Tutorial1Scheme/EQT101/20172018/ pg. 1
TUTORIAL 1: ANSWER SCHEME
Question 1
a)
2 2
1
2 3
2 3 2 3
2 3 2 3 13
3arg tan 56.31
2
z i
z i i
z i
z
b)
2
4
4 4
4 4 4
arg 90 / 270
z i
z i i
z i
z
c)
2
16
16 16
16 16 16
arg 0 / 360
z
z
z
z
Tutorial1Scheme/EQT101/20172018/ pg. 2
d)
2 2
1
3
3 3
3 3 1 10
1arg tan 18.43 / 341.57
3
z i
z i i
z i
z
e)
2 2
1
1 5
1 5 1 5
1 5 1 5 26
1arg tan 101.31 / 258.69
5
z i
z i i
z i
z
f)
22
1
1 3
1 3 1 3
1 3 1 3 2
3arg tan 120
1
z i
z i i
z i
z
Tutorial1Scheme/EQT101/20172018/ pg. 3
Question 2
a) 2 3 2 1 3p q i i i
b) 4 2 5 4 7z w i i i
c) 2 3 2 2 3 4 2 16 4p z i i i
Tutorial1Scheme/EQT101/20172018/ pg. 4
d) 3 3 2 3 5 12 2q iw i i i i
Question 3
a) 2 5 3 15 6zw i i i
b) 2 5 3 5 2
3 3 3 3
z i ii
w i i
c) 3 3 2 5 15 6
2 5 2 5 29 292 5
w i i ii
i iz i
d)
3 3 2 5
2 5 2 5 2 5
15 6 15 6
29 29 29
w i i i
z i i i
ii
e) 4 2 5 4 2 5 3 5 2
3 3 3 3 3
z i i ii
w i i i
f) 5 5 3 2 5 5 31
2 5 2 5 29 29
w i ii
z i i
g)
22 2 5 3
2 53 3
58 58
3 3
z i iz i
w i i
i
Question 4
a)
5
1 2
5 3
3 3
3 1
2 2
pi i
i
i i
i
b)
2
3 2
4
3 2 15 8
15 8 15 8
29 54
289 289
iq
i
i i
i i
i
Tutorial1Scheme/EQT101/20172018/ pg. 5
c)
7
3
4
2 3
2 3
1 2 3
2 3 2 3
2 3
13 13
iw
i i
i
i
i
i i
i
d) 4
5
1
5
1 1
5
1
5
is
i
i
i
e)
6 3
5 2 1
6 5 2 3 1
5 2 5 2 1 1
30 121 2
29 29
59 46
29 29
iz
i i
i i i
i i i i
i i
i
Question 5
a) 2 3iz z i
2 3
2 3
2 2 3
Re : Im :
2 2 3
6 3,
5 5
6 3
5 5
i a bi a bi i
i a bi a b i
b ai a b i
z z
b a a b
a b
z i
b) 5 42
z iz
5 42
15 4 4
2 2
Re : Im :
15 4 4
2 2
10 1,
9 9
10 1
9 9
a bi ia bi
a ba bi i
z z
a ba b
a b
z i
Tutorial1Scheme/EQT101/20172018/ pg. 6
c) 2 2 5z i iz
2 2 5
3 2 5 5
Re : Im :
3 2 5 5
1 5,
14 14
1 5
14 14
a bi a bi i a bi
a bi b ai
z z
a b b a
a b
z i
d) 3
2z i
i z
32
3 2 2 1
Re : Im :
2 3 2 2
10,
3
1
3
a bi i
i a bi
a b i a b i
z z
a a b b
a b
z i
Question 6
a) 2z i
1
0.1476
5
1arg tan
2
26.57
0.1476
: 5 cos 26.57 sin 26.57
: 5 i
z
z
PF z i
EF z e
b) 3 2z i
1
0.8128 1.1872
13
2arg tan
3
146.31 / 213.69
0.8128 /1.1872
: 13 cos146.31 sin146.31
13 cos 213.69 sin 213.69
: 13 / 13i i
z
z
PF z i
z i
EF z e z e
c) 4 7z i
1
0.3348 1.6652
65
7arg tan
4
60.26 / 299.74
0.3348 /1.6652
: 65 cos60.26 sin 60.26
65 cos 299.74 sin 299.74
: 65 / 65i i
z
z
PF z i
z i
EF z e z e
d) 3z i
2
3
arg 90
2
: 3 cos90 sin 90
: 3i
z
z
PF z i
EF z e
Tutorial1Scheme/EQT101/20172018/ pg. 7
Question 7
a)
6 cos30 sin 30
3 cos90 sin 90
iz
i
6 cos 30 90 sin 30 90
3
6 cos 120 sin 120
3
1 1
6 2
iz
i
i
b) 2 2 cos45 sin 45 cos180 sin180z i i
2 2 cos 45 180 sin 45 180
2 2 cos 135 sin 135
