question bank complex numberp1_answ.rtf

8/17/2019 question bank complex numberp1_answ.rtf

1/31

1. METHOD 1

z = (2 – i)(z + 2) M1 = 2z + 4 – iz – 2iz (1 – i) = – 4 + 2i

z = i1

i24

−

+−

A1

z = i1i1

i1i24

++

×−+−

M1= – 3 – i A1

METHOD 2

let z = a + ib

2i

i

+++ba

ba

= 2 – i M1a + ib = (2 – i)((a + 2) + ib)a + ib = 2(a + 2) + 2bi – i(a + 2) + ba + ib = 2a + b + 4 + (2b – a – 2)iattempt to equate real and imaginary parts M1

a = 2a + b + 4(⇒ a + b + 4 = 0)and b = 2b – a – 2(⇒ – a + b – 2 = 0) A1

Note: Award A1 for two correct equations.

b = – 1;a = – z = – 3 – i A1

[4]

2. (a) (cos θ + i sin θ ) = cos

θ + cos

2 θ (i sin θ ) + cos θ (isin θ )

2 + (isin θ )

(M1)

= cos θ – 3 cos θ sin2 θ + i( cos2 θ sin θ – sin θ ) A1

(!) from "e Moi#re’s theorem(cos θ + i sin θ ) = cos θ + i sin θ (M1)cos θ + i sin θ = (cos

θ – 3 cos θ sin2 θ ) + i( cos2 θ sin θ – sin θ )

equating real parts

cos θ = cos θ

– 3 cosθ sin

2 θ M1

= cos θ – 3 cos θ (1 – cos2 θ ) A1

= cos θ – 3 cos θ + cos θ

= 4 cos θ – 3 cos θ A$

Note: "o not award mar%s if part (a) is not used.

IB Questionbank Mathematics Higher Level 3rd edition 1


2/31

(c) (cos θ + i sin θ )& =

cos& θ + & cos

4 θ (i sin θ ) + 10 cos

θ (i sin θ )

2 + 10 cos

2 θ (i sin θ )

+ &cos θ (i sin θ )4 + (i sin θ )

&(A1)

from "e Moi#re’s theorem

cos 5θ = cos& θ – 10 cos θ sin2 θ + & cos θ sin4 θ M1 = cos

& θ – 10 cos θ (1 – cos2 θ ) + &cos θ (1 – cos2 θ )2 A1

= cos& θ – 10 cos θ + 10 cos& θ + & cos θ – 10 cos θ + & cos& θ

∴ cos &θ = 1' cos& θ – 20 cos θ + & cos θ A$

Note: f compound angles used in (!) and (c) t*en mar%s can !eallocated in (c) only.

(d) cos &θ + cos θ + cos θ

= (1' cos& θ – 20 cos θ + & cos θ ) + (4 cos θ – 3 cos θ ) + cos θ = 0 M1

1' cos& θ – 16 cos θ + cosθ = 0 A1cos θ (1' cos

4 θ – 16 cos2 θ + ) = 0

cos θ (4 cos2 θ – 3)(4 cos2 θ – 1) = 0 A1

2

1;

2

3;0cos ±±=∴ θ

A1

2

π;

3

π;

6

π±±±=∴θ

A2

IB Questionbank Mathematics Higher Level 3rd edition


3/31

(e) cos &θ = 0

&θ = ...

...;2

7π;

2

5π;

2

3π;

2

π

(M1)

θ = ...

...;10

7π;

10

5π;

10

3π;

10

π

(M1)

Note: *ese mar%s can !e awarded for #erifications later in t*e question.

now consider 1' cos& θ – 20 cos

θ + & cos θ = 0 M1

cos θ (1' cos4 θ – 20 cos2 θ + &) = 0

cos2θ = 32

)5)(16(440020 −±

, cos θ = 0 A1

cos θ = 32

)5)(16(440020 −±±

32)5)(16(440020

10πcos −+±=

since ma- #alue of cosine ⇒ angleclosest to ero /1

855

8.4

)5(42545.4

10π

cos +

=−+

= A1

8

55

10

7πcos

−−=

A1A1[22]

3. (a) A =22 )32(1 −+

M1

= 3488− A1

= 322 − A1



4/31

(!) METHOD 1

3

πr!"

