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Countesthorpe Community College
Q1. The cubic equation
z3 – 4iz2 + qz – (4 – 2i) = 0
where q is a complex number, has roots α, β and γ.
(a) Write down the value of:
(i) α + β + γ;(1)
(ii) αβγ.(1)
(b) Given that α = β + γ, show that:
(i) α = 2i;(1)
(ii) βγ = –(l + 2i);(2)
(iii) q = – (5 + 2i).(3)
(c) Show that β and γ are the roots of the equation
z2 – 2iz – (1 + 2i) = 0(2)
(d) Given that β is real, find β and γ.(3)
(Total 13 marks)
Q2. The cubic equation
x3 + px2 + qx + r = 0
where p, q and r are real, has roots α, β and γ .
(a) Given that
α + β + γ = 4 and α2 + β2 + γ2 = 20
find the values of p and q.(5)
(b) Given further that one root is 3 + i, find the value of r.(5)
(Total 10 marks)
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Countesthorpe Community College
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Countesthorpe Community College
Q3. The cubic equation
z3 + pz2 + 6z + q = 0
has roots α, β, and γ.
(a) Write down the value of αβ + βγ + γα.(1)
(b) Given that p and q are real and that α2 + β2 + γ2 = –12:
(i) explain why the cubic equation has two non-real roots and one real root;(2)
(ii) find the value of p.(4)
(c) One root of the cubic equation is – 1 + 3i.
Find:
(i) the other two roots;(3)
(ii) the value of q.(2)
(Total 12 marks)
Q4. The cubic equation
z3 + qz + (18 – 12i) = 0
where q is a complex number, has roots α, β and γ.
(a) Write down the value of:
(i) αβγ;(1)
(ii) α + β + γ.(1)
(b) Given that β + γ = 2, find the value of:
(i) α;(1)
(ii) βγ;(2)
(iii) q.(3)
(c) Given that β is of the form ki, where k is real, find β and γ.(4)
Page 3
Countesthorpe Community College(Total 12 marks)
Q5. The cubic equation
z3 + iz2 + 3z – (1 + i) = 0
has roots α, β and y.
(a) Write down the value of:
(i) α + β + γ;(1)
(ii) αβ + βγ + γα;(1)
(iii) αβγ.(1)
(b) Find the value of:
(i) α2 + β2 + γ2;(3)
(ii) α2β2 + β2γ2 + γ2α2;(4)
(iii) α2β2γ2.(2)
(c) Hence write down a cubic equation whose roots are α2, β2 and γ2.(2)
(Total 14 marks)
Q6. The cubic equation
z3 + pz2 + 25z + q = 0
where p and q are real, has a root α = 2 – 3i.
(a) Write down another non-real root, β, of this equation.(1)
(b) Find:
(i) the value of αβ;(1)
(ii) the third root, γ, of the equation;(3)
(iii) the values of p and q.(3)
(Total 8 marks)
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Countesthorpe Community College
Q7. It is given that α, β and γ satisfy the equations
α + β + γ = 1
α2 + β2 + γ2 = – 5
α3 + β3 + γ3 = – 23
(a) Show that αβ + βγ + γα = 3.(3)
(b) Use the identity
(α + β + γ)(α2 + β2 + γ2 – αβ – βγ – γα) = α3 + β3 + γ3 – 3αβγ
to find the value of αβγ.(2)
(c) Write down a cubic equation, with integer coefficients, whose roots are α, β and γ.(2)
(d) Explain why this cubic equation has two non-real roots.(2)
(e) Given that α is real, find the values of α, β and γ.(4)
(Total 13 marks)
Q8. The roots of the cubic equation
z3 – 2z2 + pz + 10 = 0
are α, β and γ.
It is given that α3 + β3 + γ3 = –4.
(a) Write down the value of α + β + γ.(1)
(b) (i) Explain why α3 – 2α2 + pα + 10 = 0.(1)
(ii) Hence show that
α2 + β2 + γ2 = p + 13(4)
(iii) Deduce that p = –3.(2)
(c) (i) Find the real root α of the cubic equation z3 – 2z2 – 3z + 10 = 0.
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Countesthorpe Community College(2)
(ii) Find the values of β and γ.(3)
(Total 13 marks)
Q9. (a) (i) Show that is a root of the equation z7 = 1.(1)
(ii) Write down the five other non-real roots in terms of ω.(2)
(b) Show that
1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)
(c) Show that:
(i) ;(3)
(ii) .(4)
(Total 12 marks)
Q10. The cubic equation
2z3 + pz2 + qz + 16 = 0
where p and q are real, has roots α, β and γ.
