Higher Engineering Mathematics, Sixth Edition

218 Higher Engineering Mathematics

(b) ( 1 +j 2 )(− 2 −j 3 )=a+jb

− 2 −j 3 −j 4 −j^26 =a+jb

Hence 4−j 7 =a+jb

Equating real and imaginary terms gives:

a= 4 andb=− 7

Problem 8. Solve the equations:

(a) ( 2 −j 3 )=

√

(a+jb)

(b) (x−j 2 y)+(y−j 3 x)= 2 +j 3

(a) ( 2 −j 3 )=

√

(a+jb)

Hence ( 2 −j 3 )^2 =a+jb,

i.e. ( 2 −j 3 )( 2 −j 3 )=a+jb

Hence 4−j 6 −j 6 +j^29 =a+jb

and − 5 −j 12 =a+jb

Thusa=− 5 andb=− 12

(b) (x−j 2 y)+(y−j 3 x)= 2 +j 3

Hence(x+y)+j(− 2 y− 3 x)= 2 +j 3

Equating real and imaginary parts gives:

x+y=2(1)

and− 3 x− 2 y=3(2)

i.e. two simultaneous equations to solve.

Multiplying equation (1) by 2 gives:

2 x+ 2 y=4(3)

Adding equations (2) and (3) gives:

−x= 7 ,i.e.,x=− 7

From equation (1),y= 9 ,which may be checked

inequation (2).

Now try the following exercise

Exercise 87 Further problems on complex

equations

In Problems 1 to 4 solve the complex equations.

1. ( 2 +j)( 3 −j 2 )=a+jb [a= 8 ,b=−1]

2.

2 +j

1 −j

=j(x+jy)

[

x=

3

2

,y=−

1

2

]

3. ( 2 −j 3 )=

√

(a+jb) [a=− 5 ,b=−12]

4. (x−j 2 y)−(y−jx)= 2 +j [x= 3 ,y=1]


5. If Z=R+jωL+ 1 /jωC, express Z in
   (a+jb)form whenR=10,L=5,C= 0. 04
   andω=4. [Z= 10 +j 13 .75]

20.6 The polar form of a complex number

(i) Let a complex numberzbex+jyas shown in

the Argand diagram of Fig. 20.4. Let distance

OZberand the angleOZmakes with thepositive

real axis beθ.

From trigonometry, x=rcosθand

y=rsinθ

Hence Z=x+jy =rcosθ+jrsinθ

=r(cosθ+jsinθ)

Z=r(cosθ+jsinθ) is usually abbreviated to

Z=r∠θwhich is known as thepolar formof

a complex number.

Real axis

Imaginary

axis

Z

x A

r

O



jy

Figure 20.4

(ii) ris called themodulus(or magnitude) ofZand

is written as modZor|Z|.

ris determined using Pythagoras’ theorem on

triangleOAZin Fig. 20.4,

i.e. r=

√

(x^2 +y^2 )