AQA Further Paper 2 Specimen — Question 2 3 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks3
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TopicComplex Numbers Arithmetic
TypeComplex conjugate properties and proofs
DifficultyEasy -1.2 This is a straightforward proof requiring only the definition of complex conjugate and modulus. For z = a + bi, students simply compute zz* = (a+bi)(a-bi) = a² + b² and |z|² = a² + b², showing they're equal. It's direct recall and application of basic definitions with no problem-solving or insight required, making it easier than average even for Further Maths.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument
Given that \(z\) is a complex number and that \(z^*\) is the complex conjugate of \(z\) prove that \(zz^* - |z|^2 = 0\) [3 marks]
Question 2:
AnswerMarks Guidance
2Defines generalised z and z* in
Cartesian or polar formAO1.2 B1
z abi then z* abi
zz* z2 abiabi  a2b2 2
a2abiabi(bi)2  a2b2
a2b2  a2b2
0
ALT
Let
z  rei then z*  rei
zz* z 2 reireir2
r2eiir2
r2 r2
0
AnswerMarks Guidance
Expands and simplifies zz* andz 2
(at least one correct)AO1.1b M1
Completes a well-structured
argument to prove the required
result. AG
Mark awarded if they have a
completely correct solution, which
is clear, easy to follow and
AnswerMarks Guidance
contains no slipsAO2.1 R1
Total3
QMarking Instructions AO 
Question 2:
2 | Defines generalised z and z* in
Cartesian or polar form | AO1.2 | B1 | Let
z abi then z* abi
zz* z2 abiabi  a2b2 2
a2abiabi(bi)2  a2b2
a2b2  a2b2
0
ALT
Let
z  rei then z*  rei
zz* z 2 reireir2
r2eiir2
r2 r2
0
Expands and simplifies zz* and |z|2
(at least one correct) | AO1.1b | M1
Completes a well-structured
argument to prove the required
result. AG
Mark awarded if they have a
completely correct solution, which
is clear, easy to follow and
contains no slips | AO2.1 | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
Given that $z$ is a complex number and that $z^*$ is the complex conjugate of $z$

prove that $zz^* - |z|^2 = 0$
[3 marks]

\hfill \mbox{\textit{AQA Further Paper 2  Q2 [3]}}
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