It is given that α = 2 + .
(a) (i) Write down another root, β, of the equation.(1)
(ii) Find the third root, γ.(3)
(iii) Find the values of p and q.(3)
(b) (i) Express α in the form reiθ, where r > 0 and –π < θ ≤ π.(2)
(ii) Show that
(2)
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Countesthorpe Community College(iii) Show that
where n is an integer.(3)
(Total 14 marks)
Q11. The cubic equation
z3 – 2z2 + k = 0 (k ≠ 0)
has roots α, β and γ.
(a) (i) Write down the values of α + β + γ and αβ + βγ + γα.(2)
(ii) Show that α2 + β2 + γ2 = 4.(2)
(iii) Explain why α3 – 2α2 + k = 0.(1)
(iv) Show that α3 + β3 + γ3 = 8 – 3k.(2)
(b) Given that α4 + β4 + γ4 = 0:
(i) show that k = 2;(4)
(ii) find the value of α5 + β5 + γ5.(3)
(Total 14 marks)
Q12. (a) Show that (1 + i)3 = 2i – 2.(2)
(b) The cubic equation
z3 – (5 + i)z2 + (9 + 4i)z + k(1 + i) = 0
where k is a real constant, has roots α, β and γ.
It is given that α = 1 + i.
(i) Find the value of k.(3)
(ii) Show that β + γ = 4.(1)
(iii) Find the values of β and γ.(5)
(Total 11 marks)
Q13. (a) Find the six roots of the equation z6 = 1, giving your answers in the form ei ,
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Countesthorpe Community College
where –π < ≤ π.(3)
(b) It is given that w = eiθ,where θ ≠ nπ.
(i) Show that = 2i sin θ.(2)
(ii) Show that .(2)
(iii) Show that cot θ – i.(3)
(iv) Given that z = cot θ – i, show that z + 2i = zw2.(2)
(c) (i) Explain why the equation
(z + 2i)6 = z6
has five roots.(1)
(ii) Find the five roots of the equation
(z + 2i)6 = z6
giving your answers in the form a + ib.(4)
(Total 17 marks)
Q14. It is given that z = eiθ.
(a) (i) Show that
z + = 2 cos θ(2)
(ii) Find a similar expression for
z2 + (2)
(iii) Hence show that
z2 – z + 2 – + = 4 cos2 θ – 2 cos θ(3)
(b) Hence solve the quartic equationPage 8
Countesthorpe Community College
z4 – z3 + 2z2 – z + 1 = 0
giving the roots in the form a + ib.(5)
(Total 12 marks)
Q15. The complex numbers z1 and z2 are given by
z1 = and z2 = i
(a) Show that z1 = i.(2)
(b) Show that | z1 | = | z2 |.(2)
(c) Express both z1 and z2 in the form reiθ, where r > 0 and –π < θ ≤ π.(3)
(d) Draw an Argand diagram to show the points representing z1, z2 and z1 + z2.(2)
(e) Use your Argand diagram to show that
tan π = 2 + (3)
(Total 12 marks)
Q16. (a) (i) Given that z6 – 4z3 + 8 = 0, show that z3 = 2 ± 2i.(2)
(ii) Hence solve the equation
z6 – 4z3 + 8 = 0
giving your answers in the form reiθ, where r > 0 and – π < θ ≤ π.(6)
(b) Show that, for any real values of k and θ,
(2)
(c) Express z6 – 4z3 + 8 as the product of three quadratic factors with real coefficients.(3)
(Total 13 marks)
Q17. Use De Moivre’s Theorem to find the smallest positive angle θ for which
(cos θ + i sin θ)15 = – i
Page 9
Countesthorpe Community College(Total 5 marks)
Q18. (a) Find the three roots of z3 = 1, giving the non-real roots in the form eiθ,where –π < θ ≤ π.
(2)
(b) Given that ω is one of the non-real roots of z3 = 1, show that
1 + ω + ω2 = 0(2)
(c) By using the result in part (b), or otherwise, show that:
(i) (2)
(ii) (1)
(iii) where k is an integer.(5)
(Total 12 marks)
Q19. (a) (i) Expand
(1)
(ii) Hence, or otherwise, expand
(3)
(b) (i) Use De Moivre’s theorem to show that if z = cos θ + i sin θ then
(3)
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Countesthorpe Community College
(ii) Write down a corresponding result for .(1)
(c) Hence express cos4 θ sin2 θ in the form
A cos 6θ + B cos 4θ + C cos 2θ + D
where A, B, C and D are rational numbers.(4)
(d) Find .(2)
(Total 14 marks)
Q20. (a) (i) By applying De Moivre’s theorem to (cos θ + i sin θ)3, show that
cos 3θ = cos3 θ – 3 cos θ sin2 θ(3)
(ii) Find a similar expression for sin 3θ.(1)
(iii) Deduce that
(3)
(b) (i) Hence show that tan is a root of the cubic equation
x3 – 3x2 – 3x + 1 = 0(3)
(ii) Find two other values of θ, where 0 < θ < π, for which tan θ is a root of this cubic equation.