4

π #r! 21 −=−= z

A1A1

Note: Allow 3

πn$4

π

.

Note: Allow degrees at t*is stage.

4

π

3

π%&'A −=

=)

12

π(ccet

12

π−

A1

Note: Allow for final A1.

METHOD 2

attempt to use scalar product or cosine rule M1

22

31%&'Acos +=

A1

12π

%&'A = A1

[6]

4. (a) using t*e factor t*eorem z + 1 is a factor (M1)z

+ 1 = (z + 1)(z

2 – z + 1) A1

(!) (i) METHOD 1

z = – 1 ⇒ z + 1 = (z + 1)(z 2 – z + 1) = 0 (M1)

sol#ing z 2 – z + 1 = 0 M1

z = 2

3i1

2

411 ±=

−±

A1t*erefore one cu!e root of – 1 is γ A$

METHOD 2

γ2 =

2

3i1

2

3i12

+−=

+

M1A1

γ = 4

31

2

3i1

2

3i1 −−=

+−×

+−

A1= –1 A

IB Questionbank Mathematics Higher Level 3rd edition !


5/31

METHOD 3

γ =

3π

i

23i1

e=+−

A1

γ = e

iπ – 1 A1

(ii) METHOD 1

as γ is a root of z 2 – z + 1 = 0 t*en γ2 – γ + 1 = 0 M1/1

∴ γ2 = γ – 1 A$

Note: Award M1 for t*e use of z 2 – z + 1 = 0 in any way.

Award /1 for a correct reasoned approac*.

METHOD 2

γ2 = 2

3i1+−

M1

γ – 1 = 23i1

12

3i1 +−=−

+

A1

(iii) METHOD 1

(1 – γ)' = ( – γ2)' (M1) = (γ)

12 A1

= (γ)4 (M1)

= ( –1)4

= 1 A1

METHOD 2

(1 – γ)'

= 1 – 'γ + 1&γ2 – 20γ + 1&γ4 – 6γ& + γ' M1A1

Note: Award M1 for attempt at !inomial e-pansion.

use of any pre#ious result e.g. = 1 – 6γ + 1&γ 2 + 20 – 15γ + 'γ 2 + 1M1

= 1 A1Note: As t*e question uses t*e word *hence’" other metho$s tht

$o not se re,ios res-ts re r$e$ no mr/s.

IB Questionbank Mathematics Higher Level 3rd edition "


6/31

(c) METHOD 1

A2 =

+=

2

2

10

1

10

11

0

1

γ

γ γ γ

γ

γ

γ

γ

A1

A2 – A + I =

+−

−++−

111

0

11

1

2

2

γ γ

γ γ γ γ

M1from part (!)

γ2 – γ + 1 = 0

)1(1

11 2 +−=−+ γ γ

γ γ γ

= 0 A1

)1(1111 222 +−=++ γ γ γ γ γ = 0 A1

*ence A2 – A + I = 0 A$

METHOD 2

A2 =

−−

+−

2

3i10

12

3i1

A1A1A1

Note: Award 1 mar% for eac* of t*e nonero elements e-pressed in t*is form.

#erifying A2 – A + I = 0

(d) (i) A2 = A – I

⇒ A = A2 – A M1A1 = A – I – A A1 = – I A$

Note: Allow ot*er #alid met*ods.

(ii) I = A – A2

A –1

= A –1

A – A –1 A2 M1A1⇒ A –1 = I – A A$

Note: Allow ot*er #alid met*ods.[20]

IB Questionbank Mathematics Higher Level 3rd edition #


7/31

5. METHOD 1

since b $ 0 (M1)⇒ arg(b + i) = 0 A1

b

1

= tan 0 M1A1b = 3 A2 32

METHOD 2

arg(b + i)2 = '0 ⇒ arg(b2 – 1 + 2bi) = '0 M1

)1(

22 −bb

= tan '0 = 3 M1A1323 2 −− bb = 0 A1

)3)(13( −+ bb = 0since b $ 0 (M1)

b = 3 A1 32[6]

6. (a) z =3i1

2

1642±−=

−±−

M1

–1 +⇒= θ ie3i r

r = 2 A1

θ = arctan 3

π2

1

3=

− A1

–1 – 3i = r eiθ ⇒ r = 2

θ = arctan 3

π2

1

3−=

− A1

3

2πi

e2=⇒α A1

32π

ie2 −

=⇒ β A1

IB Questionbank Mathematics Higher Level 3rd edition %


8/31

(!)

