Complex numbers

5(a) z−zn+1 zn+1 − z B1 or . 1−z z −1 1 5(b) zn =1 and z≠1 leading to z+z2 +z3 +...+zn =0 leading to M1 A1 Must see zn =1. 1+z+z2 +...+zn−1 =0 2 5(c) ∞ z cosθ+isinθ M1 A1 Applies sum to infinity and substitutes for z. zm = = 1−z 3−cosθ−isinθ m=1 ( cosθ+isinθ)( 3−cosθ+isinθ) M1 A1 Rationalises denominator. ( 3−cosθ)2 +sin2θ 3cosθ−cos2θ−sin2θ+isinθcosθ+isinθ( 3−cosθ) M1 Applies sin2θ+cos2θ=1 or eiθ+e−iθ= 2cosθ. 9−6cosθ+cos2θ+sin2θ  ∞  3cosθ−1 M1 A1 Takes the real part, AG. Re zm  = 10−6cosθ m=1  7 © UCLES 2021 Page 9 of 13

5(a) z−zn+1 zn+1 − z B1 or . 1−z z −1 1 5(b) zn =1 and z≠1 leading to z+z2 +z3 +...+zn =0 leading to M1 A1 Must see zn =1. 1+z+z2 +...+zn−1 =0 2 5(c) ∞ z cosθ+isinθ M1 A1 Applies sum to infinity and substitutes for z. zm = = 1−z 3−cosθ−isinθ m=1 ( cosθ+isinθ)( 3−cosθ+isinθ) M1 A1 Rationalises denominator. ( 3−cosθ)2 +sin2θ 3cosθ−cos2θ−sin2θ+isinθcosθ+isinθ( 3−cosθ) M1 Applies sin2θ+cos2θ=1 or eiθ+e−iθ= 2cosθ. 9−6cosθ+cos2θ+sin2θ  ∞  3cosθ−1 M1 A1 Takes the real part, AG. Re zm  = 10−6cosθ m=1  7 © UCLES 2021 Page 9 of 13

1(a) a=1, b=i B1 1 1(b) z=ei2 5 kπ M1 Finds one fifth root of unity. z=ei2 5 kπ, k =0,1,2,3,4 A1 Gives all fifth roots of unity. Accept 1 not in exponential form. A0 if r =1 not seen or implied. Argument of i is 1π. B1 2 z=ei1 6 π M1 A1 Finds one root of z3 =i. A0 if first root not given in exponential form. z=ei5 6 π, ei3 2 π A1 Finds other two roots of z3 =i. A0 if r =1 not seen or implied. Withhold final A1 if a and b reversed in part (b) but all other work correct. 6 SC If z=e i ( 1 1 0 π+2 5 kπ ) , k =0,1,2,3,4 award M1 A1 A1 only. ( ) SC If z=e i 2 3 kπ , k =0,1,2 award M1 A1 only. © UCLES 2021 Page 6 of 15

4 z−z −1 =2isinθand z+z −1 =2cosθ B1 Use of z−z −1 =2isinθ and z+z −1 =2cosθ ( z−z −1 )5 = ( z5 −z −5 ) −5 ( z3−z −3 ) +10 ( z−z −1 ) M1 A1 Expands and groups. A0 if grouping not shown. ( z+z −1 )5 = ( z5 +z −5 ) +5 ( z3 +z −3 ) +10 ( z+z −1 ) M1 A1 Expands and groups. A0 if grouping not shown. 25isin5θ 2isin5θ−5(2isin3θ)+10(2isinθ) M1 Substitutes zn +z −n =2cosnθand zn −z −n =2isinnθ. = 25cos5θ 2cos5θ+5 ( 2cos3θ)+10 ( 2cosθ) sin5θ−5sin3θ+10sinθ A1 Justifies cancellation of constants. tan5θ= . cos5θ+5cos3θ+10cosθ 7 © UCLES 2021 Page 9 of 15

6(a) ( cos5θ+isin5θ)=( c+is )5 M1 Uses binomial theorem. sin5θ=s5 −10c2s3 +5c4s A1 s5 −10 ( 1−s2 ) s3 +5 ( 1−s2 )2 s M1 Uses c2 =1−s2 or cot2θ=cosec2θ−1 after dividing numerator and denominator by s5. 16s5 −20s3 +5s A1 1 cosec5θ M1 Divides simplified numerator and denominator by s5. cosec5θ= × 16s5 −20s3 +5s cosec5θ cosec5θ A1 AG 5cosec4θ−20cosec2θ+16 6 6(b) ( ) x5 M1 Relates with equation in part (a). x5 =2 5x4 −20x2 +16 leading to =2 5x4 −20x2 +16 cosec5θ=2leading to sin5θ= 1 M1 Solves sin5θ= 1. 2 2 ( ) x=cosec 1 π A1 Gives one correct solution. 30 ( ) ( ) ( ) ( ) cosec 5 π ,cosec − 7 π ,cosec −11π ,cosec 13π A1 Gives four other solutions. Allow different values of q 30 30 30 30 as long as all five solutions are found. 9.56, 2, –1.49, –1.09, 1.02 4 © UCLES 2021 Page 10 of 13