(2)
(c) Hence show that
(2)(Total 14 marks)
Q21. (a) Express 4 + 4i in the form reiθ, where r > 0 and –π < θ ≤ π.(3)
(b) Solve the equation
Page 11
Countesthorpe Community Collegez5 = 4 + 4i
giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.(5)
(Total 8 marks)
Q22. (a) Prove by induction that, if n is a positive integer,
(cos θ + i sin θ)n = cos nθ + i sin nθ(5)
(b) Hence, given that
z = cos θ + i sin θ
show that
(3)
(c) Given further that , find the value of
(4)(Total 12 marks)
Q23. Given that satisfies the equation
where a is real:
(a) find the value of a;(3)
(b) find the other three roots of this equation, giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.
(5)(Total 8 marks)
Q24. (a) Show that
(z4 – eiθ) (z4 – e–iθ) = z8 – 2z4 cos θ + 1
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Countesthorpe Community College(2)
(b) Hence solve the equation
z8 – z4 + 1 = 0
giving your answers in the form , where –π < ≤ π.(6)
(c) Indicate the roots on an Argand diagram.(3)
(Total 11 marks)
Q25. (a) (i) Express each of the numbers 1 + i and 1 – i in the form reiθ, where r > 0.(3)
(ii) Hence express
(1 + i)8(1 – i)5
in the form reiθ, where r > 0.(3)
(b) Solve the equation
z3 = (1 + i)8 (1 – i)5
giving your answers in the form a eiθ, where a is a positive integer and –π < θ ≤ π.
(4)(Total 10 marks)
Q26. (a) (i) Show that is a root of the equation z7 = 1.(1)
(ii) Write down the five other non-real roots in terms of ω.(2)
(b) Show that
1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)
(c) Show that:
(i) ;(3)
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Countesthorpe Community College
(ii) .(4)
(Total 12 marks)
Q27. The cubic equation
2z3 + pz2 + qz + 16 = 0
where p and q are real, has roots α, β and γ.
It is given that α = 2 + .
(a) (i) Write down another root, β, of the equation.(1)
(ii) Find the third root, γ.(3)
(iii) Find the values of p and q.(3)
(b) (i) Express α in the form reiθ, where r > 0 and –π < θ ≤ π.(2)
(ii) Show that
(2)
(iii) Show that
where n is an integer.(3)
(Total 14 marks)
Q28. (a) (i) Use de Moivre’s Theorem to show that
cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ
and find a similar expression for sin 5 (5)
(ii) Deduce that
(3)
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Countesthorpe Community College
(b) Explain why is a root of the equation
t4 – 10t2 + 5 = 0
and write down the three other roots of this equation in trigonometrical form.(3)
(c) Deduce that
(5)(Total 16 marks)
Q29. (a) Express in the form reiθ, where r > 0 and –π < θ ≤ π:
(i) 4(1 + i√3);
(ii) 4(1 – i√3).(3)
(b) The complex number z satisfies the equation
(z3 – 4)2 = –48
Show that .(2)
(c) (i) Solve the equation
(z3 – 4)2 = –48
giving your answers in the form reiθ, where r > 0 and –π < θ ≤ π.(5)
(ii) Illustrate the roots on an Argand diagram.(3)
(d) (i) Explain why the sum of the roots of the equation
(z3 – 4)2 = –48
is zero.(1)
(ii) Deduce that .(3)
(Total 17 marks)
Q30. (a) Use the identity with A = (r + 1)x and B = rx to show that
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Countesthorpe Community College
(4)
(b) Use the method of differences to show that
(5)(Total 9 marks)
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Countesthorpe Community College
Q31. The diagram shows a curve which starts from the point A with coordinates (0, 2). The curve is such that, at every point P on the curve,
where s is the length of the arc AP.
