256 COMPLEX NUMBERS
Hence x=7 cos 35◦= 5. 734
and y=7 sin 35◦= 4. 015
Hence 7∠− 145 ◦=− 5. 734 −j 4. 015
Alternatively
7 ∠− 145 ◦=7 cos (− 145 ◦)+j7 sin (− 145 ◦)
=− 5. 734 −j 4. 015
23.7 Multiplication and division in
polar form
IfZ 1 =r 1 ∠θ 1 andZ 2 =r 2 ∠θ 2 then:
(i)Z 1 Z 2 =r 1 r 2 ∠(θ 1 +θ 2 ) and
(ii)
Z 1
Z 2
=
r 1
r 2
∠(θ 1 −θ 2 )
Problem 12. Determine, in polar form:
(a) 8∠ 25 ◦× 4 ∠ 60 ◦
(b) 3∠ 16 ◦× 5 ∠− 44 ◦× 2 ∠ 80 ◦
(a) 8∠ 25 ◦× 4 ∠ 60 ◦=(8×4)∠(25◦+ 60 ◦)= 32 ∠ 85 ◦
(b) 3∠ 16 ◦× 5 ∠− 44 ◦× 2 ∠ 80 ◦
=(3× 5 ×2)∠[16◦+(− 44 ◦)+ 80 ◦]= 30 ∠ 52 ◦
Problem 13. Evaluate in polar form
(a)
16 ∠ 75 ◦
2 ∠ 15 ◦
(b)
10 ∠
π
4
× 12 ∠
π
2
6 ∠−
π
3
(a)
16 ∠ 75 ◦
2 ∠ 15 ◦
=
16
2
∠(75◦− 15 ◦)= 8 ∠ 60 ◦
(b)
10 ∠
π
4
× 12 ∠
π
2
6 ∠−
π
3
=
10 × 12
6
∠
(π
4
+
π
2
−
(
−
π
3
))
= 20 ∠
13 π
12
or 20 ∠−
11 π
12
or
20 ∠ 195 ◦or 20 ∠− 165 ◦
Problem 14. Evaluate, in polar form
2 ∠ 30 ◦+ 5 ∠− 45 ◦− 4 ∠ 120 ◦.
Addition and subtraction in polar form is not possible
directly. Each complex number has to be converted
into cartesian form first.
2 ∠ 30 ◦=2(cos 30◦+jsin 30◦)
=2 cos 30◦+j2 sin 30◦= 1. 732 +j 1. 000
5 ∠− 45 ◦=5(cos(− 45 ◦)+jsin(− 45 ◦))
=5 cos(− 45 ◦)+j5 sin(− 45 ◦)
= 3. 536 −j 3. 536
4 ∠ 120 ◦=4( cos 120◦+jsin 120◦)
=4 cos 120◦+j4 sin 120◦
=− 2. 000 +j 3. 464
Hence 2∠ 30 ◦+ 5 ∠− 45 ◦− 4 ∠ 120 ◦
=(1. 732 +j 1 .000)+(3. 536 −j 3 .536)
−(− 2. 000 +j 3 .464)
= 7. 268 −j 6 .000, which lies in the
fourth quadrant
=
√
[(7.268)^2 +(6.000)^2 ]∠tan−^1
(
− 6. 000
7. 268
)
= 9. 425 ∠− 39. 54 ◦or 9. 425 ∠− 39 ◦ 32 ′
Now try the following exercise.
Exercise 103 Further problems on polar
form
- Determine the modulus and argument of
(a) 2+j4 (b)− 5 −j2 (c)j(2−j).
⎡
⎢
⎣
(a) 4.472, 63◦ 26 ′
(b) 5.385,− 158 ◦ 12 ′
(c) 2.236, 63◦ 26 ′
⎤
⎥
⎦
In Problems 2 and 3 express the given Cartesian
complex numbers in polar form, leaving answers
in surd form.
- (a) 2+j3 (b)−4 (c)− 6 +j
[
(a)
√
13 ∠ 56 ◦ 19 ′ (b) 4∠ 180 ◦
(c)
√
37 ∠ 170 ◦ 32 ′
]
- (a)−j3 (b) (− 2 +j)^3 (c)j^3 (1−j)
[
(a) 3∠− 90 ◦ (b)
√
125 ∠ 100 ◦ 18 ′
(c)
√
2 ∠− 135 ◦
]