A1A1

(c) cos nθ + i sin nθ = (cos θ + i sin θ )n

et n = 1eft *and side = cos 1θ + i sin 1θ = cos θ + i sin θ

/ig*t *and side = (cos θ + i sin θ )1 = cos θ + i sin θ

5ence true for n = 1 M1A1 Assume true for n = k M1

cos k θ + i sin k θ = (cos θ + i sin θ )k

⇒ cos(k + 1)θ + i sin(k + 1)θ = (cosθ + i sin θ )k (cos θ + i sin θ ) M1A1= (cos k θ + i sin k θ )(cos θ + i sin θ )= cos k θ cosθ – sin k θ sin θ + i(cos k θ sinθ + sin k θ cosθ ) A1= cos(k + 1)θ + i sin(k + 1)θ A1

5ence if true for n = k true for n = k + 15owe#er if it is true for n = 1⇒ true for n = 2 etc& /1⇒ *ence pro#ed !y induction

(d)

34π

i

34π

i

i2π

2

3

e2

e4

e8==

−β

α

A1

= 3

4πsini2

3

π4cos2 +

(M1)

=3i1

2

3i2

2

2−−=−−

A1A1

(e) a = 6e

i2π A1

β = 6e

–i2π A1

7ince e2π

and e –2π

are t*e same α = β

/1

IB Questionbank Mathematics Higher Level 3rd edition '


9/31

(f) EITHER

α = – 1 + 3i β = –1 – 3i

α8 = – 1 – 3i β 8 = – 1 + 3i A1

αβ 8 = (

–1 +

3i ) ( –1 +

3i ) = 1

– 2

3i

– 3 = 2 – 2

3i M1A1

βα8 = ( –1 – 3i )( –1 – 3i ) = 1 + 2 3i – 3 = –2 + 2 3i A1⇒ αβ 8 + βα8 = – 4 A1

OR

7ince α8 = β and β ( = α

αβ 8 =3

4πi

3

2πi

3

2πi

e4e2e2 =× M1A1

βα8 =3π4

i3π2

i3π2

ie4e2e2−−−

=× A1

αβ 8 + βα8 =

+

−3

4πi

3

4πi

ee4

=

−++

3

4πsini

3

π4cos

3

4πsini

3

π4cos4

A1

=4

21

83π4

cos8 −=−×= A1

(g) αn =

3π

i2e2

n

nM1A1

*is is real w*en n is a multiple of /1

i.e. n = ) w*ere ) ∈ +

[31]

7. (a) z =θ

θ

i

2i

e

e2

1

(M1)

z =

θ ie21

A1 32

(!) z 21

= A2

z 1 A

IB Questionbank Mathematics Higher Level 3rd edition *


10/31

(c) 9sing = r

a

−1 (M1)

=

θ

θ

i

i

e2

11

e

− A1 32

(d) (i) =θ

θ

θ

θ

cis2

11

cis

e2

11

e

i

i

−=

−(M1)

)sini(cos2

11

sinicos

θ θ

θ θ

+−

+

(A1)

Also = eiθ + ...e4

1

e2

1 3i2i

++ θ θ

= cis θ +...cis3

4

1cis2

2

1++ θ θ

(M1)