(a) (i) Show that
(3)
(ii) Hence show that
(4)
(iii) Hence find the cartesian equation of the curve.(3)
(b) Show that
y2 = 4 + s2
(2)(Total 12 marks)
Q32. (a) (i) Show that is a root of the equation z7 = 1.(1)
(ii) Write down the five other non-real roots in terms of ω.(2)
(b) Show that
1 + ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0(2)
(c) Show that:
(i) ;
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Countesthorpe Community College(3)
(ii) .(4)
(Total 12 marks)
Q33. (a) Express
5 sinh x + cosh x
in the form Aex + Be–x, where A and B are integers.(2)
(b) Solve the equation
5 sinh x + cosh x + 5 = 0
giving your answer in the form ln a, where a is a rational number.(4)
(Total 6 marks)
Q34. (a) Show that
9 sinh x – cosh x = 4ex – 5e–x
(2)
(b) Given that
9 sinh x – cosh x = 8
find the exact value of tanh x.(7)
(Total 9 marks)
Q35. (a) Show that
9 sinh x – cosh x = 4ex – 5e–x
(2)
(b) Given that
9 sinh x – cosh x = 8
find the exact value of tanh x.(7)
(Total 9 marks)
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Countesthorpe Community College
M1. (a) (i)
B11
(ii)
B11
(b) (i)
AG
B11
(ii)
Some method must be shown, eg
M1
AG
A12
(iii) q = αβ + βγ + γα
M1
= α(β + γ) + βγOr α2 + βγ , ie suitable grouping
M1
= 2i.2i − 2i − 1 = −2i − 5
AG
A13
(c) Use of β + γ = 2i and βγ = −2i −1
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Countesthorpe Community CollegeM1
z2 −2iz − (1 + 2i) = 0
Elimination of say to arrive at − (1 + 2i) = 0 M1A0 unless also some reference to γ being a root
AG
A12
(d) f (−1) = 1 + 2i − 1 − 2i = 0
For any correct method
M1
β = −1, γ = 1 + 2i
A1 for each answer
A1A13
[13]
M2. (a) p = −4
B1
(α + β + γ)2 = Σα2 + 2Σαβ
M1
16 = 20 + 2Σαβ
A1
Σαβ = −2
A1F
q = −2
A1F5
Page 20
Countesthorpe Community College(b) 3 − i is a root
B1
Third root is −2
B1F
αβγ = (3 + i) (3 − i)(−2)
M1
= −20
Real αβγ
A1F
r = +20
Real r
A1F5
Alternative to (b)Substitute 3 + i into equation
M1
(3 + i)2 = 8 + 6i
B1
(3 + i)3 = 18 + 26i
B1
r = 20
Provided r is real
A2,1,0[10]
M3. (a)
B11
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Countesthorpe Community College
(b) (i) Sum of squares < 0 ∴ not all real
E1
Coefficients real ∴ conjugate pair
E12
(ii)
A1 for numerical values inserted
M1A1
A1F
p = 0
cao
A1F4
(c) (i) −1 − 3i is a root
B1
Use of appropriate relationship eg
M0 if used unless the root 2 is checked
M1
Third root 2
incorrect p
A1F3
(ii) q = − (−1 − 3i)(−1 + 3i)2
allow even if sign error
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Countesthorpe Community CollegeM1
= −20
ft incorrect 3rd root
A1F2
[12]
M4. (a) (i) αβγ = – 18 + 12i
accept – (18 – 12i)
B11
(ii) α + β + γ = 0
B11
(b) (i) α = –2
B1F1
(ii)
ft sign errors in (a) or (b)(i) or slips suchas miscopy
M1A1F2
(iii)
M1
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Countesthorpe Community College= –2 × 2 + 9 – 6i
ft incorrect βγ or α
A1F
= 5 – 6i
A1F3
(c) β = ki, γ = 2 – ki
B1
ki(2 – ki) = 9 – 6i
M1
2k = – 6 (k2 = 9) k = –3
imaginary parts
m1
β = –3i, γ = 2 + 3i
A14
[12]
M5. (a) (i)
B11
(ii)
B11
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Countesthorpe Community College
(iii)
B11
(b) (i) used
Allow if sign error or 2 missing
M1
= (– i)2 – 2 × 3
A1F
= –7
ft errors in (a)
A1F3
(ii)
Allow if sign error in 2 missing
M1
A1
= 9 – 2(1 + i)(–i)
ft errors in (a)
A1F
= 7 + 2i
ft errors in (a)
A1F4
(iii) α2 β2 γ2 = (1 +i)2 = 2i
M1
ft sign error in αβγ
A1F2
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Countesthorpe Community College