4(a) ei2 5 kπ, k =0,1,2,3,4 B2 OE. B1 for 1 correct fifth root of unity. B2 for exactly 5 distinct, correct roots. 2 4(b) ( c+is )4 =c4 +4c3(is)+6c2(is)2 +4c(is)3+(is)4 M1 A1 Uses binomial expansion. Can also go RHS to LHS using 2cosθ=z+z −1. cos4θ=c4 −6c2s2 +s4 =c4 −6c2(1−c2)+(1−c2)2 M1 Applies s2 =1−c2 or 2cosθ=z+z −1. 8cos4θ−8cos2θ+1 A1 AG. Can get final A1 without first. SC: Using double angle formulae scores 1/4. 4 © UCLES 2021 Page 8 of 14 4(c) 8x9 −8x7 +x5 −8x4 +8x2 −1=(x5 −1)(8x4 −8x2 +1) B1 cos4θ=0 leading to4θ= 1(2k+1)π M1 A1 Solves cos4θ=0. A1 for 4θ= 1π,3π,5π,7π OE. 2 2 2 2 2 cos0,cos1π,cos3π,cos5π,cos7π A1 Gives exactly 5 distinct, real roots. 8 8 8 8 Accept 1,±cos1π,±cos3π. 8 8 4 Question Answer Marks Guidance

6(a) ( cos5θ+isin5θ)=( c+is )5 M1 Uses binomial theorem. sin5θ=s5 −10c2s3 +5c4s A1 s5 −10 ( 1−s2 ) s3 +5 ( 1−s2 )2 s M1 Uses c2 =1−s2 or cot2θ=cosec2θ−1 after dividing numerator and denominator by s5. 16s5 −20s3 +5s A1 1 cosec5θ M1 Divides simplified numerator and denominator by s5. cosec5θ= × 16s5 −20s3 +5s cosec5θ cosec5θ A1 AG 5cosec4θ−20cosec2θ+16 6 6(b) ( ) x5 M1 Relates with equation in part (a). x5 =2 5x4 −20x2 +16 leading to =2 5x4 −20x2 +16 cosec5θ=2leading to sin5θ= 1 M1 Solves sin5θ= 1. 2 2 ( ) x=cosec 1 π A1 Gives one correct solution. 30 ( ) ( ) ( ) ( ) cosec 5 π ,cosec − 7 π ,cosec −11π ,cosec 13π A1 Gives four other solutions. Allow different values of q 30 30 30 30 as long as all five solutions are found. 9.56, 2, –1.49, –1.09, 1.02 4 © UCLES 2021 Page 10 of 13

7(a) sin7Imcis7 M1 Uses binomial theorem. sin7s7 21c2s535c4s37c6s A1 sin7s7 21  1s2 s5 35  1s22 s37  1s23 s M1 Uses c2 1s2. sin7s7 21  1s2 s5 35  12s2 s4 s37  13s2 3s4 s6 s sin764s7 112s556s37s A1 1 M1 Divides numerator and denominator by s7. cosec7 64s7 112s5 56s37s cosec7 A1 AG cosec7 CWO 7cosec656cosec4112cosec264 6 7(b) x7 M1 A1 Relates with equation in part (a). x7 2  7x6 56x4 112x2 64  leading to 2 7x6 56x4 112x2 64 cosec72 leading to sin7 1 M1 Solves sin7 1. 2 2 xcosec 1 π A1 Gives one correct solution. 42 xcosecqπ, q19,11, 7 , 5 ,13,17 A1 Gives six other solutions. Allow different values of q as 42 42 42 42 42 42 long as all seven solutions are found. 5 © UCLES 2022 Page 11 of 13

7(a) sin7Imcis7 M1 Uses binomial theorem. sin7s7 21c2s535c4s37c6s A1 sin7s7 21  1s2 s5 35  1s22 s37  1s23 s M1 Uses c2 1s2. sin7s7 21  1s2 s5 35  12s2 s4 s37  13s2 3s4 s6 s sin764s7 112s556s37s A1 1 M1 Divides numerator and denominator by s7. cosec7 64s7 112s5 56s37s cosec7 A1 AG cosec7 CWO 7cosec656cosec4112cosec264 6 7(b) x7 M1 A1 Relates with equation in part (a). x7 2  7x6 56x4 112x2 64  leading to 2 7x6 56x4 112x2 64 cosec72 leading to sin7 1 M1 Solves sin7 1. 2 2 xcosec 1 π A1 Gives one correct solution. 42 xcosecqπ, q19,11, 7 , 5 ,13,17 A1 Gives six other solutions. Allow different values of q as 42 42 42 42 42 42 long as all seven solutions are found. 5 © UCLES 2022 Page 11 of 13

1 z3 7 37i14ei1 6 π B1 Converts 7 37i to exponential form. z 14 1 3ei 1 1 8 π M1 A1 Finds one root. 1 z 14 1 3ei 1 1 8 1π, z 14 1 3ei 1 1 8 3π A1FT A1FT Finds other two roots. FT on modulus. Others given 2 3 scores A1 A0. 5 Question Answer Marks Guidance

7(a) wn −1 B1 w−1 1 7(b) (1+itan)k =seck(cos+isin)k =seck(cosk+isink) M1 A1 Applies de Moivre’s theorem, AG. 2 7(c) n−1 (1+itan)n −1 M1 A1 Applies part (a). (1+itan)k = itan k=0 −cotsecn(icosn−sinn)+icot A1 n−1 M1 A1 Takes imaginary part, AG. secksink=cot ( 1−secncosn ) k=0 5 7(d) 6m−1 M1 A1 Sets = 1π. =1π 2ksin (1kπ ) =1 3 ( 1−26mcos(2mπ)) =1 3 ( 1−26m) 3 3 3 3 3 k=0 2 © UCLES 2022 Page 11 of 13