=

++++

+++ ...3sin

4

12sin

2

1sini...3cos

4

12cos

2

1cos θ θ θ θ θ θ

A1

IB Questionbank Mathematics Higher Level 3rd edition 1,


11/31

(ii) a%ing real parts

+−

+=+++

)sini(cos2

11

sinicose...3cos

4

12cos

2

1cos

θ θ

θ θ θ θ θ

A1

=

+−

+−×

−−

+

θ θ

θ θ

θ θ

θ θ

sini21

cos21

1

sini21

cos21

1

sini21

cos21

1

)sini(cose

M1

=

θ θ

θ θ θ

22

22

sin41

cos21

1

sin21

cos21

cos

+

−

−−

A1

=)cos(sin

41

cos1

21cos

22 θ θ θ

θ

++−

−

A1

=)cos45(2

)1cos2(4

4)1cos44(

2)1cos2(

θ

θ

θ

θ

−−

=÷+−

÷−

A1

= θ

θ

cos45

2cos4

−−

A1A$ 30[25]

8. iz 1 + 2z 2 = ⇒ z 2 = 2

3i

2

11 +− z

z 1 + (1 – i)z 2 = 4

⇒ z 1 + (1 – i)

+−

2

3i

2

11 z

= 4 M1A1

⇒ z 1 i

2

3i

2

1

2

3i

2

11

21 −++− z z

= 4

⇒ i

23

25i

21

21

11 +=− z z

⇒ z 1 – iz 1 = & + i A1



12/31

EITHER

et z 1 = - + i. (M1)

⇒ - + i. – i - – i2. = & + i:quate real and imaginary parts M1

⇒ - + . = & –x + . = 2. = 6

. = 4 ⇒ - = 1 i.e. z 1 = 1 + 4i A1A1

z 2 = 2

34i)i(1

2

1++−

M1

z 2 = 2

3i2i

2

1 2 +−−

z 2 =

i2

1

2

7−

A1

OR

z 1 = i1i35

−+

M1

z 1 =

−+=

+−++

2

3i85

i)i)(11(

i)i)(135(

M1A1

z 1 = 1 + 4i A1

z 2 = – 21

i(1 + 4i) + 2

3

M1

z 2 = 2

3i2i

2

1 2 +−−

z 2 =i

2

1

2

7−

A1

[9]

9. METHOD 1

20 + 10bi = (1 – bi)( –7 + i) (M1)20 + 10bi = ( –7 + b) + (; +


13/31

METHOD 2

=10

i7

i)i)(11(

)ii)(12( +−=

+−++bb

bb

(M1)

10

i7

1

i322

2 +−=

++−b

bb

A1

:quate real and imaginary parts (M1)

10

7

1

22

2

−=+−b

b

:quation A

10

13

2 =

+ bb

:quation

rom equation A

20 – 10b2 = –7 – 7b2

b2 = 2<

b = 3 A1

rom etion %30b = ; + ;b2

b2 – 10b + = 0

y factorisation or using t*e quadratic formula

b = 3

1

or A17ince is t*e common solution to !ot* equations b = /1

[6]

10. (a) sin (2n + 1) - cos - – cos (2n + 1) - sin - = sin (2n + 1) - – - M1A1= sin 2n- A$



14/31

(!) if n = 1 M157 = cos -

/57 = x

x x

x

x

sin2

cossin2

sin2

2sin=

= cos - M1so 57 = /57 and t*e statement is true for n = 1 /1assume true for n = k M1

Note: nly award M1 if t*e word true appears."o not award M1 for *-et n = k ’ on-9.:seent mr/s re in$een$ent o< this M1&

so cos - + cos - + cos & - + ... + cos(2k – 1) - = xkx

sin2

2sin

if n = k + 1 t*encos - + cos - + cos & - + ... + cos(2k – 1) - + cos(2k + 1) - M1

= x

kx

sin2

2sin

cos (2k + 1) - A1

= x

x xk kx

sin2

sin)12cos(22sin ++

M1

= x

x xk x xk x xk

sin2

sin)12cos(2sin)12cos(cos)12sin( +++−+

M1

= x

x xk x xk

sin2

sin)12cos(cos)12sin( +++

A1

= x

xk

sin2)22sin( +

M1

= x

xk

sin2

)1(2sin +

A1so if true for n = k t*en also true for n = k + 1

as true for n = 1 t*en true for all n ∈ + /1

Note: inal /1 is independent of pre#ious wor%.

IB Questionbank Mathematics Higher Level 3rd edition 1!