1 12 2
2 2
2 2
z i
i
i
i
c) 2
2
5
i
i
ez
e
2
2
2
5
2
5
i
i
z e
e
Transform the equation into the polar
form:
2
cos90 sin 905
2
5
z i
i
Write in the standard form:
2
5z i
d) 4 2
5 3.7i i
z e e
4 2
5 3
22
15
7
7
i
i
z e
e
Transform the equation into polar
form:
7 cos 264 sin 264
0.7315 6.9617
z i
i
Write in the standard form:
0.7315 6.9617z i
Tutorial1Scheme/EQT101/20172018/ pg. 8
Question 8
a) 1
3 34 , ?z i z
11
33
4 , 17, arg 14.04
3, 0,1,2
14.04 360 14.04 36017 cos sin
3 3k
w i w w
n k
k kz i
0
11
330
1
3
1
3
0 :
14.04 14.0417 cos sin
3 3
17 cos 4.68 sin 4.68
17 0.9967 0.0816
1.5982 0.1308
1.6035z
k
z i
i
i
i
r
1
11
331
1
3
1
3
1:
14.04 360 14.04 36017 cos sin
3 3
17 cos 115.32 sin 115.32
17 0.4277 0.9039
0.6858 1.4495
1.6035z
k
z i
i
i
i
r
1
11
332
1
3
1
3
2 :
14.04 720 14.04 72017 cos sin
3 3
17 cos 235.32 sin 235.32
17 0.569 0.8224
0.9124 1.3187
1.6035z
k
z i
i
i
i
r
Tutorial1Scheme/EQT101/20172018/ pg. 9
b) 1
2 22 3 , ?z i z
11
22
2 3 , 13, arg 123.69
2, 0,1
123.69 360 123.69 36013 cos sin
3 3k
w i w w
n k
k kz i
0
11
220
1
2
1
2
0 :
123.69 123.6913 cos sin
2 2
13 cos 61.85 sin 61.85
13 0.4719 0.8817
0.896 1.6741
1.5334z
k
z i
i
i
i
r
1
11
221
1
2
1
2
1:
123.69 360 123.69 36013 cos sin
2 2
13 cos 241.845 sin 241.845
13 0.4719 0.8817
0.896 1.6741
1.5334z
k
z i
i
i
i
r
c) 1
4 45 0, ?z i z
1 1
4 4
5 , 5, arg 90
4, 0,1,2,3
90 360 90 3605 cos sin
4 4k
w i w w
n k
k kz i
Tutorial1Scheme/EQT101/20172018/ pg. 10
0
1 1
4 40
1
4
1
4
0 :
90 905 cos sin
4 4
5 cos 22.5 sin 22.5
5 0.9239 0.3827
1.3815 0.5722
1.4953z
k
z i
i
i
i
r
1
1 1
4 41
1
4
1
4
1:
90 360 90 3605 cos sin
4 4
5 cos 67.5 sin 67.5
5 0.3827 0.9239
0.5722 1.3815
1.4953z
k
z i
i
i
i
r
2
1 1
4 42
1
4
1
4
2 :
90 720 90 7205 cos sin
4 4
5 cos 157.5 sin 157.5
5 0.9239 0.3827
1.3815 0.5722
1.4953z
k
z i
i
i
i
r
3
1 1
4 43
1
4
1
4
3:
90 1080 90 10805 cos sin
4 4
5 cos 247.5 sin 247.5
5 0.3827 0.9239
0.5722 1.3815
1.4953z
k
z i
i
i
i
r
Tutorial1Scheme/EQT101/20172018/ pg. 11
d) 1
3 32 0, ?z z
1 13 3
2, 2, arg 0
3, 0,1,2
0 360 0 3602 cos sin
3 3k
w w w
n k
k kz i
0
1 13 30
1
3
1
3
0 :
0 02 cos sin
3 3
2 cos 0 sin 0
2 1
1.2599
1.2599z
k
z i
i
r
1
1 13 3
1
1
3
1
3
1:
0 360 0 3602 cos sin
3 3
2 cos 120 sin 120
1 32
2 2
0.63 1.0911
1.2599z
k
z i
i
i
i
r
2
1 13 32
1
3
1
3
2 :
0 720 0 7202 cos sin
3 3
2 cos 240 sin 240
1 32
2 2
0.63 1.0911
1.2599z
k
z i
i
i
i
r
Tutorial1Scheme/EQT101/20172018/ pg. 12
Question 9
Multiple angles n
Theorem 6:
12cos
12 sin
n
n
n
n
z nz
z i nz
a) 3cos
From theorem 6, let 1:n
1
2cos 1zz