5(a) 1,i B1 Accept exponential form e0,eiπ 2,eiπ,ei3 2 π or trigonometric form. Four values specified. 1 © UCLES 2022 Page 9 of 15 5(b) cos4=Re(cos+isin)4 =cos4−6cos2sin2+sin4 M1 A1 Expands and takes real part. cos4−6cos2 ( 1−cos2 ) + ( 1−2cos2+cos4 ) M1 Applies sin2=1−cos2. 8cos4−8cos2+1 A1 AG. Can award final A1 without previous A1 if penultimate line seen. SC. Applying cos2A=2cos2 A−1 twice scores 1/4. 4 5(c) ( 8x4 −8x2 +1 )4 = 9 , x=cos M1 Applies identify given in (b). 16 cos4=1 3 A1 Need  and exact value for A1. 2 4=1π+2kπ 4=5π+2kπ M1 Solves cos4=1 3 or cos4=−1 3 6 6 2 2 cos ( 1 π ) A1 Gives one correct solution. 24 cos(11π),cos(13π),cos(23π), A1 Gives other solutions. Accept 24 24 24 cos ( 1 π ) ,cos(11π),cos(5 π),cos(7 π) cos(5 π),cos(7 π),cos(17π),cos(19π), 24 24 24 24 24 24 24 24 5 © UCLES 2022 Page 10 of 15

7(a) wn −1 B1 w−1 1 7(b) (1+itan)k =seck(cos+isin)k =seck(cosk+isink) M1 A1 Applies de Moivre’s theorem, AG. 2 7(c) n−1 (1+itan)n −1 M1 A1 Applies part (a). (1+itan)k = itan k=0 −cotsecn(icosn−sinn)+icot A1 n−1 M1 A1 Takes imaginary part, AG. secksink=cot ( 1−secncosn ) k=0 5 7(d) 6m−1 M1 A1 Sets = 1π. =1π 2ksin (1kπ ) =1 3 ( 1−26mcos(2mπ)) =1 3 ( 1−26m) 3 3 3 3 3 k=0 2 © UCLES 2022 Page 11 of 13

3(a) zz1 2cos B1 Use of zz1 2cos.  zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. M1 A0 for no clear grouping. Correct substitution of zn cosnisinn for each term scores M1 A1. 2cos4 2cos442cos26 M1 Substitutes zn zn 2cosn. If zn cosnisinn used then must cancel sin. cos41cos44cos23 A1 AG. SC B1 expands cosisin4 and uses 8 trigonometric identities. 5 3(b)  1 2 1x23 2dx 1 6 π cos4d 11sin42sin23 1 6 π M1 A1 Applies substitution (M1) gets to cos4d,   0 0 8 4 0 changes limits, integration correct (A1). 1  9 3π  A1 16 4 3 © UCLES 2023 Page 8 of 15

3(a) zz1 2cos B1 Use of zz1 2cos.  zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. M1 A0 for no clear grouping. Correct substitution of zn cosnisinn for each term scores M1 A1. 2cos4 2cos442cos26 M1 Substitutes zn zn 2cosn. If zn cosnisinn used then must cancel sin. cos41cos44cos23 A1 AG. SC B1 expands cosisin4 and uses 8 trigonometric identities. 5 3(b)  1 2 1x23 2dx 1 6 π cos4d 11sin42sin23 1 6 π M1 A1 Applies substitution (M1) gets to cos4d,   0 0 8 4 0 changes limits, integration correct (A1). 1  9 3π  A1 16 4 3 © UCLES 2023 Page 8 of 15

3 zz1 2isinand zz1 2cos B1 Use of zz1 2isin and zz1 2cos  zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups.  zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. Only withhold one A1 mark (this or the previous) for no clear grouping. 24cos4 2cos442cos26 M1 Substituteszn zn 2cosn once in LHS.  24sin4 2cos442cos26 cos44cos23 A1 cot4 cos44cos23 7 Partial Question Answer Guidance Marks

2 1 π B1 Finds modulus and argument of 4+4i 3. (z+5i)3 =4+4i 3=8e3  1 1  1  1  M1 A1 Finds one root. z =2cos π+isin π−5i = 2cos π+i2sin π−5 1  9 9  9  9  7  7  13  13  A1 FT Finds other two roots. FT on their modulus. z =2cos π+i  2sin π−5 , z =2cos π+i  2sin π−5  2 9  9  3 9  9  A1 FT 5 © UCLES 2023 Page 6 of 14