15/31

(c) 2

1

sin2

4sin=

x

x

M1A1sin 4 - = sin -

4 - = - ⇒ - = 0 !ut t*is is impossi!le

4 - = π – - 5π

=⇒ x A1

4 - = 2π + - 3π2

=⇒ x A1

4 - = π – - 5π3

=⇒ x A1

for not including any answers outside t*e domain /1

Note: Award t*e first M1A1 for correctly o!taining 6 cos - – 4 cos - – 1 = 0

or equi#alent and su!sequent mar%s as appropriate including t*e

answers arccos

±

− 451

"2

1

.[20]

11. 6i =

+ π22

πi

e8n

(M1)

or n = 0

6π

i31

e2i)8( = (M1)

= 6

πsini2

6

πcos2 +

A1

= 3 + i A1

or n = 1

65π

sini26π5

cos2i)8( 31

+=M1

= – 3 + i A1

or n = 2

23π

sini22π3

cos2i)8( 31

+=M1

= – 2i A1[8]

IB Questionbank Mathematics Higher Level 3rd edition 1"


16/31

12. &zz( + 10 = (' – 18i)z 8 M1et z = a + ib& 10 + 10 = (6 – 18i)(a – bi) (= 'a – 'bi – 16ai – 18b) M1A1:quate real and imaginary parts (M1)⇒ 'a – 16b = '0 and 'b + 16a = 0

⇒ a = 1 and b = – A1A1z = 1 – 3i A1

[7]

13. 2 + i is a root ⇒ 2 – i is root 1

> - – (2 + i)> ? - – (2 – i)? re


17/31

(!) b is a root/ (b) = 0

b& = 1 M1

b& – 1 = 0 A1

(b – 1)(b4 + b

+ b

2 + b + 1) = 0

b 1 11 + b + b2 + b + b4 = 0 as s*own. A$

(c) (i) u + v = b4 + b

+ b

2 + b = – 1 A1

uv = (b + b4)(b

2 + b

) = b

+ b

4 + b

' + b

< A1

3ow b& = 1 (A1)

5ence uv = b + b

4 + b + b

2 = – 1 A1

5ence u + v = uv = – 1 A$

(ii) (u – v )2 = (u2 + v 2) – 2uv (M1)= ((u + v )

2 – 2uv ) – 2uv (= (u + v )2 – 4uv ) (M1)A1

$i#en u – v @ 0

u – v = uvvu 4)(2 −+

=)1(4)1( 2 −−−

= 41+ A1

= 5 A$

Note: Award A0 unless an indicator is gi#en t*at u – v = – 5 is in#alid.[13]

IB Questionbank Mathematics Higher Level 3rd edition 1%


18/31

16. (a) z =4

1

i)1( −

et 1 – i = r (cos θ + i sin θ )2=⇒ r A1

θ = 4π− A1

z =

41

4

πisin

4

πcos2

−+

−

M1

=

41

π24

πsiniπ2

4

πcos2

+−+

+− nn

=

+−+

+−

2π

16π

sini2π

16π

cos2 81

nn

M1

=

−+

−

16π

sini16π

cos2 81

Note: Award M1 a!o#e for t*is line if t*e candidate *as forgotten toadd 2π and no ot*er solution gi#en.

=

+

167π

sini167π

cos2 81

=

+

1615π

sini16

15πcos2 8

1

=

−+

−

16π

sini16π

cos2 81

A2

Note: Award A1 for 2 correct answers. Accept any equi#alent form.

IB Questionbank Mathematics Higher Level 3rd edition 1'


19/31

(!)

A2

Note: Award A1 for roots !eing s*own equidistant from t*e origin and one in eac* quadrant. A1 for correct angular positions. t is not necessary to see written e#idence of angle !ut must agree wit* t*e diagram.