Equation 1 to the power of 3:
3
312cos 2z
z
From (2):
RHS: 3 32cos 8cos
LHS:
3 2 3
3 3 2 3 3
1 2 3
3
3
3
3
1 1 1 1
1 13 3
1 13
2cos3 3 2cos
2cos3 6cos
z z C z C z Cz z z z
z zz z
z zz z
Equate LHS and RHS 3
3
8cos 2cos3 6cos
1 3cos cos3 cos
4 4
b) 4cos
From theorem 6, let 1:n
1
2cos 3zz
Equation 3 to the power of 4:
4
412cos 4z
z
Tutorial1Scheme/EQT101/20172018/ pg. 13
From (4):
RHS: 4 42cos 16 cos
LHS:
4 2 3 4
4 4 3 4 3 4 4
1 2 3 4
4 2
2 4
24
4 2
1 1 1 1 1
1 14 6 4
1 14 6
2cos 4 4 2cos 2 6
2cos 4 8cos 2 6
z z C z C z C z Cz z z z z
z zz z
z zz z
Equate LHS and RHS 4
4
16cos 2cos 4 8cos 2 6
1 1 3cos cos 4 cos 2
8 2 8
c) 5cos
From theorem 6, let 1:n
1
2cos 5zz
Equation 5 to the power of 5:
5
512cos 6z
z
From (6):
RHS: 5 52cos 32 cos
LHS:
5 2 3 4 5
5 5 4 5 3 5 2 5 5
1 2 3 4 5
4 3
3 5
35
5 3
1 1 1 1 1 1
1 1 15 10 10 5
1 1 15 10
2cos5 5 2cos3 10 2cos
2cos5 10cos 2 20cos
z z C z C z C z C z Cz z z z z z
z z zz z z
z z zz z z
Equate LHS and RHS 5
5
32cos 2cos5 10cos3 20cos
1 5 5cos cos5 cos3 cos
16 16 8
Tutorial1Scheme/EQT101/20172018/ pg. 14
d) 3sin
From theorem 6, let 1:n
1
2 sin 7z iz
Equation 7 to the power of 3:
3
312 sin 8z i
z
From (8):
RHS: 3 32 sin 8 sini i
LHS:
3 2 3
3 3 2 3 3
1 2 3
3
3
3
3
1 1 1 1
1 13 3
1 13
2 sin 3 3 2 sin
2 sin 3 6 sin
z z C z C z Cz z z z
z zz z
z zz z
i i
i i
Equate LHS and RHS 3
3
8 sin 2 sin 3 6 sin
1 3sin sin 3 sin
4 4
i i i
e) 4sin
From theorem 6, let 1:n
1
2 sin 9z iz
Equation 9 to the power of 4:
4
412 sin 10z i
z
From (10):
RHS: 4 42 sin 16sini
Tutorial1Scheme/EQT101/20172018/ pg. 15
LHS:
4 2 3 4
4 4 3 4 3 4 4
1 2 3 4
4 2
2 4
24
4 2
1 1 1 1 1
1 14 6 4
1 14 6
2cos 4 4 2cos 2 6
2cos 4 8cos 2 6
z z C z C z C z Cz z z z z
z zz z
z zz z
Equate LHS and RHS 4
4
16sin 2cos 4 8cos 2 6
1 1 3sin cos 4 cos 2
8 2 8
f) 5sin
From theorem 6, let 1:n
1
2 sin 11z iz
Equation 11 to the power of 5:
5
512 sin 12z i
z
From (12):
RHS: 5 52 sin 32 sini i
LHS:
5 2 3 4 5
5 5 4 5 3 5 2 5 5
1 2 3 4 5
5 3
3 5
35
5 3
1 1 1 1 1 1
1 1 15 10 10 5
1 1 15 10
2 sin 5 5 2 sin 3 10 2 sin
2 sin 5 10 sin 2 2
z z C z C z C z C z Cz z z z z z
z z zz z z
z z zz z z
i i i
i i
0 sini
Equate LHS and RHS 5
5
32 sin 2 sin 5 10 sin 3 20 sin
1 5 5sin sin 5 sin 3 sin
16 16 8
i i i i
Tutorial1Scheme/EQT101/20172018/ pg. 16
Question 10
De Moivre’s Theorem (DMT)
cos sin cos sinn
i n i n
a) From the DMT, when n =3:
3
cos sin cos3 sin 3 13i i
LHS, let:
cos
sin
a
b i
3 3
3 3 2 3 2 3 3
1 2 3
3 2 2 3
3 2 2 3
3 2 2 3
cos sin
3 3
cos 3 cos sin 3cos sin sin
cos 3cos sin 3cos sin sin
i a b