8(a) zn −1 B1 z−1 1 8(b) zn −1 cosn−1+isinn B1 = z−1 cos−1+isin (cosn−1+isinn)(cos−1−isin) M1 Multiplies numerator and denominator by complex conjugate. (cos−1+isin)(cos−1−isin) zn −1 cosncos+sinnsin−cosn−cos+1 M1 Takes real part. Re = cos(n−1)=cosncos+sinnsin  z−1  (cos−1)2 +sin2 cosncos+sinnsin−cosn 1 A1 = + 2(1−cos) 2 cosn(cos−1)+sinnsin 1 M1 Factorises. = + 2(1−cos) 2 © UCLES 2023 Page 12 of 14 8(b) 1 sinnsin M1 A1 Divides through by denominator. AG. =  1−cosn+  2 1−cos  Alternative method for question 8(b) zn −1 ein−1 B1 = z−1 ei−1 e i(n−1 2 ) −e−i1 2  cos(n−1)+isin(n−1)−cos1+isin(1) M1 = 2 2 2 2 ei1 2 −e−i1 2  2isin1 2 zn −1 sin(n−1)+sin1 M1 Takes real part Re = 2 2 z−1 2sin1   2 sin(n−1) 1 A1 = 2 + 2sin1 2 2 sinncos1−cosnsin1 1 M1 Uses compound angle identity = 2 2 + 2sin1 2 2 sinncos1 1 1 M1 Divides through by denominator. = 2 − cosn+ 2sin1 2 2 2 sinnsin 1 1 sinnsin 1 1 A1 AG. sin=2sin1cos1 and − cosn+ = − cosn+ 2 2 4sin2 1 2  2 2 2(1−cos) 2 2 2sin2 1=1−cos. 2 7 © UCLES 2023 Page 13 of 14 8(c) 1 1 1 1 1 2 1 n−1 M1 A1 Forms sum of areas of rectangles given in the diagram.  cosxdx + cos + cos + cos A0 if comparison with integral missing or unclear. 0 n n n n n n n  n 1 M1 A1 1 sin sin Applies result from part (b) with = . AG. 1 1 2 n−1 1  n n n  n =  1+cos +cos ++cos  = 1−cos +  n n n n  2n  n 1−cos 1   n  4 8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison  cosxdx cos + cos + cos with integral missing or unclear. 0 n n n n n n  1  1 A1 sin1sin sin1sin 1  n  1 1 1  n  = 1−cos1+ + cos1− = cos1−1+  2n 1 n n 2n 1  1−cos   1−cos   n   n  3 © UCLES 2023 Page 14 of 14

3(a) cos5=Re(cos+isin)5 =cos5−10sin2cos3+5sin4cos M1 A1 Expands and takes real part. Accept 1 RHS to LHS using 2cos=z+ . z =cos5−10cos3 ( 1−cos2 ) +5cos ( 1−2cos2+cos4 ) M1 Applies sin2=1−cos2. =16cos5−20cos3+5cos A1 AG 4 3(b) x=cos, cos5=1 2 M1 Applies identify given in (b). 2 5=1+2k M1 Solves cos5=1 2 4 2 cos ( 1 π ) A1 Gives one correct solution. Accept 20 q= 1 . 20 cos(9 π),cos(17π),cos(25π),cos(33π) A1 Gives other solutions. OE. A0 for 20 20 20 20 repeated roots. 4 © UCLES 2023 Page 8 of 15

2 1 π B1 Finds modulus and argument of 4+4i 3. (z+5i)3 =4+4i 3=8e3  1 1  1  1  M1 A1 Finds one root. z =2cos π+isin π−5i = 2cos π+i2sin π−5 1  9 9  9  9  7  7  13  13  A1 FT Finds other two roots. FT on their modulus. z =2cos π+i  2sin π−5 , z =2cos π+i  2sin π−5  2 9  9  3 9  9  A1 FT 5 © UCLES 2023 Page 6 of 14

8(a) zn −1 B1 z−1 1 8(b) zn −1 cosn−1+isinn B1 = z−1 cos−1+isin (cosn−1+isinn)(cos−1−isin) M1 Multiplies numerator and denominator by complex conjugate. (cos−1+isin)(cos−1−isin) zn −1 cosncos+sinnsin−cosn−cos+1 M1 Takes real part. Re = cos(n−1)=cosncos+sinnsin  z−1  (cos−1)2 +sin2 cosncos+sinnsin−cosn 1 A1 = + 2(1−cos) 2 cosn(cos−1)+sinnsin 1 M1 Factorises. = + 2(1−cos) 2 © UCLES 2023 Page 12 of 14 8(b) 1 sinnsin M1 A1 Divides through by denominator. AG. =  1−cosn+  2 1−cos  Alternative method for question 8(b) zn −1 ein−1 B1 = z−1 ei−1 e i(n−1 2 ) −e−i1 2  cos(n−1)+isin(n−1)−cos1+isin(1) M1 = 2 2 2 2 ei1 2 −e−i1 2  2isin1 2 zn −1 sin(n−1)+sin1 M1 Takes real part Re = 2 2 z−1 2sin1   2 sin(n−1) 1 A1 = 2 + 2sin1 2 2 sinncos1−cosnsin1 1 M1 Uses compound angle identity = 2 2 + 2sin1 2 2 sinncos1 1 1 M1 Divides through by denominator. = 2 − cosn+ 2sin1 2 2 2 sinnsin 1 1 sinnsin 1 1 A1 AG. sin=2sin1cos1 and − cosn+ = − cosn+ 2 2 4sin2 1 2  2 2 2(1−cos) 2 2 2sin2 1=1−cos. 2 7 © UCLES 2023 Page 13 of 14 8(c) 1 1 1 1 1 2 1 n−1 M1 A1 Forms sum of areas of rectangles given in the diagram.  cosxdx + cos + cos + cos A0 if comparison with integral missing or unclear. 0 n n n n n n n  n 1 M1 A1 1 sin sin Applies result from part (b) with = . AG. 1 1 2 n−1 1  n n n  n =  1+cos +cos ++cos  = 1−cos +  n n n n  2n  n 1−cos 1   n  4 8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison  cosxdx cos + cos + cos with integral missing or unclear. 0 n n n n n n  1  1 A1 sin1sin sin1sin 1  n  1 1 1  n  = 1−cos1+ + cos1− = cos1−1+  2n 1 n n 2n 1  1−cos   1−cos   n   n  3 © UCLES 2023 Page 14 of 14

1 z3 216cos5π i216sin5π B1 Finds modulus and argument of 108 3108i 6 6 z 6cos5π i6sin5π M1 A1FT Finds one root. FT on their modulus. For M1 1 18 18 must have divided their argument of 108 3108i by 3. z 6cos17π i6sin17π,z 6cos29π i6sin29π M1 A1 Finds other two roots. 2 18 18 3 18 18 5 © Cambridge University Press & Assessment 2024 Page 6 of 14