(c)

+

+

=

167πsini16π7cos2

1615π

sini16

π15cos2

8

1

81

1

2

z

z

M1A1

= 2

πsini

2

πcos +

(A1)= i A1 32

(⇒ a = 0 b = 1)[12]

17. (a) EITHER5

5

5

2πsini

5

2πcos

+=w

(M1)

= cos 2p + i sin 2p A1

= 1 A1

5ence 0 is a root of z & 1 = 0 A$

IB Questionbank Mathematics Higher Level 3rd edition 1*


20/31

OR

7ol#ing z & = 1 (M1)

z = cos.4"3"2"1"0"

5

π2sini

5

π2=+ nnn

A1

n = 1 gi#es cos 5

π2sini

5

π2+

w*ic* is 0 A1

(!) (0 1)(1 + 0 + 0 2 + 0

+ 0

4) = 0 + 0

2 + 0

+ 0

4 + 0

& 1

0 0 2 0

0

4M1

= 0 & 1 A1

7ince 0 & 1 = 0 and 0 ≠ 1 0 4 + 0 + 0 2 + 0 + 1 = 0. /1

(c) 1 + 0 + 0 2 + 0

+ 0

4 =

+

++++

2

5

π2sini

5

π2cos

5

π2sini

5

π2cos1

43

5

π2sini

5

π2cos

5

π2sini

5

π2cos

++

+

(M1)

+++++5

4sini

5

4cos

5

2sini

5

2cos1

π π π π

5

π8sini

5

π8cos

5

π6sini

5

π6cos +++

M1

+++++5

π4sini

5

π4cos

5

π2sini

5

π2cos1

5

π2sini

5

π2cos

5

π4sini

5

π4cos −+−

M1A1A1

Notes: Award M1 for attempting to replace 'p and 6p !y 4p and 2p Award A1 for correct cosine terms and A1 for correct sine terms.

05

2πcos25

4πcos21 =++= A1

Note: orrect met*ods in#ol#ing equating real parts use of conBugates or reciprocals are also accepted.

2

1

5

4πcos

5

2πcos −=+

A$

Note: 9se of cis notation is accepta!le t*roug*out t*is question.[12]

18. (a) r = 3

1−

(A1)

IB Questionbank Mathematics Higher Level 3rd edition ,


21/31

31

1

27

+=∞S

M1

( )25.204

81==∞S

A1 31

(!) Attempting to s*ow t*at t*e result is true for n = 1 M1

57 = a and /57 =

( )a

r

r a=

−−

1

1

A1

5ence t*e result is true for n = 1

Assume it is true for n = k

( )r

r aar ar ar a

k k

−−

=++++ −1

1... 12

M1onsider n = k + 1C

( ) k k k k ar

r

r aar ar ar ar a +

−−

=+++++ −1

1... 12

M1

( ) ( )r

r ar r a k k

−−+−

=1

11

=r

ar ar ar ak k k

−−+− +

1

1

A1

Note: Award A1 for an equi#alent correct intermediate step.

r

ar ak

−−

=+

1

1

=

( )r

r a k

−− +

1

1 1

A1

Note: llogical attempted proofs t*at use t*e result to !e pro#edwould gain M1A0A0 for t*e last t*ree a!o#e mar%s.

*e result is true for n = k

⇒ it is true for n = k + 1 and as it is

true for n =1 t*e result is pro#ed !y mat*ematical induction. /1 30

Note: o o!tain t*e final /1 mar% a reasona!le attempt must*a#e !een made to pro#e t*e k + 1 step.

[10]



22/31

19. METHOD 1

3

π"2 −== θ r

(A1)(A1)

( )

3

333πsini3πcos23i1

−

−−

−+ −=−∴ M1

( )πsiniπcos8

1+=

(M1)

8

1−=

A1

METHOD 2

(1 i 3 )(1 i 3 ) = 1 2i 3 (= 2 2i 3 ) (M1)A1

( 2 2i 3 )(1 i 3 ) = 6 (M1)(A1)

( ) 81

3i1

13 −=

−∴

A1

METHOD 3

Attempt at inomial e-pansion M1

(1 i 3 ) = 1 + (i 3 ) + (i 3 )2 + (i 3 ) (A1)

= 1 i 3 ; + i 3

(A1)= 6 A1

( ) 81

3i1

13 −=

−∴

M1[5]

IB Questionbank Mathematics Higher Level 3rd edition


23/31

20. EITHER

c*anging to modulusargument formr = 2

θ = arctan 3

π3 =

(M1)A1

+=+⇒

3

πisin

3

πcos231 nnnn

M1

if sin⇒= 0

3

πn

n = D0 3" 6"...B (M1)A1 C2

OR

θ = arctan 3

π3 =

(M1)(A1)

M1

n ∈ ∈=⇒ k k

n"π

3

π

M1∈=⇒ k k n "3 A1 32

[5]

21. (a) (i)2222

iii

i11

y x

y

y x

x

y x

y x

y x z +−

+=

−−

×+

=(M1)A1

(ii) z +

+−+

++=

2222 i

1

y x

y y

y x

x x

z = k (A1)

for k to !e real . – ⇒=+ 022 y x

y

. ( - 2 + .

