OCR MEI Further Pure Core AS 2022 June — Question 3 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeLinear equations in z and z*
DifficultyModerate -0.3 This is a straightforward Further Maths question requiring substitution of z = a + bi and z* = a - bi, then equating real and imaginary parts to solve simultaneous equations. While it involves complex conjugates, the algebraic manipulation is routine with no conceptual difficulty beyond the basic technique.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02f Convert between forms: cartesian and modulus-argument
3 The complex number \(z\) satisfies the equation \(5 ( z - \mathrm { i } ) = ( - 1 + 2 \mathrm { i } ) z ^ { * }\).
Determine \(z\), giving your answer in the form \(\mathrm { a } + \mathrm { bi }\), where \(a\) and \(b\) are real.
Question 3:
AnswerMarks
35[a+(b−1)i]=(−1+2i)(a−bi)
=−a+2ia+bi−2bi2
 5a = −a + 2b, 5b − 5 = 2a + b
 b = 3a, 12a − 5 = 2a [so a = ½ ,b = 3/2]
1 3
z= + i
AnswerMarks
2 2M1
A1
M1
M1
A1cao
AnswerMarks
[5]1.2
1.1
1.1
1.1
AnswerMarks
1.1z* = a −bi
expanding brackets correctly
equating Re and Im parts
solving simultaneous eqns by elimination or substitution
Alternative solution
5 ( z − i) 5 ( z − i) ( − 1 − 2 i)
z * = =
AnswerMarks Guidance
− 1 + 2 i ( − 1 + 2 i) ( − 1 − 2 i)M1 M1
5 ( − z + i − 2 z i + 2 i 2 )
=
AnswerMarks
5A1
 z* = −z − 2 + i − 2zi
AnswerMarks Guidance
 a − bi = −a − bi − 2 + i − 2ai + 2bM1 z* = a −bi
 a = −a + 2b − 2, −b = −b + 1 − 2aM1 equating Re and Im parts
 a = ½ , b = 3/2
1 3
 z = + i
AnswerMarks
2 2A1cao
[5]
Question 3:
3 | 5[a+(b−1)i]=(−1+2i)(a−bi)
=−a+2ia+bi−2bi2
 5a = −a + 2b, 5b − 5 = 2a + b
 b = 3a, 12a − 5 = 2a [so a = ½ ,b = 3/2]
1 3
z= + i
2 2 | M1
A1
M1
M1
A1cao
[5] | 1.2
1.1
1.1
1.1
1.1 | z* = a −bi
expanding brackets correctly
equating Re and Im parts
solving simultaneous eqns by elimination or substitution
Alternative solution
5 ( z − i) 5 ( z − i) ( − 1 − 2 i)
z * = =
− 1 + 2 i ( − 1 + 2 i) ( − 1 − 2 i) | M1 | M1 | realising denominator | realising denominator
5 ( − z + i − 2 z i + 2 i 2 )
=
5 | A1
 z* = −z − 2 + i − 2zi
 a − bi = −a − bi − 2 + i − 2ai + 2b | M1 | z* = a −bi
 a = −a + 2b − 2, −b = −b + 1 − 2a | M1 | equating Re and Im parts
 a = ½ , b = 3/2
1 3
 z = + i
2 2 | A1cao
[5]
3 The complex number $z$ satisfies the equation $5 ( z - \mathrm { i } ) = ( - 1 + 2 \mathrm { i } ) z ^ { * }$.\\
Determine $z$, giving your answer in the form $\mathrm { a } + \mathrm { bi }$, where $a$ and $b$ are real.

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2022 Q3 [5]}}
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