1 z3 216cos5π i216sin5π B1 Finds modulus and argument of 108 3108i 6 6 z 6cos5π i6sin5π M1 A1FT Finds one root. FT on their modulus. For M1 1 18 18 must have divided their argument of 108 3108i by 3. z 6cos17π i6sin17π,z 6cos29π i6sin29π M1 A1 Finds other two roots. 2 18 18 3 18 18 5 © Cambridge University Press & Assessment 2024 Page 6 of 14

6(a) z4 z4n 1  M1 Uses sum of geometric series. z4 z4(z4)z4(z4)n1  z4 1 z4 z4n 1  z2 z4n2 z2 A1 Divides numerator and denominator by z2, AG.   z4 1 z2 z2 z2 2 6(b) z2 z2 2isin2 B1 Simplifies denominator. z4n2 z2 cos(4n2)isin(4n2)cos2isin2 M1 A1 Applies de Moivre’s theorem to numerator.  (Used at least once for M1.) Accept z2 z2 2isin2 z2 cos22isincossin2. n cos(4n2)cos2 M1 Equates imaginary parts. sin4r 2sin2 r1 n cos2cos(4n2) A1 AG. Must see evidence of eliminating i from the sin4r denominator. 2sin2 r1 5 © Cambridge University Press & Assessment 2024 Page 11 of 15

4(a) cos6=Re(cos+isin)6 =cos6−15cos4sin2+15cos2sin4−sin6 M1A1 Expands and takes real part. sin6=Im(cos+isin)6 =6cos5sin−20cos3sin3+6cossin5 M1A1 Takes imaginary part cos6−15cos4sin2+15cos2sin4−sin6 sin−6 M1 Divides numerator and denominator by cot6=  6cos5sin−20cos3sin3+6cossin5 sin−6 sin6. cot6−15cot4+15cot2−1 A1 AG cot6= 6cot5−20cot3+6cot 6 4(b) x=cot, cot6=1 B1 Applies identity given in (b). 6= 1+k M1 Solves cot6=1 4 cot ( 1 π ) A1 Gives one correct solution. 24 7.60 cot ( 5 π ) ,cot (3π ) ,cot (13π ) ,cot (17π ) ,cot (7π ) A1 Gives other five solutions. 24 8 24 24 8 1.30, 0.41, –0.13, –0.77, –2.41 Withhold A1 mark for repeated roots. 4 © Cambridge University Press & Assessment 2024 Page 8 of 13

1 z3 =27−27i=27 2e−i1 4 π B1 Modulus and argument of z3 (does not need to be in exponential form). z =32 1 6e−i 1 1 2 π M1 A1 Finds one root. Cube root must be taken for M1. 1 z =32 1 6ei 1 7 2 π, z =32 1 6e−i 1 9 2 π A1FT A1FT Finds other two roots. FT on modulus. Withhold one 2 3 answer mark for omission of i. 5 Question Answer Marks Guidance

3 z−z−1=2isin B1 Use of z−z−1=2isin ( z−z−1)5 = ( z5 −z−5) −5 ( z3−z−3) +10 ( z−z−1) M1 A1 Expands and groups. No clear grouping is M1 A0. 25isin5=2isin5−5(2isin3)+10(2isin) M1 A1FT Substituteszn −z−n =2isinn on RHS. FT on signs of coefficients only. Substitutingzn −z−n =isinnon RHS is M1 A0. Omission of i is M0. 16 A1 Cancelation of i justified.  cosec5= . sin5−5sin3+10sin 6 Question Answer Marks Guidance

1 z3 =27−27i=27 2e−i1 4 π B1 Modulus and argument of z3 (does not need to be in exponential form). z =32 1 6e−i 1 1 2 π M1 A1 Finds one root. Cube root must be taken for M1. 1 z =32 1 6ei 1 7 2 π, z =32 1 6e−i 1 9 2 π A1FT A1FT Finds other two roots. FT on modulus. Withhold one 2 3 answer mark for omission of i. 5 Question Answer Marks Guidance

3 z−z−1=2isin B1 Use of z−z−1=2isin ( z−z−1)5 = ( z5 −z−5) −5 ( z3−z−3) +10 ( z−z−1) M1 A1 Expands and groups. No clear grouping is M1 A0. 25isin5=2isin5−5(2isin3)+10(2isin) M1 A1FT Substituteszn −z−n =2isinn on RHS. FT on signs of coefficients only. Substitutingzn −z−n =isinnon RHS is M1 A0. Omission of i is M0. 16 A1 Cancelation of i justified.  cosec5= . sin5−5sin3+10sin 6 Question Answer Marks Guidance

5(a) (cos5+isin5)=(c+is)5 M1 Uses binomial theorem. cos5=c5−10s2c3+5s4c A1 =c5−10 ( 1−c2) c3+5 ( 1−c2)2 c M1 Uses s2 =1−c2. =16c5−20c3+5c A1 1 sec5 M1 Divides simplified numerator and sec5=  16c5 −20c3+5c sec5 denominator by c5. sec5 A1 AG. = 5sec4−20sec2+16 6 5(b) x5 = 2 ( 5x4 −20x2 +16 )  x5 = 2 B1 Relates with equation in part (a). Can be 3 5x4 −20x2 +16 3 implied by M1 mark. sec5= 2 cos5= 3 M1 Solves cos5= 3. 3 2 2 x=sec ( 1 π ) A1 Gives one correct solution. 30 sec (11π ) ,sec (13π ) ,sec (23π ) ,sec (25π ) A1 Gives four other solutions. Allow different 30 30 30 30 values of q as long as all five solutions are found. -1.35, -1.15, 1.01, 2.46, 4.81 4 © Cambridge University Press & Assessment 2025 Page 13 of 17