(c) z3 + 7z2 + (7 + 2i)z – 2i = 0
Correct numbers in correct places
B1F
Correct signs
B1F2
[14]
M6. (a) 2 + 3i
B11
(b) (i) αβ = 13
B11
(ii) αβ + βγ + γα = 25
M1A0 for –25 (no ft)
M1
γ(α + β) = 12
A1F
γ = 3
ft error in αβ
A1F3
Alternative
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Countesthorpe Community College
Attempt at (z – 2 + 3i)(z – 2 – 3i)
(M1)
z2 – 4z + 13
(A1)
cubic is (z2 – 4z + 13)(z – 3) γ = 3
(A1)(3)
(iii)
M1 for a correct method for either p or q
M1A1F
ft from previous errorsp and q must be realfor sign errors in p and q allow M1 but A0
A1F3
AlternativeMultiply out or pick out coefficients
(M1)
p = –7, q = –39
(A1,A1)(3)
[8]
M7. (a) Use of
M1
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Countesthorpe Community College
A1
AG
A13
(b) 1(–5 – 3) = –23 – 3αβγFor use of identity
M1
αβγ = –5
A12
(c) z3 – z2 + 3z + 5 = 0
M1
For correct signs and “= 0”
A1F2
(d) α2 + β2 + γ2 < 0 non real roots
B1
Coefficients real conjugate pair
B12
(e) f(–1) = 0 z + 1 is a factor
M1A1
(z + 1)(z2 – 2z + 5) = 0
A1
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Countesthorpe Community College
z = –1, 1 ± 2i
A14
[13]
M8. (a) α + β + γ = 2
B11
(b) (i) α is a root and so satisfies the equation
E11
(ii)
M1A1
Substitution for and
m1
AG
A14
(iii)
do not allow this M mark if used in (b)(ii)
M1
p = –3
AG
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Countesthorpe Community CollegeA1
2
(c) (i) f(–2) = 0
M1
α = –2
A12
(ii) (z + 2)(z2 – 4z + 5) = 0
For attempting to find quadratic factor
M1
Use of formula or completing the squarem0 if roots are not complex
m1
= 2 ± i
CAO
A13
[13]
M9. (a) (i)
B11
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Countesthorpe Community College
(ii) Roots are ω2, ω3, ω4, ω5, ω6
OE; M1A0 for incomplete setSC B1 for a set of correct roots in terms of eiθ
M1A12
(b) Sum of roots considered
M1
= 0
or
A12
(c) (i)
M1
Or
A1
AG
A13
(ii)
Allow these marks if seen earlier in the solution
B1,B1
Using part (b)
M1
ResultPage 31
Countesthorpe Community CollegeAG
A14
[12]
M10. (a) (i)
B11
(ii) αβγ = –8
Allow for +8 but not ±16
M1
αβ = 16
B1
A13
(iii)
SC if failure to divide by 2 throughout,allow M1A1 for either p or q correct ft
M1
p = –7, q = 28
ft incorrect γ
A1F,A1F3
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Countesthorpe Community College
Alternative to (a)(ii) and (a)(iii)(z2 – 4z + 16)(az + b)
(M1)
αβ = 16
(B1)
a = 2, b = +1, γ =
(A1)
Equating coefficients
(M1)
p = –7
(A1F)
q = 28
(A1F)
(b) (i)
B1,B12
(ii)
M1
AG
A12
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Countesthorpe Community College
(iii)
B1
M1
AG
A13
[14]
M11. (a) (i)
B1
B12
(ii)
Used. Watch (M1A0)
M1
= 4
AG
A12
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Countesthorpe Community College
(iii) Clear explanation
eg α satisfies the cubic equation since it is a root.Accept z = α
E11
(iv)
Or
M1
= 8 – 3kAG
A12
(b) (i) α 4 = 2α 3 – kα
B1
Or
M1
= 2 (8 – 3k) – 2k
ft on
A1
k = 2
AG
A14
(ii)
M1
Page 35
Countesthorpe Community CollegeSubstitution of values
A1
= – 8
A13
[14]
M12. (a) (1 + i)2 = 2i or (1 + i) =
B1
2i(1 + i) = 2i – 2
AG
Alternative method:
(1 + i)3 = 1 + 3i2 + 3i3 + i3 B1
= 2i − 2 B1
B12
(b) (i) Substitute z = 1 + i
M1
Correct expansion
allow for correctly picking out either the realor the imaginary parts
A1
k = −5
A13
(ii) β + γ = 5 + i − α = 4
AG
B1
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Countesthorpe Community College1
(iii) αβγ = 5(1 + i)
allow if sign error
M1
βγ = 5
ft incorrect k
A1F
z2 − 4z + 5 = 0
M1
