IB Questionbank Mathematics Higher Level 3rd edition 1

Chapter 10 IB HL Complex Numbers Worksheet MARKSCHEME

1. METHOD 1

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

z = A1

z = M1

= – 3 – i A1

METHOD 2

let z = a + ib

= 2 – i M1

a + ib = (2 – i)((a + 2) + ib) a + ib = 2(a + 2) + 2bi – i(a + 2) + b a + ib = 2a + b + 4 + (2b – a – 2)i attempt 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 = –3 z = – 3 – i A1

[4]

2. METHOD 1

(A1)(A1)

M1

(M1)

A1

METHOD 2

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

(− 2 − 2i )(1 − i ) = −8 (M1)(A1)

A1

i1i24

−+−

i1i1

i1i24

++×

−+−

2ii++

+baba

⇒⇒

3π,2 −== θr

( )3

33

3πsini

3πcos23i1

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛−+⎟

⎠⎞⎜

⎝⎛−=−∴

( )πsiniπcos81 +=

81−=

3 3 3 3

3 3

( ) 81

3i1

13

−=−

∴

IB Questionbank Mathematics Higher Level 3rd edition 2

METHOD 3

Attempt at Binomial expansion M1

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

= 1 − 3i − 9 + 3i (A1)

= −8 A1

M1

[5]

3. (a) AB = M1

= A1

= A1

(b) METHOD 1

A1A1

Note: Allow .

Note: Allow degrees at this stage.

= A1

Note: Allow FT for final A1.

METHOD 2

attempt to use scalar product or cosine rule M1

A1

A1

[6]

4. (a) METHOD 1

= i

z + i = iz + 2i M1 (1 – i)z = i A1

z = A1

EITHER

3 3 3 3

3 3

( ) 81

3i1

13

−=−

∴

22 )32(1 −+

3488−

322 −

3πarg,

4π z arg 2 1 −=−= z

3πand

4π

4π

3πBOA −=

)12π(accept

12π −

2231BOAcos +=

12πBOA =

2i

++zz

i1i−

IB Questionbank Mathematics Higher Level 3rd edition 3

z = M1

z = A1A1

OR

z = M1

z = A1A1

METHOD 2

i = M1

x + i(y + 1) = –y + i(x + 2) A1 x = –y; x + 2 = y + 1 A1

solving, x = A1

z =

z = A1A1

Note: Award A1 fort the correct modulus and A1 for the correct argument, but the final answer must be in the form r cis θ. Accept 135° for the argument.

(b) substituting z = x + iy to obtain w = (A1)

use of (x + 2) – yi to rationalize the denominator M1

ω = A1

= AG

(c) Re ω = = 1 M1

x2 + 2x + y2 + y = x2 + 4x + 4 + y2 A1 y = 2x + 4 A1

which has gradient m = 2 A1

(d) EITHER

⎟⎠⎞⎜

⎝⎛

⎟⎠⎞⎜

⎝⎛

43πcis2

2πcis

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

43πcis

21or

43πcis

22

⎟⎠⎞⎜

⎝⎛ +−=+− i

21

21

2i1

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

43πcis

21or

43πcis

22

yxyxi2)1i(

++++

21;

21 =− y

i21

21 +−

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

43πcis

21or

43πcis

22

i)2(i)1(yx

yx++++

22)2())2)(1(i()1()2(

yxxyxyyyxx

+++++−++++

22

22

)2()22i()2(

yxyxyyxx

++++++++

22

22

)2(2

yxyyxx

+++++

⇒⇒

IB Questionbank Mathematics Higher Level 3rd edition 4

arg (z) = x = y (and x, y > 0) (A1)

ω =

if arg(ω) = θ (M1)

M1A1

OR

arg (z) = x = y (and x, y > 0) A1

arg (w) = x2 + 2x + y2 + y = x + 2y + 2 M1

solve simultaneously M1 x2 + 2x + x2 + x = x + 2x + 2 (or equivalent) A1

THEN

x2 = 1 x = 1 (as x > 0) A1

Note: Award A0 for x = ±1.

