OCR MEI Further Pure Core 2023 June — Question 1 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeModulus-argument form conversion
DifficultyModerate -0.8 This is a routine complex numbers question testing standard techniques: conjugate notation, real part extraction, division by multiplying by conjugate, and conversion to modulus-argument form. All parts follow textbook procedures with no problem-solving or novel insight required, making it easier than average but not trivial due to the algebraic manipulation involved.
Spec4.02a 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)4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02f Convert between forms: cartesian and modulus-argument

1
  1. The complex number \(\mathrm { a } + \mathrm { ib }\) is denoted by \(z\).
    1. Write down \(z ^ { * }\).
    2. Find \(\operatorname { Re } ( \mathrm { iz } )\).
  2. The complex number \(w\) is given by \(w = \frac { 5 + \mathrm { i } \sqrt { 3 } } { 2 - \mathrm { i } \sqrt { 3 } }\).
    1. In this question you must show detailed reasoning. Express \(w\) in the form \(\mathrm { x } + \mathrm { iy }\).
    2. Convert \(w\) to modulus-argument form.

Question 1:
AnswerMarks Guidance
1(a) (i)
[1]1.2
1(a) (ii)
so Re(iz) = –bM1
A1
AnswerMarks
[2]1.1
1.1
AnswerMarks Guidance
1(b) (i)
5+√3i (5+√3i)(2+√3i)
=
2−√3i (2−√3i)(2+√3i)
7+7√3i
=
7
AnswerMarks
= 1+√3iM1
A1
AnswerMarks Guidance
[2]1.1a
1.1an intermediate step must be seen before final answer
1(b) (ii)
𝜋
arg(1+√3i)=
3
(cid:3095) (cid:3095)
so 𝑤 = 2(cos +i sin )
AnswerMarks
(cid:2871) (cid:2871)B1ft
B1
AnswerMarks
[2]1.1
1.1for either their modulus or argument
soi
cao. Allow 60°
Question 1:
1 | (a) | (i) | a – ib | B1
[1] | 1.2
1 | (a) | (ii) | iz = –b +ai
so Re(iz) = –b | M1
A1
[2] | 1.1
1.1
1 | (b) | (i) | DR
5+√3i (5+√3i)(2+√3i)
=
2−√3i (2−√3i)(2+√3i)
7+7√3i
=
7
= 1+√3i | M1
A1
[2] | 1.1a
1.1 | an intermediate step must be seen before final answer
1 | (b) | (ii) | (cid:3627)1+√3i(cid:3627)= 2
𝜋
arg(1+√3i)=
3
(cid:3095) (cid:3095)
so 𝑤 = 2(cos +i sin )
(cid:2871) (cid:2871) | B1ft
B1
[2] | 1.1
1.1 | for either their modulus or argument
soi
cao. Allow 60°
1
\begin{enumerate}[label=(\alph*)]
\item The complex number $\mathrm { a } + \mathrm { ib }$ is denoted by $z$.
\begin{enumerate}[label=(\roman*)]
\item Write down $z ^ { * }$.
\item Find $\operatorname { Re } ( \mathrm { iz } )$.
\end{enumerate}\item The complex number $w$ is given by $w = \frac { 5 + \mathrm { i } \sqrt { 3 } } { 2 - \mathrm { i } \sqrt { 3 } }$.
\begin{enumerate}[label=(\roman*)]
\item In this question you must show detailed reasoning.

Express $w$ in the form $\mathrm { x } + \mathrm { iy }$.
\item Convert $w$ to modulus-argument form.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q1 [7]}}