OCR MEI FP1 2013 January — Question 2 4 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeReal and imaginary part expressions
DifficultyModerate -0.3 This is a straightforward algebraic manipulation requiring students to divide a complex number by its conjugate, then rationalize and identify real/imaginary parts. While it's a Further Maths topic (making it inherently more advanced), the technique is standard and mechanical with no conceptual insight needed—just multiply by conjugate over conjugate and simplify. Slightly easier than average due to its routine nature.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)
2 Given that \(z = a + b \mathrm { j }\), find \(\operatorname { Re } \left( \frac { z } { z ^ { * } } \right)\) and \(\operatorname { Im } \left( \frac { z } { z ^ { * } } \right)\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{z}{z^*} = \frac{a+bj}{a-bj} = \frac{(a+bj)^2}{(a-bj)(a+bj)}\)M1 Multiply top and bottom by \(a+bj\) and attempt to simplify
\(= \frac{a^2 + 2abj - b^2}{a^2 + b^2}\)M1 Using \(j^2 = -1\)
\(\Rightarrow \text{Re}\!\left(\frac{z}{z^*}\right) = \frac{a^2 - b^2}{a^2 + b^2}\) and \(\text{Im}\!\left(\frac{z}{z^*}\right) = \frac{2ab}{a^2 + b^2}\)A1, A1 cao correctly labelled; cao correctly labelled
[4] 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{z}{z^*} = \frac{a+bj}{a-bj} = \frac{(a+bj)^2}{(a-bj)(a+bj)}$ | M1 | Multiply top and bottom by $a+bj$ and attempt to simplify |
| $= \frac{a^2 + 2abj - b^2}{a^2 + b^2}$ | M1 | Using $j^2 = -1$ |
| $\Rightarrow \text{Re}\!\left(\frac{z}{z^*}\right) = \frac{a^2 - b^2}{a^2 + b^2}$ and $\text{Im}\!\left(\frac{z}{z^*}\right) = \frac{2ab}{a^2 + b^2}$ | A1, A1 | cao correctly labelled; cao correctly labelled |
| | **[4]** | |

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2 Given that $z = a + b \mathrm { j }$, find $\operatorname { Re } \left( \frac { z } { z ^ { * } } \right)$ and $\operatorname { Im } \left( \frac { z } { z ^ { * } } \right)$.

\hfill \mbox{\textit{OCR MEI FP1 2013 Q2 [4]}}
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