│z│ = A1

Note: A1low FT from incorrect values of x. [19]

5. (a) (A1)

arg or arg (1 − i) = (A1)

A1

A1

arg (z1) = m arctan A1

arg (z2) = n arctan (−1) = A1 N2

(b) (M1)A1

M1A1

⇒4π

2222

2

)2()2i(3

)2(32

xxx

xxxx

+++

+++

+

xxx3223tan

2 ++=⇒ θ

13223

2=

++xx

x

⇒4π

⇒4π

2

2i1or23i1 =−=+

( )3π3i1 =+

4π− ⎟

⎠⎞⎜

⎝⎛

47πaccept

mz 21 =

nz 22 =

3π3 m=

4π−n ⎟

⎠⎞⎜

⎝⎛

47πaccept n

mnnm 222 =⇒=

integeran is where,π24π

3π kknm +−=

knm π24π

3π =+⇒

IB Questionbank Mathematics Higher Level 3rd edition 5

(M1)

A1

The smallest value of k such that m is an integer is 5, hence

m = 12 A1

n = 24 . A1 N2 [14]

6. (a) z = Let 1 – i = r(cos θ + i sin θ)

A1

θ = A1

z = M1

=

= M1

=

Note: Award M1 above for this line if the candidate has forgotten to add 2π and no other solution given.

=

=

= A2

Note: Award A1 for 2 correct answers. Accept any equivalent form.

(b)

kmm π24π2

3π =+⇒

km π2π65 =

km512=⇒

41

i)1( −

2=⇒ r

4π−

41

4πisin

4πcos2 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛−+⎟

⎠⎞⎜

⎝⎛−

41

π24πsiniπ2

4πcos2 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛ +−+⎟

⎠⎞⎜

⎝⎛ +− nn

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛ +−+⎟

⎠⎞⎜

⎝⎛ +−

2π

16πsini

2π

16πcos2 8

1 nn

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛−+⎟

⎠⎞⎜

⎝⎛−

16πsini

16πcos2 8

1

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

167πsini

167πcos2 8

1

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

1615πsini

1615πcos2 8

1

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛−+⎟

⎠⎞⎜

⎝⎛−

169πsini

169πcos2 8

1

IB Questionbank Mathematics Higher Level 3rd edition 6

A2

Note: Award A1 for roots being shown equidistant from the origin and one in each quadrant. A1 for correct angular positions. It is not necessary to see written evidence of angle, but must agree with the diagram.

(c) M1A1

= (A1)

= i A1 N2 ( a = 0, b = 1)

[12]

7. (a) z = (M1)

z = A1 N2

(b) │z│ A2

│z│< 1 AG

(c) Using S∞ = (M1)

S∞ = A1 N2

(d) (i) S∞ = (M1)

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

=

167πsini

16π7cos2

1615πsini

16π15cos2

81

81

1

2

zz

2πsini

2πcos +

⇒

θ

θ

i

2i

e

e21

θie21

21=

ra−1

θ

θ

i

i

e211

e

−

θ

θθ

θ

cis211

cis

e211

ei

i

−=

−

IB Questionbank Mathematics Higher Level 3rd edition 7

(A1)

Also S∞ = eiθ +

= cis θ + (M1)

S∞ = A1

(ii) Taking real parts,

A1

= M1

= A1

= A1

= A1

= A1AG N0

[25]

8. EITHER

changing to modulus-argument form r = 2

θ = arctan (M1)A1

M1

if sin n = {0, ±3, ±6,...} (M1)A1 N2

OR

)sin i(cos211

sin icos

θθ

θθ

+−

+

...e41e

21 3i2i ++ θθ

...cis341cis2

21 ++ θθ

⎟⎠⎞⎜

⎝⎛ ++++⎟

⎠⎞⎜

⎝⎛ +++ ...3sin

412sin

21sini...3cos

412cos

21cos θθθθθθ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

+=+++)sin i(cos

211

sin icosRe...3cos412cos

21cos

θθ

θθθθθ

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞⎜

⎝⎛ +−

+−×

⎟⎠⎞⎜

⎝⎛ −−

+

θθ

θθ

θθ

θθ

sin i21cos

211

sin i21cos

211

sin i21cos

211

)sin i(cosRe

θθ

θθθ

22

22

sin41cos

211

sin21cos

21cos

+⎟⎠⎞⎜

⎝⎛ −

−−

)cos(sin41cos1

21cos

22 θθθ

θ

++−

⎟⎠⎞⎜

⎝⎛ −

)cos45(2)1cos2(4

4)1cos44(2)1cos2(

θθ

θθ

−−

=÷+−

÷−

θθcos45

2cos4−

−

3π3 =

⎟⎠⎞⎜

⎝⎛ +=+⇒

3πisin

3πcos231 nnnn

⇒= 03πn

IB Questionbank Mathematics Higher Level 3rd edition 8

θ = arctan (M1)(A1)