Use of formula or (z – 2)2 = –1
No ft for real roots if error in k
A1F
z = 2 ± i
A1F5
NB allow marks for (b) in whatever order they appear [11]
M13. (a)
M1
OE
M1A1 only if:
(1) range for k is incorrect e.g. 0,1,2,3,4,5
(2) i is missing
A2,1,03
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Countesthorpe Community College
(b) (i)
AG
M1A12
(ii)
M1
AG
A12
(iii)
Or for
M1
A1
= cotθ − i
AG
A13
(iv)
ie any correct method
M1
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Countesthorpe Community College
z + 2i = zw2
AG
A12
(c) (i) No coefficient of z6
E11
(ii) (w2)6 = 1 w2 =
B1
z = cot −i, k = ± 1, ± 2, 3
M1A2, 1, 0
Alternatively:
z + 2i = z B1
z = M1
roots A2, 1, 0
(NB roots are ± − i; ± − i; − i)4
[17]
M14. (a) (i) z + = cos θ + i sin θ + cos (−θ) + i sin (−θ)
Or z + = eiθ + e−iθ
M1
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Countesthorpe Community College
= 2cos θAG
A12
(ii) z2 + = cos 2θ + i sin 2θ + cos(−2θ) + i sin(−2θ)
M1
= 2 cos 2θOE
A12
(iii) z2 − z + 2 − +
= 2cos 2θ − 2cosθ + 2
M1
Use of cos 2θ = 2 cos2θ − 1
m1
= 4cos2 θ − 2cosθAG
A13
(b) z + = 0 z = ±i
M1A1
z + =1 z2 − z + l = 0
M1A1
z +
A1F
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Countesthorpe Community College
Alternative:cosθ = 0 θ = ± π M1
z = ±i A1
cosθ = θ = ± π M1
z = A1A15
Accept solution to (b) if done otherwise
AlternativeIf θ = + π θ = π
M1
z = i z =
A1
Or any correct z values of θ
M1
Any 2 correct answers
A1
One correct answer only
B1[12]
M15. (a)
AG
M1A12
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Countesthorpe Community College
(b) |z2| = = 1 = |z1|
M1A12
(c) r = 1
PI
B1
Deduct 1 mark if extra solutions
B1B13
(d)
Positions of the 3 points, relative to each other, must be approximately correct
B2,1F2
(e) Arg(z1+z2) =
Clearly shown
B1
tan =
Page 42
Countesthorpe Community CollegeAllow if B0 earned
M1
= 2 +
AG must earn B0 for this
B13
[12]
M16. (a) (i)
M1
= 2 ± 2i
AG
A12
(ii) 2 + 2i = 2 , 2 − 2i = 2
M1 for either result or for one of r = 2 , θ =
M1
A1A1
or
M1 for either
M1
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Countesthorpe Community College
allow A1 for any 3 correct ft errors in
A2,1,0 F6
(b) Multiplication of brackets
M1
Use of eiθ + e–iθ = 2 cos θAG
A12
(c)
PI
M1A1F
Product is
(or z2 + 2z + 2)
A1F3
(z2 − 2 cos z + 2)[13]
Page 44
Countesthorpe Community College
M17. (cos θ + i sin θ)15 = cos15 θ + i sin15 θor = e15i θ
M1
cos15 θ = 0
sin15 θ = −1
or −i =
m1A1
m1 for both R&I parts written down
A1F
ft provided the value of 15θ is a correct value
A1F5
SCcos15 θ + i sin15 θ = i
(M1)
sin15 θ = −1
or for cos15 θ = 0
(B1)
(B1)(3)
[5]
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Countesthorpe Community College
M18. (a) 1,
M1 for any method which would lead tothe correct answers
Accept e0 or e0i
Also accept answers written down correctly
M1A12
(b) Any correct method
M1
Shown for one root
AG
A12
(c) (i)
ie use of result in (b)
AG
A12
(ii)
AG
A11
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Countesthorpe Community College
(iii)
M1A1
Use of
m1
A1
AG
A15
[12]
M19. (a) (i)
B11
(ii)
Alternatives for M1A1:
Page 47
Countesthorpe Community CollegeM1A1
CAO (not necessarily in this form)
A13
(b) (i)
M1A1
= 2 cos nθAGSC: if solution is incomplete and(cos θ + i sin θ)–n is written ascos nθ – i sin nθ, award M1A0A1
A13
(ii) zn – z–n = 2i sin nθ
B11
(c) RHS = 2 cos 6θ + 4 cos 4θ – 2 cos 2θ – 4
ft incorrect values in (a)(ii) provided theyare cosines
M1A1F
LHS = –64 cos4 θ sin2 θ
M1
cos4 θ sin2 θ
A14