2 – 1) = 0 M1A1

*ence . = 0 or - 2 + .

2 – 1 = 0 ⇒ - 2 + . 2 = 1 A$

(iii) w*en - 2 + .

2 = 1 z + z

1

= 2 - (M1)A1 - D 1 1⇒k E 2 A$



24/31

(!) (i) 0 – n

= cos( – nθ ) + i sin( – nθ ) = cos nθ – i sin nθ M1A1⇒ 0 n + 0 – n = (cos nθ + i sin nθ ) + (cos nθ – i sin nθ ) = 2 cos nθ M1A$

(ii) (rearranging)

(0 2 + 0 –2) – (0 + 0 –1) + 2 = 0 (M1)⇒ (2 cos 2θ ) – 2 cos θ + 2 = 0 A1⇒ 2( cos 2θ – cos θ + 1) = 0⇒ (2 cos2 θ – 1) – cos θ + 1 = 0 M1⇒ ' cos2 θ – cos θ – 2 = 0 A1⇒ ( cos θ – 2)(2 cos θ + 1) = 0 M1

2

1cos"

3

2cos −==∴ θ θ

A1A1

3

5sin

3

2cos ±=⇒= θ θ

A1

2

3sin

2

1cos ±=⇒−= θ θ

A1

2

3i

2

1"

3

5i

3

2±−±=∴w

A1A1

Note: Allow from incorrect cos θ andFor sin θ .[22]

22. (a) any appropriate form e.g. (cos θ + i sin θ )n = cos (nθ ) + i sin (nθ ) A1

(!) z n = cos nθ + i sin nθ A1

n z

1

= cos( – nθ ) + i sin( – nθ ) (M1)= cos nθ – i sin (nθ ) A1

t*erefore z n –

n z

1

= 2i sin (nθ ) A$

(c)

543

2

2

345

511

4

51

3

51

2

51

1

51

−+

−

+

−

+

−

+

−

+=

−

z z z

z z

z z

z z z

z z

(M1)(A1)

= z & – 5z + 10z –

53

1510

z z z −+

A1

IB Questionbank Mathematics Higher Level 3rd edition !


25/31


26/31

(g) 15

8$cos2

π

0

5 =∫ θ θ wit* appropriate reference to symmetry and grap*s.A1/1/1

Note: Award first /1 for partially correct reasoning e.g. s%etc*esof grap*s of sin and cos.

Award second /1 for fully correct reasoning in#ol#ing sin& and cos&.[22]

23. (a) 1 – 3i A1

(!) EITHER

(z – (1 + 3i ))(z – (1 – 3i )) = z 2 – 2z + 4 (M1)A1 (z ) = (z – 2)(z

2 – 2z + 4) (M1)

= z – 4z 2 + 6z – 8 A1

there


27/31

(iii) L( –1 + i) = 43π

i2-n + A1 31

(c) for comparing t*e product of two of t*e a!o#e results wit* t*e t*ird M1

for stating t*e result –1 + i = –1 (1 – i) n$ L ( –1 + i) L ( –1) + L (1 – i)1hence" the roert9 L(z 1z 2) = L(z 1) + L(z 2)does not *old for all #alues of z 1 and z 2 A$ 30

[9]

25. (a) z = 5 and 0 =24 a+

0 = 2z

524 2 =+ aattempt to sol#e equation M1

Note: Award M0 if modulus is not used.

a = ±4 A1A1 30

(!) z0 = (2 – 2a) + (4 + a)i A1forming equation 2 – 2a = 2 (4 + a) M1

a = 2

3−

A1 30[6]