a C a b C ab C b
a a b ab b
i i
i
Equate LHS and RHS
3 2 2 3
3 2
2 3
cos3 sin 3 cos 3cos sin 3cos sin sin
Re : cos3 cos 3cos sin
Im : sin 3 3cos sin sin
i i
z
z
b) From the DMT, when n =4:
4
cos sin cos 4 sin 4 14i i
LHS, let:
cos
sin
a
b i
4 4
4 4 3 4 2 2 4 3 4 4
1 2 3 4
4 3 2 2 3 4
4 3 2 2 3 4
4 2 2 4 3 3
cos sin
4 6 4
cos 4 cos sin 6 cos sin 4 cos sin sin
cos 6 cos sin sin 4 cos sin 4 cos sin
i a b
a C a b C a b C ab C b
a a b a b ab b
i i
i
Tutorial1Scheme/EQT101/20172018/ pg. 17
Equate LHS and RHS
4 2 2 4 3 3
4 2 2 4
3 3
cos 4 sin 4 cos 6 cos sin sin 4 cos sin 4 cos sin
Re : cos 4 cos 6 cos sin sin
Im : sin 4 4 cos sin 4 cos sin
i i
z
z
c) From the DMT, when n =5:
5
cos sin cos5 sin 5 15i i
LHS, let:
cos
sin
a
b i
5 5
5 5 4 5 3 2 5 2 3 5 4 5 5
1 2 3 4 5
5 4 3 2 2 3 4 5
5 4 3 2 2 3 4 5
5 3 2 4 4 2
cos sin
5 10 10 5
cos 5 cos sin 10 cos sin 10 cos sin 5cos sin sin
cos 10 cos sin 5cos sin 5cos sin 10 cos si
i a b
a C a b C a b C a b C ab C b
a a b a b a b ab b
i i i
i
3 5n sin
Equate LHS and RHS
5 3 2 4 4 2 3 5
5 3 2 4
4 2 3 5
cos5 sin 5 cos 10 cos sin 5cos sin 5cos sin 10 cos sin sin
Re : cos5 cos 10 cos sin 5cos sin
Im : sin 5 5cos sin 10 cos sin sin
i i
z
z
d) From the DMT, when n =6:
6
cos sin cos6 sin 6 16i i
LHS, let:
cos
sin
a
b i
6 6
6 6 5 6 4 2 6 3 3 6 2 4 6 5 6 6
1 2 3 4 5 6
6 5 4 2 3 3 2 4 5 6
6 5 4 2 3 3
2 4 5 6
6 4 2
cos sin
6 15 20 15 6
cos 6 cos sin 15cos sin 20 cos sin
15cos sin 6 cos sin sin
cos 15cos sin 15co
i a b
a C a b C a b C a b C a b C ab C b
a a b a b a b a b ab b
i i
i
2 4
5 3 3 5
s sin
6 cos sin 20 cos sin 6 cos sini
Tutorial1Scheme/EQT101/20172018/ pg. 18
Equate LHS and RHS
6 4 2 2 4 6
5 3 3 5
6 4 2 2 4 6
5 3 3 5
cos6 sin 6 cos 15cos sin 15cos sin sin
6 cos sin 20 cos sin 6 cos sin
Re : cos6 cos 15cos sin 15cos sin sin
Im : sin 6 6 cos sin 20 cos sin 6 cos sin
i
i
z
z
Question 11
a) 4 2 3z i
2 2
2 2
4 2 3
4 2 3
4 2 3
4 2 9
x iy i
x i y
x y
x y
A circle with radius, r =3 and centre, C(4,-2).
b) 3 5z z
22 2 2
22 2 2
2 2 2 2
2 2
2 2
2
2
3 5
3 5
3 5
9 5
9 9 10 25
8 10 25 8 0
5 250
4 4
5 225
8 64
x iy x iy
x iy x iy
x y x y
x y x y
x y x x y
x x y
x x y
x y
A circle with radius, r = 15
8and centre,
5,0
8C
.
Tutorial1Scheme/EQT101/20172018/ pg. 19
c) 3 2z i z
2 22 2
2 22 2
2 2 2 2
3 2
3 2
3 2
3 2
6 9 4 4
6 4 5
2 5
3 6
x iy i x iy
x y i x iy
x y x y
x y x y
x y y x x y
y x
y x
A
straight line with slope, 2
3m .
d) arg4
z i
1
arg4
arg 14
1tan
4
1tan
4
11
1
x iy i
x y i
y
x
y
x
y
x
y x
A half line from (0,1) with gradient, 1m
Top Related