms with $1,3,3,1$ and correct powers. Condone missing brackets. Or by $(2+3\text{i})^2\times(2+3\text{i})$ but marks only awarded once all binomial brackets expanded. |
| $2^3+3\times2^2\times3\text{i}-3\times2\times3^2-3^3\text{i}$ or better | A1 | All correct and $\text{i}^2=-1$ twice. Must see evidence of $\text{i}^2$ becoming $-1$. |
| $-46+9\text{i}$ | A1 | SC if only working seen is $(-5+12\text{i})(2+3\text{i})=-46+9\text{i}$ award B1 |
**[3 marks total]**
---
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2+3\text{i})^2=-5+12\text{i}$ | B1 | May be seen in (i). If only seen in (i) and not implied by working for this part award B0. |
| $3(-46+9\text{i})-8(-5+12\text{i})+23(2+3\text{i})+52$ | M1 | Attempt to substitute their $z^2$ and $z^3$ into $3z^3-8z^2+23z+52$. $3(2+3\text{i})^3-8(2+3\text{i})^2+23(2+3\text{i})+52$ enough for M1. |
| $=-138+27\text{i}+40-96\text{i}+46+69\text{i}+52$ $=-138+138+96\text{i}-96\text{i}=0$ | A1 (AG) | Convincingly cancels to 0. Must show some collection/cancellation. Could be gathering real and imaginary terms. |
**[3 marks total]**
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2-3\text{i}$ is also a root | B1 | Seen or implied |
| $(z-(2+3\text{i}))(z-(2-3\text{i}))$ | M1 | ...is the required quadratic factor |
| $=z^2-4z+13$ | A1 | |
| $(z^2-4z+13)(3z+4)$ | A1 | Can be deduced by inspection. Must be written as product of linear factor and quadratic factor. Condone "$=0$" present. |
**Alt:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2-3\text{i}$ is also a root | B1 | Seen or implied |
| $z-2=\pm3\text{i}$, $(z-2)^2=-9$ | M1 | Using the two roots to get an equation in $z^2$ |
| $z^2-4z+13\ (=0)$ | A1 | |
**[4 marks total]**5 In this question you must show detailed reasoning.\\
(i) Express $( 2 + 3 \mathrm { i } ) ^ { 3 }$ in the form $a + \mathrm { i } b$.\\
(ii) Hence verify that $2 + 3$ i is a root of the equation $3 z ^ { 3 } - 8 z ^ { 2 } + 23 z + 52 = 0$.\\
(iii) Express $3 z ^ { 3 } - 8 z ^ { 2 } + 23 z + 52$ as the product of a linear factor and a quadratic factor with real coefficients.
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q5 [10]}}