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Countesthorpe Community College
(d)
ft incorrect coefficients but not letters A, B, C, D
M1A1F2
[14]
M20. (a) (i) cos 3θ + i sin 3θ = (cos θ + i sin θ)3
M1
= cos3 θ + 3 i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
A1
Real parts: cos 3θ = cos3 θ – 3 cos θ sin2 θAG
A13
(ii) Imaginary parts:sin 3θ = 3 cos2 θ sin θ – sin3 θ
A1F1
(iii)
Used
M1
Error in sin 3θ
A1F
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Countesthorpe Community College
AG
A13
(b) (i)
Used (possibly implied)
B1
Must be hence
M1
x3 – 3x2 – 3x + 1 = 0
A13
(ii) Other roots are tan , tan
B1B12
(c)
Must be hence
M1
A12
[14]
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Countesthorpe Community College
M21. (a) Any method for finding r or θ
M1
A1A13
(b)
M1 needs some reference to a +2kπi
M1A1FA1F
Accept r in any form eg
Correct but some answers outside rangeallow A1ft incorrect r, θ in part (a)
A2,1,0F5
[8]
M22. (a) (cos θ + i sin θ)k+1 =(cos kθ + i sin kθ)(cos θ + i sin θ)
M1
Page 51
Countesthorpe Community CollegeMultiply out
Any form
A1
= cos(k + 1)θ + i sin(k + 1)θClearly shown
A1
True for n = 1 shown
B1
P(k) P(k + 1) and P(1) true
provided previous 4 marks earned
E15
(b)
M1A1
AG
A13
(c)
M1
A1
Page 52
Countesthorpe Community College
M0 for merely writing
M1
= 0
A1F4
[12]
M23. (a)
Allow M1 if
M1
OE could be or
A1
ft errors in 24
A1F3
(b) For other roots, r = 2
for realising roots are of form 2 × eiθ.M1 for strictly correct θ
B1
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Countesthorpe Community College
i.e must be
M1A1
Roots are
ft error in or r
A2,1,0F5
[8]
M24. (a) Correct multiplication of brackets
M1
eiθ + e–iθ = 2 cos θClearly shown
A12
(b) 2 cos θ = 1
SC If ‘hence’ not used and, say,z8 – z4 + 1 = 0 is solved by formula, lose
M1
M1A1, but then continue M1m1 etc if is obtained
A1
Page 54
Countesthorpe Community College
M1
m1
A1 if 3 roots correct
A2,1,0F6
(c)
B1 for 4 roots indicated correctly on a circle.CAO
B2,1,0
Indication that r = 1
B13
[11]
M25. (a) (i)
B1 both correct
B1
Page 55
Countesthorpe Community College
OE
B1B13
(ii) or equivalent single expression
No decimals; must include fractional powers
B1F
Raising and adding powers of e
M1
or equivalent angle
Denominators of angles must be different
A1F3
(b)
M1
CAO
B1
Correct answers outside range: deduct 1 mark only
A2,1F4
[10]
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Countesthorpe Community College
M26. (a) (i)
B11
(ii) Roots are ω2, ω3, ω4, ω5, ω6
OE; M1A0 for incomplete setSC B1 for a set of correct roots in terms of eiθ
M1A12
(b) Sum of roots considered
M1
= 0
or
A12
(c) (i)
M1
Or
A1
AG
A13
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Countesthorpe Community College
(ii)
Allow these marks if seen earlier in the solution
B1,B1
Using part (b)
M1
Result
AG
A14
[12]
M27. (a) (i)
B11
(ii) αβγ = –8
Allow for +8 but not ±16
M1
αβ = 16
B1
A13
(iii)
Page 58
Countesthorpe Community College
SC if failure to divide by 2 throughout,allow M1A1 for either p or q correct ft
M1
p = –7, q = 28
ft incorrect γ
A1F,A1F3
Alternative to (a)(ii) and (a)(iii)(z2 – 4z + 16)(az + b)
(M1)
αβ = 16
(B1)
a = 2, b = +1, γ =
(A1)
Equating coefficients
(M1)
p = –7
(A1F)
q = 28
(A1F)
(b) (i)
B1,B12
Page 59
Countesthorpe Community College
(ii)
M1
AG
A12
(iii)
B1
M1
AG
A13
[14]
M28. (a) (i) cos 5θ + i sin 5θ = (cosθ + i sinθ )5
Attempt to expand 3 correct terms
M1
Expansion in any form
Correct simplification
Page 60
Countesthorpe Community CollegeA1
