| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring complex number multiplication and application of the sum of two squares identity to prime factorization. Part (i) is routine complex multiplication, but part (ii) requires recognizing that |6+5i|²=61, |7+5i|²=74, and connecting these via norms to factor 4514=2×17×7²×61, which demands non-trivial insight into the relationship between complex numbers and prime factorization—more sophisticated than standard A-level but still a guided multi-step problem. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument |
| Answer | Marks | Guidance |
|---|---|---|
| \((6 + 5i)(7 + 5i) = 6 \times 7 + 6 \times 5i + 5i \times 7 + 5 \times 5 \times i^2 = 17 + 65i\) | M1 A1 | Expanding brackets with \(i^2 = -1\) If no working M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \((6 - 5i)(7 - 5i) = 17 - 65i\) | B1ft | Conjugate of (i) |
| \((17 + 65i)(17 - 65i) = 17^2 + 65^2 = 4514\) | M1 | |
| \((6 + 5i)(6 - 5i)\) & \((7 + 5i)(7 - 5i)\) \(4514 = 61 \times 74\) so expressed as a product of prime factors \(4514 = 2 \times 37 \times 61\) | M1 M1 A1 A1 | Products considered separately Answer with no working M0M0A0. |
## (i)
$(6 + 5i)(7 + 5i) = 6 \times 7 + 6 \times 5i + 5i \times 7 + 5 \times 5 \times i^2 = 17 + 65i$ | M1 A1 | Expanding brackets with $i^2 = -1$ If no working M0A0
[2]
## (ii)
$(6 - 5i)(7 - 5i) = 17 - 65i$ | B1ft | Conjugate of (i)
$(17 + 65i)(17 - 65i) = 17^2 + 65^2 = 4514$ | M1 |
$(6 + 5i)(6 - 5i)$ & $(7 + 5i)(7 - 5i)$ $4514 = 61 \times 74$ so expressed as a product of prime factors $4514 = 2 \times 37 \times 61$ | M1 M1 A1 A1 | Products considered separately Answer with no working M0M0A0.
[4]
---In this question you must show detailed reasoning.
\begin{enumerate}[label=(\roman*)]
\item Express $(6+5i)(7+5i)$ in the form $a+bi$. [2]
\item You are given that $17^2 + 65^2 = 4514$. Using the result in part (i) and by considering $(6-5i)(7-5i)$ express $4514$ as a product of its prime factors. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q8 [6]}}