Higher Engineering Mathematics

(Greg DeLong) #1
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



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


  1. 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.


  1. (a) 2+j3 (b)−4 (c)− 6 +j
    [
    (a)



13 ∠ 56 ◦ 19 ′ (b) 4∠ 180 ◦

(c)


37 ∠ 170 ◦ 32 ′

]


  1. (a)−j3 (b) (− 2 +j)^3 (c)j^3 (1−j)
    [
    (a) 3∠− 90 ◦ (b)



125 ∠ 100 ◦ 18 ′

(c)


2 ∠− 135 ◦

]
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