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.
- ( 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
]
- ( 2 −j 3 )=
√
(a+jb) [a=− 5 ,b=−12]
- (x−j 2 y)−(y−jx)= 2 +j [x= 3 ,y=1]
- 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 )