5(a) sin7=Im ( (c+is)7) M1 Uses binomial expansion and takes imaginary part. =−s7 +21c2s5−35c4s3+7c6s A1 =−s7 +21 ( 1−s2) s5−35 ( 1−s2)2 s3+7 ( 1−s2)3 s M1 Applies c2 =1−s2. ( 1−s2)3 =−s6 +3s4 −3s2 +1 B1 No sight of ( 1−s2)3 fully expanded is B0. sin7=−64s7 +112s5−56s3+7s A1 AG. 5 5(b) x=sin sin0 sin7=0 M1 Solves sin7=0 and excludes sin=0. For M1 the roots from sin=0 must be excluded at some stage of their working. (This could be implied by their final answers.) x=sin (1π ) A1 One correct root. 7 sin ( −1π ) , sin ( 2π ) , sin ( 3π ) A1 Exactly five other correct roots. Allow different 7 7 7 values of q which give these roots. Accept −sin (1π ) , sin (2π ) , sin (3π ) . 7 7 7 Just listing correct values of q without any reference of x is M1 SCB1 (2/3). 3 © Cambridge University Press & Assessment 2025 Page 13 of 17

8(a) zn+1−z B1 z−1 1 8(b) (1)n+1 cos(n+1)−1cos+i ( (1)n+1 sin(n+1)−1sin ) B1 zn+1−z 2 2 2 2 = z−1 1cos−1+i1sin 2 2 (1)n+1 cos(n+1)−1cos+i ( (1)n+1 sin(n+1)−1sin ) *M1 Multiplies numerator and denominator by complex 2 2 2 2 1cos−1−i1sin conjugate of z–1.  2 2 1cos−1+i1sin 1cos−1−i1sin 2 2 2 2 −1sin ( (1)n+1 cos(n+1)−1cos ) +(1cos−1) ( (1)n+1 sin(n+1)−1sin ) DM1 A1 Takes imaginary part. 2 2 2 2 2 2 5−cos 4 −(1)n+2 sincos(n+1)+(1)n+2 cossin(n+1)−(1)n+1 sin(n+1)+ 1sin M1 A1 Uses sin(A−B)=sinAcosB−cosAsinB in 2 2 2 2 numerator, AG. 5 −cos 4 (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin = 2 2 2 5 −cos 4 © Cambridge University Press & Assessment 2025 Page 17 of 19 8(b) Alternative method for question 8(b) z=1ei B1 2 zn+1−z (1)n+1 ei(n+1)−(1)ei M1 A1 = 2 2 z−1 1ei−1 2 (1)n+1 ei(n+1)−(1)ei 1e−i−1 M1 Multiplies numerator and denominator by complex 2 2  2 conjugate. 1ei−1 1e−i−1 2 2 (1)n+2 ein−(1)n+1 ei(n+1)+ 1ei−1  M1 A1 Takes imaginary part Im 2 2 2 4  5 −cos   4  (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin = 2 2 2 5 −cos 4 6 8(c)  n (1)m mcosm= d   (1 2 )n+2 sinn−(1 2 )n+1 sin(n+1)+1 2 sin   M1 Differentiates the result from (b). 2 d 5−cos  m=1  4  (5 −cos) ( (1)n+2 ncosn−(1)n+1 (n+1)cos(n+1)+ 1cos ) M1 A1 4 2 2 2 = (5 −cos)2 4 ( (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin ) sin 2 2 2 − (5 −cos)2 4 3 © Cambridge University Press & Assessment 2025 Page 18 of 19 8(d)   (1)m mcosm= (5 4 −cos)(1 2 cos)−(1 2 sin)sin M1 n(1 2 )n →0 as n→. 2 (5 −cos)2 m=1 4 5cos−1 A1 = 8 2 (5 −cos)2 4 2 © Cambridge University Press & Assessment 2025 Page 19 of 19

2 z4 =8−8i 3=16e−i1 3 π B1 Finds the modulus and argument of z4. Does not have to be in exponential form. z =2e−i 1 1 2 π M1 A1 Finds one root. Allow for the method mark for 1  them to use their r and their . 4 z =2ei 1 5 2 π A1FT Finds another root. Follow through only on the 2 modulus z =2e−i 1 7 2 π, z =2ei 1 1 2 1π A1FT Finds other two roots. Follow through only on 3 4 the modulus. Withhold 1 answer mark for omission of i. 5 © Cambridge University Press & Assessment 2025 Page 9 of 17

4(a) z+z−1 =2cos B1 SOI ( z+z−1)6 = ( z6 +z−6) +6 ( z4 +z−4) +15 ( z2 +z−2) +20 M1 A1 Expands and groups. 26cos6=2cos6+12cos4+30cos2+20 M1 Substitutes zn +z−n =2cosn. cos6= 1 cos6+ 3 cos4+15cos2+ 5 A1 OE. Fractions do not need to be simplified. 32 16 32 16 5 4(b)  1 2 π cos62xdx= 1 π( 1 cos6+ 3 cos4+15cos2+ 5 )d M1 Applies (a). Must have the 1 and the correct 0 2 0 32 16 32 16 2 limits for this mark. π A1 = 1  1 sin6+ 3 sin4+15sin2+ 5  2 192 64 64 16 0 = 5 π A1 32 3 © Cambridge University Press & Assessment 2025 Page 11 of 17