M1

n M1

A1 N2 [5]

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

(b) zn = cos nθ + i sin nθ A1

= cos(–nθ) + i sin(–nθ) (M1)

= cos nθ – i sin (nθ) A1

therefore zn – = 2i sin (nθ) AG

(c)

(M1)(A1)

= z5 – 5z3 + 10z – A1

(d) M1A1

(2i sin θ)5 = 2i sin 5θ – 10i sin 3θ + 20i sin θ M1A1 16 sin5 θ = sin 5θ – 5 sin 3θ + 10 sin θ AG

(e) 16 sin5 θ = sin 5θ – 5 sin 3θ + 10 sin θ

LHS =

=

= A1

3π3 =

∈ ∈=⇒ kkn ,π3π

∈=⇒ kkn ,3

nz1

nz1

5432

2345

5 11451

351

251

151

⎟⎠⎞⎜

⎝⎛−+⎟

⎠⎞⎜

⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞⎜

⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎠⎞⎜

⎝⎛ −

zzz

zz

zz

zzz

zz

53

1510zzz

−+

⎟⎠⎞⎜

⎝⎛ −+⎟

⎠⎞⎜

⎝⎛ −−−=⎟

⎠⎞⎜

⎝⎛ −

zz

zz

zz

zz 1101511

33

55

5

5

4πsin16 ⎟⎠⎞⎜

⎝⎛

5

2216 ⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2422

IB Questionbank Mathematics Higher Level 3rd edition 9

RHS =

= M1A1

Note: Award M1 for attempted substitution.

= A1

hence this is true for θ = AG

(f) M1

= A1

= A1

= A1

(g) , with appropriate reference to symmetry and graphs.A1R1R1

Note: Award first R1 for partially correct reasoning e.g. sketches of graphs of sin and cos. Award second R1 for fully correct reasoning involving sin5 and cos5.

[22]

10. (a) (i) ω3 =

= (M1)

= cos 2π + i sin 2π A1 = 1 AG

(ii) 1 + ω + ω2 = 1 + M1A1

= 1 + A1

= 0 AG

(b) (i)

⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛−⎟

⎠⎞⎜

⎝⎛

4πsin10

4π3sin5

4π5sin

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛−−

2210

225

22

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2422

4π

∫∫ +−= 2π

02π

0

5 d)sin103sin55(sin161dsin θθθθθθ

2π

0

cos1033cos5

55cos

161

⎥⎦⎤

⎢⎣⎡ −+− θθθ

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞⎜

⎝⎛ −+−− 10

35

510

161

158

158dcos2

π

0

5 =∫ θθ

3

32πisin

3π2cos ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

⎟⎠⎞⎜

⎝⎛ ×+⎟

⎠⎞⎜

⎝⎛ ×

32π3isin

3π23cos

⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛

34πsin i

34πcos

32πsin i

3π2cos

23i

21

23i

21 −−+−

⎟⎠⎞⎜

⎝⎛ +⎟

⎠⎞⎜

⎝⎛ +

++ 34πi

32πi

i eeeθθ

θ

IB Questionbank Mathematics Higher Level 3rd edition 10

= (M1)

=

= eiθ(1 + ω + ω2) A1 = 0 AG

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

++ 34πi

i32πi

ii eeeee θθθ

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛++

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛

34πi

32πi

i ee1e θ

IB Questionbank Mathematics Higher Level 3rd edition 11

(ii)

A1A1

Note: Award A1 for one point on the imaginary axis and another point marked with approximately correct modulus and argument. Award A1 for third point marked to form an equilateral triangle centred on the origin.

(c) (i) attempt at the expansion of at least two linear factors (M1) (z – 1)z2 – z(ω + ω2) + ω3 or equivalent (A1) use of earlier result (M1) F(z) = (z – 1)(z2 + z + 1) = z3 – 1 A1

(ii) equation to solve is z3 = 8 (M1) z = 2, 2ω, 2ω2 A2

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