OCR MEI Further Pure Core 2020 November — Question 6 4 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeQuadratic equations involving z² and z*
DifficultyStandard +0.8 This Further Maths question requires substituting z = a + bi and z* = a - bi into a non-standard quadratic equation, then solving a system of simultaneous equations from equating real and imaginary parts. It demands algebraic manipulation beyond typical A-level and systematic problem-solving rather than formula application, placing it moderately above average difficulty.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02i Quadratic equations: with complex roots

6 The complex number \(z\) satisfies the equation \(z ^ { 2 } - 4 \mathrm { i } z ^ { * } + 11 = 0\).
Given that \(\operatorname { Re } ( z ) > 0\), find \(z\) in the form \(a + b \mathrm { i }\), where \(a\) and \(b\) are real numbers. Section B (108 marks)
Answer all the questions.

Question 6:
AnswerMarks
6a2 ​− b2 ​+ 2abi − 4i(a − ib) + 11 = 0
​ ​ ​ ​ ​ ​ ​ ​ ​
⇒ a2 ​− b2 ​− 4b + 11 = 0, 2ab − 4a = 0
​ ​ ​ ​ ​ ​ ​ ​ ​ ​
⇒ b = 2
​ ​
a2 ​= 1, a = 1 so z = 1 + 2i
AnswerMarks
​ ​ ​ ​​B1
M1
B1
A1
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
3.2asubstitution for z and z* soi
put Re and Im parts equal to 0
Question 6:
6 | a2 ​− b2 ​+ 2abi − 4i(a − ib) + 11 = 0
​ ​ ​ ​ ​ ​ ​ ​ ​
⇒ a2 ​− b2 ​− 4b + 11 = 0, 2ab − 4a = 0
​ ​ ​ ​ ​ ​ ​ ​ ​ ​
⇒ b = 2
​ ​
a2 ​= 1, a = 1 so z = 1 + 2i
​ ​ ​ ​​ | B1
M1
B1
A1
[4] | 3.1a
1.1
1.1
3.2a | substitution for z and z* soi
put Re and Im parts equal to 0
6 The complex number $z$ satisfies the equation $z ^ { 2 } - 4 \mathrm { i } z ^ { * } + 11 = 0$.\\
Given that $\operatorname { Re } ( z ) > 0$, find $z$ in the form $a + b \mathrm { i }$, where $a$ and $b$ are real numbers.

Section B (108 marks)\\
Answer all the questions.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q6 [4]}}