Higher Engineering Mathematics

252 COMPLEX NUMBERS

number and its complex conjugate is always a

real number.

For example,

(3+j4)(3−j4)= 9 −j 12 +j 12 −j^216

= 9 + 16 = 25

[(a+jb)(a−jb) may be evaluated ‘on sight’ as

a^2 +b^2 ].

(iii)Division of complex numbersis achieved by
multiplying both numerator and denominator
by the complex conjugate of the denominator.

For example,

2 −j 5

3 +j 4

=

2 −j 5

3 +j 4

×

(3−j4)

(3−j4)

=

6 −j 8 −j 15 +j^220

32 + 42

=

− 14 −j 23

25

=

− 14

25

−j

23

25

or− 0. 56 −j 0. 92

Problem 5. IfZ 1 = 1 −j3,Z 2 =− 2 +j5 and

Z 3 =− 3 −j4, determine ina+jbform:

(a)Z 1 Z 2 (b)

Z 1

Z 3

(c)

Z 1 Z 2

Z 1 +Z 2

(d)Z 1 Z 2 Z 3

(a)Z 1 Z 2 =(1−j3)(− 2 +j5)

=− 2 +j 5 +j 6 −j^215

=(− 2 +15)+j(5+6), sincej^2 =−1,

= 13 +j 11

(b)

Z 1

Z 3

=

1 −j 3

− 3 −j 4

=

1 −j 3

− 3 −j 4

×

− 3 +j 4

− 3 +j 4

=

− 3 +j 4 +j 9 −j^212

32 + 42

=

9 +j 13

25

=

9

25

+ j

13

25

or 0. 36 +j 0. 52

(c)

Z 1 Z 2

Z 1 +Z 2

=

(1−j3)(− 2 +j5)

(1−j3)+(− 2 +j5)

=

13 +j 11

− 1 +j 2

, from part (a),

=

13 +j 11

− 1 +j 2

×

− 1 −j 2

− 1 −j 2

=

− 13 −j 26 −j 11 −j^222

12 + 22

=

9 −j 37

5

=

9

5

−j

37

5

or 1. 8 −j 7. 4

(d) Z 1 Z 2 Z 3 =(13+j11)(− 3 −j4), since

Z 1 Z 2 = 13 +j11, from part (a)

=− 39 −j 52 −j 33 −j^244

=(− 39 +44)−j(52+33)

= 5 −j 85

Problem 6. Evaluate:

(a)

2

(1+j)^4

(b) j

(

1 +j 3

1 −j 2

) 2

(a) (1+j)^2 =(1+j)(1+j)= 1 +j+j+j^2

= 1 +j+j− 1 =j 2

(1+j)^4 =[(1+j)^2 ]^2 =(j2)^2 =j^24 =− 4

Hence

2

(1+j)^4

=

2

− 4

=−

1

2

(b)

1 +j 3

1 −j 2

=

1 +j 3

1 −j 2

×

1 +j 2

1 +j 2

=

1 +j 2 +j 3 +j^26

12 + 22

=

− 5 +j 5

5

=− 1 +j 1 =− 1 +j

(

1 +j 3

1 −j 2

) 2

=(− 1 +j)^2 =(− 1 +j)(− 1 +j)

= 1 −j−j+j^2 =−j 2