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]
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