8(a) zn+1−z B1 z−1 1 8(b) (1)n+1 cos(n+1)−1cos+i ( (1)n+1 sin(n+1)−1sin ) B1 zn+1−z 2 2 2 2 = z−1 1cos−1+i1sin 2 2 (1)n+1 cos(n+1)−1cos+i ( (1)n+1 sin(n+1)−1sin ) *M1 Multiplies numerator and denominator by complex 2 2 2 2 1cos−1−i1sin conjugate of z–1.  2 2 1cos−1+i1sin 1cos−1−i1sin 2 2 2 2 −1sin ( (1)n+1 cos(n+1)−1cos ) +(1cos−1) ( (1)n+1 sin(n+1)−1sin ) DM1 A1 Takes imaginary part. 2 2 2 2 2 2 5−cos 4 −(1)n+2 sincos(n+1)+(1)n+2 cossin(n+1)−(1)n+1 sin(n+1)+ 1sin M1 A1 Uses sin(A−B)=sinAcosB−cosAsinB in 2 2 2 2 numerator, AG. 5 −cos 4 (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin = 2 2 2 5 −cos 4 © Cambridge University Press & Assessment 2025 Page 17 of 19 8(b) Alternative method for question 8(b) z=1ei B1 2 zn+1−z (1)n+1 ei(n+1)−(1)ei M1 A1 = 2 2 z−1 1ei−1 2 (1)n+1 ei(n+1)−(1)ei 1e−i−1 M1 Multiplies numerator and denominator by complex 2 2  2 conjugate. 1ei−1 1e−i−1 2 2 (1)n+2 ein−(1)n+1 ei(n+1)+ 1ei−1  M1 A1 Takes imaginary part Im 2 2 2 4  5 −cos   4  (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin = 2 2 2 5 −cos 4 6 8(c)  n (1)m mcosm= d   (1 2 )n+2 sinn−(1 2 )n+1 sin(n+1)+1 2 sin   M1 Differentiates the result from (b). 2 d 5−cos  m=1  4  (5 −cos) ( (1)n+2 ncosn−(1)n+1 (n+1)cos(n+1)+ 1cos ) M1 A1 4 2 2 2 = (5 −cos)2 4 ( (1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin ) sin 2 2 2 − (5 −cos)2 4 3 © Cambridge University Press & Assessment 2025 Page 18 of 19 8(d)   (1)m mcosm= (5 4 −cos)(1 2 cos)−(1 2 sin)sin M1 n(1 2 )n →0 as n→. 2 (5 −cos)2 m=1 4 5cos−1 A1 = 8 2 (5 −cos)2 4 2 © Cambridge University Press & Assessment 2025 Page 19 of 19

3(a) sin5=Im(cos+isin)5 =sin5−10sin3cos2+5sincos4 M1 A1 Expands and clearly taking imaginary part, enough terms for M1. Accept RHS 1 to LHS using 2isin= z− . z =sin5−10sin3 ( 1−sin2 ) +5sin ( 1−sin2 )2 M1 Applies cos2=1−sin2. =16sin5−20sin3+5sin A1 AG, intermediate line of working needed, e.g. expanding ( 1−sin2 )2 =1−2sin2+sin4. 4 3(b) M1 Applies identify given in (b). x=sin, sin5= 1 3 2 5= 1π+2kπ or 5= 2π+2kπ M1 Solves sin5= 1 3. 3 3 2 sin( 1 π) A1 Gives one correct solution. Accept 15 q= 1 . 15 sin(2 π),sin(− 5 π),sin(− 4 π),sin(7 π) A1 Gives other solutions. OE. A0 for 15 15 15 15 repeated roots. sin( 2 π)=sin(13π) 15 15 sin(7 π)=sin(8 π) 15 15 sin(− 5 π)=sin(−1π)=sin(5π)= 15 3 3 sin(25π) 15 sin(− 4 π)=sin(−11π)=sin(19π) 15 15 15 4 © Cambridge University Press & Assessment 2025 Page 12 of 19

7(a) z − 1 − z n z + 1 or z n + z 1 − − 1 z or z ( z z n − − 1 1 ) B1 1 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) © Cambridge University Press & Assessment 2026 Page 16 of 18 e 2i7 π B1 States one root of z 7 = 1 , z  1 . ei2 7 kπ k = 2 , , 6 B1 Five other roots. No repeats or incorrect roots, z = 1 must be omitted. 2 7(c) EITHER: 1 1 1 e 1 e i 1 i   m  = z m = z − z = − − + − + (*M1 A1 Applies sum to infinity and substitutes for z. Expect e e i e i   − . ( 1 e 1 e i 1 ( i 1 ) ( 1 e 1 e i 1 ) i ) 1 e e 1 i 1 i e e 1 i 2 e 2        = − − + − + − − − − − − = − − − − + − − − + − + − DM1 Rationalises denominator. Expect e e i e i e e e e i i e 2 e 1 e( e i i e 1 i ) 1        −  − − − −  = − + + − − + . e 1 1 ( c o 2 s e 1 c i o s s i n ) e e 2 2    = − − − + + − − − e ( c e o 2 s 2 e i s c o i s n ) 1 1     = − + + −  DM1 A1 Applies ei+e−i=2cos R e 1 1 e 2 1 e c o 1 s c o s e 2 e 2 e 2 e c 2 o e s c o s 1 1      m  = z m  = − − − − + − − = − − + DM1 A1) Takes the real part, AG. OR: Alternative method for question 7(c) 1 1 e e i e i ( e c o s c o s i) s i n i s i n       m  = z m = z − z = − = − + − (*M1 A1 Applies sum to infinity and substitutes for z. Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 17 of 18 ( e e c o s c o c o s s i s ) c i o n i s i 2 s n ( e ( e i s i n ( e c o s c o s c o s c o s )) 2 ) + + i i i s s s s i n i n i n i n 2 ( e c o s ) s i n 2                  = − + − − +  − − − + +  − − DM1 Rationalises denominator. e c e o 2 s 2 e 1 c o i s e s i n 1    = − − + + DM1 A1 Applies s i n 2 c o s 2 1   + = .    ecos−1 DM1 A1) Takes the real part, AG. Re zm =   e2 −2ecos+1 m=1 Available marks 7 Question Answer Marks Guidance

