| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Square roots with follow-up application |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard algebraic methods for finding square roots of complex numbers (equating real and imaginary parts) followed by a routine calculation. While it requires careful algebra and is harder than typical A-level pure maths, it's a textbook exercise with no novel insight required, making it slightly easier than average for Further Maths content. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02h Square roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR \((a+bi)^2 = a^2-b^2+2abi\) | B1 | Seen or implied in solution |
| \(a^2-b^2=-16\) and \(2ab=30\) (where \(a\) and \(b\) are real) | M1 | Comparing real and imaginary parts (no \(i\) unless later recovered) from a 3 (or 4) term expansion. Allow sign slips |
| \(b=\dfrac{15}{a} \Rightarrow a^2 - \left(\dfrac{15}{a}\right)^2 = -16 \Rightarrow a^4+16a^2-225=0\); \(a\) (\(b\)) real so \(a^2=9\) (or \(b^2=25\)) only | M1, A1 | Eliminating \(b\) or \(a\) to obtain 3 term quadratic in \(a^2\) or \(b^2\). Unknowns must not be in denominator. Must be an equation. Rogue solutions: \(a^2=-25\), \(b^2=-9\). Factored forms: \((b^4-16b^2-225=0)\); \((a^2-9)(a^2+25)\); \((b^2-25)(b^2+9)\) |
| \(3+5i\) and \(-3-5i\) | A1 | Both roots. Can be \(\pm(3+5i)\) but not \(\pm3\pm5i\). Note: 4/5 possible following B0 |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3+5i)^3 = (-16+30i)(3+5i) = -198+10i\) | M1 1.1 | Can be awarded for cubing an incorrect answer to 5(a). Could be done by expansion either in two steps or binomial: \(3^3 + 3\times3^2\times5i + 3\times3\times(5i)^2 + (5i)^3\). Do not need to see intermediate step before correct answer. If incorrect need to see proof of expanding three brackets. Or \((-3-5i)^3 = 198-10i...\). For binomial want to see 4 terms and either 2nd or 3rd term correct (up to sign error) |
| \(= \frac{-99+5i}{4} \times 2^3\) so \(\frac{3}{2}+\frac{5i}{2}\) | A1 [2] 3.1a | Correct answer must follow a correct root. \(= \frac{-99+5i}{4}\times(-2)^3\) |
# Question 5:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** $(a+bi)^2 = a^2-b^2+2abi$ | B1 | Seen or implied in solution |
| $a^2-b^2=-16$ and $2ab=30$ (where $a$ and $b$ are real) | M1 | Comparing real and imaginary parts (no $i$ unless later recovered) from a 3 (or 4) term expansion. Allow sign slips |
| $b=\dfrac{15}{a} \Rightarrow a^2 - \left(\dfrac{15}{a}\right)^2 = -16 \Rightarrow a^4+16a^2-225=0$; $a$ ($b$) real so $a^2=9$ (or $b^2=25$) only | M1, A1 | Eliminating $b$ or $a$ to obtain 3 term quadratic in $a^2$ or $b^2$. Unknowns must not be in denominator. Must be an equation. Rogue solutions: $a^2=-25$, $b^2=-9$. Factored forms: $(b^4-16b^2-225=0)$; $(a^2-9)(a^2+25)$; $(b^2-25)(b^2+9)$ |
| $3+5i$ and $-3-5i$ | A1 | Both roots. Can be $\pm(3+5i)$ but not $\pm3\pm5i$. Note: 4/5 possible following B0 |
| **[5]** | | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3+5i)^3 = (-16+30i)(3+5i) = -198+10i$ | M1 1.1 | Can be awarded for cubing an incorrect answer to 5(a). Could be done by expansion either in two steps or binomial: $3^3 + 3\times3^2\times5i + 3\times3\times(5i)^2 + (5i)^3$. Do not need to see intermediate step before correct answer. If incorrect need to see proof of expanding three brackets. Or $(-3-5i)^3 = 198-10i...$. For binomial want to see 4 terms and either 2nd or 3rd term correct (up to sign error) |
| $= \frac{-99+5i}{4} \times 2^3$ so $\frac{3}{2}+\frac{5i}{2}$ | A1 [2] 3.1a | Correct answer must follow a correct root. $= \frac{-99+5i}{4}\times(-2)^3$ |
---5 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Use an algebraic method to find the square roots of $- 16 + 30 \mathrm { i }$.
\item By finding the cube of one of your answers to part (a) determine a cube root of $\frac { - 99 + 5 i } { 4 }$.
Give your answer in the form $a + b \mathrm { i }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2022 Q5 [7]}}