| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Division plus modulus/argument |
| Difficulty | Moderate -0.8 This is a straightforward Further Maths question testing basic complex number operations: subtraction (trivial), division by conjugate multiplication (standard technique), and conversion to modulus-argument form (routine calculation). While it's Further Maths content, these are foundational skills with no problem-solving or insight required, making it easier than average overall. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02f Convert between forms: cartesian and modulus-argument |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | (i) |
| [1] | 1.1 | cao |
| 2 | (a) | (ii) |
| Answer | Marks |
|---|---|
| 5 5 | M1* |
| Answer | Marks |
|---|---|
| A1 | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | numerator and denominator by (−2+i) or (2−i) |
| Answer | Marks | Guidance |
|---|---|---|
| = −2𝑎−2𝑏i−𝑎i+𝑏 | M1* | Expanding (−2−i)(𝑎+𝑏i). Allow one slip only. |
| −1 = −2𝑎+𝑏 and 1 = −2𝑏−𝑎 | M1dep | Equating both real and imaginary coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| 5 5 𝑣 5 5 | A1 | 1−3i |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | √2 |
| Answer | Marks |
|---|---|
| 4 4 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Correct modulus seen |
Question 2:
2 | (a) | (i) | 1 + 2i | B1
[1] | 1.1 | cao
2 | (a) | (ii) | DR
u ( − 1 + i ) ( − 2 + i )
=
v ( − 2 − i ) ( − 2 + i )
2 − 2 − i i + i 2
=
5
1 3
= − i
5 5 | M1*
M1dep
A1 | 1.1a
2.1
1.1 | numerator and denominator by (−2+i) or (2−i)
Numerator expanded to include at least three terms with
correct denominator seen (must be expanded but need not
be simplified). Allow one sign slip in numerator only.
1−3i
Or simplified equivalent, e.g.
5
Alternative method
−1+i = (−2−i)(𝑎+𝑏i)
= −2𝑎−2𝑏i−𝑎i+𝑏 | M1* | Expanding (−2−i)(𝑎+𝑏i). Allow one slip only.
−1 = −2𝑎+𝑏 and 1 = −2𝑏−𝑎 | M1dep | Equating both real and imaginary coefficients
1 3 𝑢 1 3
𝑎 = , 𝑏 = − ⇒ = − i
5 5 𝑣 5 5 | A1 | 1−3i
Or simplified equivalent, e.g.
5
[3]
2 | (b) | √2
3
𝜋
4
3𝜋 3𝜋
√2(cos +isin )
4 4 | B1
B1
B1
[3] | 1.1
1.1
1.1 | Correct modulus seen
Correct argument seen. Condone 135o
Must be exact; condone 135°2 Two complex numbers are given by $u = - 1 + \mathrm { i }$ and $v = - 2 - \mathrm { i }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { u } - \mathrm { v }$ in the form $\mathrm { a } + \mathrm { bi }$, where $a$ and $b$ are real.
\item In this question you must show detailed reasoning.
Find $\frac { \mathrm { u } } { \mathrm { v } }$ in the form $\mathrm { a } + \mathrm { bi }$, where $a$ and $b$ are real.
\end{enumerate}\item Express $u$ in exact modulus-argument form.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q2 [7]}}