Higher Engineering Mathematics

COMPLEX NUMBERS 257

E

In Problems 4 and 5 convert the given polar com-

plex numbers into (a+jb) form giving answers

correct to 4 significant figures.

4. (a) 5∠ 30 ◦ (b) 3∠ 60 ◦ (c) 7∠ 45 ◦
   ⎡

⎣

(a) 4. 330 +j 2. 500

(b) 1. 500 +j 2. 598

(c) 4. 950 +j 4. 950

⎤

⎦

5. (a) 6∠ 125 ◦ (b) 4∠π (c) 3. 5 ∠− 120 ◦
   ⎡

⎣

(a)− 3. 441 +j 4. 915

(b)− 4. 000 +j 0

(c)− 1. 750 −j 3. 031

⎤

⎦

In Problems 6 to 8, evaluate in polar form.

6. (a) 3∠ 20 ◦× 15 ∠ 45 ◦

(b) 2. 4 ∠ 65 ◦× 4. 4 ∠− 21 ◦

[(a) 45∠ 65 ◦(b) 10. 56 ∠ 44 ◦]

7. (a) 6. 4 ∠ 27 ◦÷ 2 ∠− 15 ◦

(b) 5∠ 30 ◦× 4 ∠ 80 ◦÷ 10 ∠− 40 ◦

[(a) 3. 2 ∠ 42 ◦(b) 2∠ 150 ◦]

8. (a) 4∠

π

6

+ 3 ∠

π

8

(b) 2∠ 120 ◦+ 5. 2 ∠ 58 ◦− 1. 6 ∠− 40 ◦

[(a) 6. 986 ∠ 26 ◦ 47 ′(b) 7. 190 ∠ 85 ◦ 46 ′]

23.8 Applications of complex numbers

There are several applications of complex numbers
in science and engineering, in particular in electrical
alternating current theory and in mechanical vector
analysis.
The effect of multiplying a phasor byjis to rotate
it in a positive direction (i.e. anticlockwise) on an
Argand diagram through 90◦without altering its
length. Similarly, multiplying a phasor by−jrotates
the phasor through− 90 ◦. These facts are used in
a.c. theory since certain quantities in the phasor dia-
grams lie at 90◦to each other. For example, in the
R−Lseries circuit shown in Fig. 23.8(a),VLleads
Iby 90◦(i.e.IlagsVLby 90◦) and may be written
asjVL, the vertical axis being regarded as the imagi-
nary axis of an Argand diagram. ThusVR+jVL=V
and sinceVR=IR,V=IXL(whereXLis the induc-
tive reactance, 2πfLohms) andV=IZ(whereZis
the impedance) thenR+jXL=Z.

R L

V

l

VR VL

R C

V

l

VR VC

VL V

VR l

(a)

Phasor diagram

VR l

VC

(b) V

Phasor diagram

θ

φ

Figure 23.8

Similarly, for the R−C circuit shown in

Fig. 23.8(b),VClagsIby 90◦(i.e.IleadsVCby

90 ◦) andVR−jVC=V, from whichR−jXC=Z

(whereXCis the capacitive reactance

1

2 πfC

ohms).

Problem 15. Determine the resistance and

series inductance (or capacitance) for each of the

following impedances, assuming a frequency of

50 Hz:

(a) (4. 0 +j 7 .0) (b)−j 20 

(c) 15∠− 60 ◦

(a) Impedance,Z=(4. 0 +j 7 .0)hence,

resistance= 4. 0

 and reactance= 7. 00 .

Since the imaginary part is positive, the reac-

tance is inductive,

i.e.XL= 7. 0 

SinceXL= 2 πfLtheninductance,

L=

XL

2 πf

=

7. 0

2 π(50)

= 0 .0223 Hor 22 .3mH

(b) Impedance,Z=j20, i.e.Z=(0−j20)hence

resistance= 0 and reactance= 20 . Since the

imaginary part is negative, the reactance is cap-

acitive, i.e.,XC= 20 and sinceXC=

1

2 πfC

then:

capacitance,C=

1

2 πfXC

=

1

2 π(50)(20)

F

=

106

2 π(50)(20)

μF= 159. 2 μF