IB Questionbank Mathematics Higher Level 3rd edition %


28/31

26. (a) METHOD 1

2

i

++

z

z

= iz + i = iz + 2i M1

(1 – i)z = i A1

z = i1

i

− A1

EITHER

z =

43π

cis2

2π

cis

M1

z =

4

3πcis

2

1or

4

3πcis

2

2

A1A1

OR

z =

+−=

+−i

2

1

2

1

2

i1

M1

z =

4

3πcis

2

1or

4

3πcis

2

2

A1A1

METHOD 2

i = y x

y x

i2

)1i(

++++

M1 - + i(. + 1) = – . + i( - + 2) A1 - = – . , - + 2 = . + 1 A1

sol#ing - = 2

1;

2

1=− y

A1

z =i

2

1

2

1+−

z =

4

3π

cis2

1

or 4

3π

cis2

2

A1A1

Note: Award A1 fort t*e correct modulus and A1 for t*e correct argument!ut t*e final answer must !e in t*e form r cis θ . Accept 1&


29/31

(!) su!stituting z = - + i. to o!tain 0 =i)2(

i)1(

y x

y x

++++

(A1)use of ( - + 2) – . i to rationalie t*e denominator M1

ω = 22)2(

))2)(1(i()1()2(

y x

x y xy y y x x

++

+++−++++

A1

=22

22

)2(

)22i()2(

y x

y x y y x x

++++++++

A$

(c) /e ω =22

22

)2(

2

y x

y y x x

+++++

= 1 M1⇒ - 2 + 2 - + . 2 + . = - 2 + 4 - + 4 + . 2 A1⇒ . = 2 - + 4 A1

w*ic* *as gradient m = 2 A1

(d) EITHER

arg (z ) =⇒

4

π

- = . (and - . @ 0) (A1)

ω =2222

2

)2(

)2i(3

)2(

32

x x

x

x x

x x

+++

+++

+

if arg(ω) = θ x x

x

32

23tn

2 ++

=⇒ θ

(M1)1

32

232

=++

x x

x

M1A1

OR

arg (z ) =⇒

4

π

- = . (and - . @ 0) A1

arg (0 ) =⇒

4

π

- 2 + 2 - + .

2 + . = - + 2. + 2 M1

sol#e simultaneously M1

- 2 + 2 - + -

2 + - = - + 2 - + 2 (or equi#alent) A1

THEN

- 2 = 1

- = 1 (as - @ 0) A1

Note: Award A0 for - = 1.

z = 2 A1

Note: A1low from incorrect #alues of - .[19]

IB Questionbank Mathematics Higher Level 3rd edition *


30/31

27. (a) (i) ω =

3

3

2πisin

3

π2cos

+

=

×+

×

3

2π3isin

3

π23cos

(M1)= cos 2π + i sin 2π A1= 1 A$

(ii) 1 + ω + ω2 = 1 +

+

+

+

3

4πsini

3

4πcos

3

2πsini

3

π2cos

M1A1

= 1 + 2

3i

2

1

2

3i

2

1−−+−

A1= 0 A$

(!) (i)

+

+ ++ 34π

i3

2π

ii eeeθ θ θ

=

++ 34π

ii3

2πi

ii eeeee θ θ θ (M1)

=

++

34π

i32π

ii ee1e θ

= eiθ (1 + ω + ω

2) A1

= 0 A$

(ii)

A1A1

Note: Award A1 for one point on t*e imaginary a-is and anot*er point mar%ed wit* appro-imately correct modulus and argument. Award A1 for t*ird point mar%ed to form an equilateral trianglecentred on t*e origin.

IB Questionbank Mathematics Higher Level 3rd edition 3,


31/31

(c) (i) attempt at t*e e-pansion of at least two linear factors (M1)

(z – 1)z 2 – z (ω + ω2) + ω or equi#alent (A1)use of earlier result (M1)

2 (z ) = (z – 1)(z 2 + z + 1) = z – 1 A1

(ii) equation to sol#e is z = 6 (M1)

z = 2 2ω 2ω2

A2

Note: Award A1 for 2 correct solutions.[16]

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question bank complex numberp1_answ.rtf