7(a) z − 1 − z n z + 1 or z n + z 1 − − 1 z or z ( z z n − − 1 1 ) B1 1 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) © Cambridge University Press & Assessment 2026 Page 16 of 18 e 2i7 π B1 States one root of z 7 = 1 , z  1 . ei2 7 kπ k = 2 , , 6 B1 Five other roots. No repeats or incorrect roots, z = 1 must be omitted. 2 7(c) EITHER: 1 1 1 e 1 e i 1 i   m  = z m = z − z = − − + − + (*M1 A1 Applies sum to infinity and substitutes for z. Expect e e i e i   − . ( 1 e 1 e i 1 ( i 1 ) ( 1 e 1 e i 1 ) i ) 1 e e 1 i 1 i e e 1 i 2 e 2        = − − + − + − − − − − − = − − − − + − − − + − + − DM1 Rationalises denominator. Expect e e i e i e e e e i i e 2 e 1 e( e i i e 1 i ) 1        −  − − − −  = − + + − − + . e 1 1 ( c o 2 s e 1 c i o s s i n ) e e 2 2    = − − − + + − − − e ( c e o 2 s 2 e i s c o i s n ) 1 1     = − + + −  DM1 A1 Applies ei+e−i=2cos R e 1 1 e 2 1 e c o 1 s c o s e 2 e 2 e 2 e c 2 o e s c o s 1 1      m  = z m  = − − − − + − − = − − + DM1 A1) Takes the real part, AG. OR: Alternative method for question 7(c) 1 1 e e i e i ( e c o s c o s i) s i n i s i n       m  = z m = z − z = − = − + − (*M1 A1 Applies sum to infinity and substitutes for z. Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 17 of 18 ( e e c o s c o c o s s i s ) c i o n i s i 2 s n ( e ( e i s i n ( e c o s c o s c o s c o s )) 2 ) + + i i i s s s s i n i n i n i n 2 ( e c o s ) s i n 2                  = − + − − +  − − − + +  − − DM1 Rationalises denominator. e c e o 2 s 2 e 1 c o i s e s i n 1    = − − + + DM1 A1 Applies s i n 2 c o s 2 1   + = .    ecos−1 DM1 A1) Takes the real part, AG. Re zm =   e2 −2ecos+1 m=1 Available marks 7 Question Answer Marks Guidance

1 © Cambridge University Press & Assessment 2026 Page 9 of 19 w 5 = − 3 2 = 3 2 e iπ B1 Converts to exponential form. z 1 = 2 e 1i5 π M1 A1 Finds one root in required form. z 2 = 2 e − 1i5 π A1 Finds another root. z 3 = 2 e 3i5 π , z 4 = 2 e − 3i5 π , z 5 = 2 e − iπ = − 2 A1 Finds other three roots. 5 Question Answer Marks Guidance

7(a) © Cambridge University Press & Assessment 2026 Page 16 of 19 c o s 7 R e ( c i s ) 7  = + *M1 Uses binomial theorem. =c7 −21s2c5+35s4c3−7s6c A1 =c7 −21c5( 1−c2) +35c3( 1−c2)2 −7c ( 1−c2)3 M1 Uses =c7 −21c5( 1−c2) +35c3( 1−2c2 +c4) −7c ( 1−3c2 +3c4 −c6) s 2 = 1 − c 2 . =64c7 −112c5 +56c3−7c A1 Must have shown expansion of ( 1 − c 2 ) 3 clearly. s e c 7 6 4 c 7 1 1 2 c 1 5 5 6 c 3 7 c s s e e c c 7 7     = − + −  DM1 Divides numerator and denominator by c7. (Must see working for this.) sec7 A1 AG. CWO sec7= 64−112sec2+56sec4−7sec6 6 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) x7 =2 ( 7x6 −56x4 +112x2 −64 )  x7 =−2 M1A1 Relates with equation in (a). M1 for setting 64−112x2 +56x4 −7x6 © Cambridge University Press & Assessment 2026 Page 17 of 19 x s e c  = and deriving sec7=a. A1 for a = − 2 . M1 can be implied from x7 =−2. 64−112x2 +56x4 −7x6 sec7=−2cos7=−1 M1 Solves 2 c o s 7 12 .  = − x = s e c ( 22 1 π ) A1 Gives one correct solution. x = s e c ( q π ) , q = − 1 62 1 , − 1 02 1 , − 42 1 , 82 1 , 1 42 1  = 23  , 22 01 A1 Gives six other solutions. Allow different values of q if all seven solutions are found. Might see q = 22 1 , 82 1 , 1 42 1 , 22 01 , 22 61 , 32 21 , 32 81 . 5 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/