Equate real parts:
m1
cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cosθ sin4 θAG
A1
Equate imaginary parts:
sin 5θ = 5 cos4 θ sinθ – 10 cos2 θ sin3 θ + sin5 θCAO
A15
(ii)
Used
M1
Division by cos5 θ or by cos4 θ
m1
AG
A13
(b)
Or for tan 4 θ – 10 tan 2 θ + 5 = 0
M1
Or for tan 5θ = 0
Page 61
Countesthorpe Community CollegeA1
Other roots
OE
B13
(c) Product of roots = 5
M1
Or
B1
A1
A1
– sign rejected with reason
E15
Alternative
Use of quadratic formula M1
A1
B1
Correct selection of +ve values E1
Multiplied together to get A1[16]
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Countesthorpe Community College
M29. (a) (i)
for either used
M1
If either r or θ is incorrect but the same value in both
(i) and (ii) allow A1 but for θ only if it is given as
A1
(ii)
A13
(b)
taking square root
M1
AG
A12
(c) (i)
for the 2; ft incorrect 8, but no decimals
Page 63
Countesthorpe Community CollegeB1F
for either, PI
M1
Allow A1 for any 2 roots not +/– each other
Allow A2 for any 3 roots not +/– each other
Allow A3 for all 6 correct roots
Deduct A1 for each incorrect root in the interval;ignore roots outside the intervalft incorrect r
A3,2,1F5
(ii)
Radius 2
clearly indicated; ft incorrect r
allow B1 for 3 correct points condone lines
B1F
Plotting roots
B2,13
(d) (i) Sum of roots = 0 as coefficient of z5 = 0
OE
E11
Page 64
Countesthorpe Community College
(ii)
M1
A1
AG
A13
[17]
M30. (a)
M1A1
Multiplying up
A1
Printed result
AG
A14
(b)
Page 65
Countesthorpe Community College
At least three lines to be shownAccept if x‘s used
M1A1
.................................
Clear cancellation
m1
Sum
A1
AG
A15
[9]
M31. (a) (i)
Allow M1 for
Page 66
Countesthorpe Community College
then A1 for
M1A1
AG
A13
(ii)
For separation of variables; allow withoutintegral sign
M1
Allow if C is missing
A1
C = 0
A1
AG if C not mentioned allow
SC incomplete proof of (a)(ii),differentiating
allow M1A1 only
A14
(iii)
M1
Page 67
Countesthorpe Community College
Allow if C is missing
A1
C = 0
Must be shown to be zero and CAO
A13
(b)
Use of cosh2 = 1 + sinh2
M1
= 4 + s2
AG
A12
[12]
M32. (a) (i)
B11
(ii) Roots are ω2, ω3, ω4, ω5, ω6
OE; M1A0 for incomplete setSC B1 for a set of correct roots in terms of eiθ
M1A12
Page 68
Countesthorpe Community College
(b) Sum of roots considered
M1
= 0
or
A12
(c) (i)
M1
Or
A1
AG
A13
(ii)
Allow these marks if seen earlier in the solution
B1,B1
Using part (b)
M1
Result
AG
A14
[12]
Page 69
Countesthorpe Community College
M33. (a)
M0 if no 2s in denominator
M1
= 3ex – 2e–x
A12
(b) 3ex – 2e–x + 5 = 03e2x + 5ex – 2 = 0
ft if 2s missing in (a)
M1
(3ex – 1)(ex + 2) = 0
A1F
ex ≠ – 2
any indication of rejection
E1
ex = x = In
provided quadratic factorises into real factors
A1F4
[6]
M34. (a)
M0 if cosh x mixed up with sinh xPage 70
Countesthorpe Community CollegeM1
= 4ex – 5e–x
AG
A12
(b) Attempt to multiply by ex
M1
4e2x – 8ex – 5 = 0
A1
(2ex – 5)(2ex + 1) = 0
ft provided quadratic factorises(or use of formula)
M1
PI but not ignored
E1F
A1F
M1 PI for attempt to use
or equivalent fraction
M1A1F7
[9]
Page 71
Countesthorpe Community College
M35. (a)
M0 if cosh x mixed up with sinh x
M1
= 4ex – 5e–x
AG
A12
(b) Attempt to multiply by ex
M1
4e2x – 8ex – 5 = 0
A1
(2ex – 5)(2ex + 1) = 0
ft provided quadratic factorises(or use of formula)
M1
PI but not ignored
E1F
A1F
M1 PI for attempt to use
or equivalent fraction
M1A1F7
[9]
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Countesthorpe Community College
Page 73
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