UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
1. CARTESIAN COMPLEX NUMBERS 1.1 INTRODUCTION Try to solve this quadratic equation : 0522 =++ xx By using quadratic formula : the discriminant , 16)5)(1(4)2(4 22 −=−=−=∆ acb
the solution : )1(216)2( −±−
=x
but it is not possible to evaluate −1 however if an operator j is defined as
then the solution may be expressed as : 12 −=j
212
42)1(2
16)2( jjx ±−=±
−=−±−
=
21 j+− and are known as COMPLEX NUMBERS . 21 j−− Both solutions are of the form : = x 1− ± 2j Complex
number Real part
Imaginarypart
z = ± a jb this form is known as the CARTESIAN COMPLEX NUMBERS ( ALGEBRAIC FORM )
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
1.2 EXAMPLES EXAMPLE 1 : Solve the quadratic equation , 042 =+x
24
042
2
jxxx
±=−=
=+
EXAMPLE 2 : Solve the quadratic equation , 2 x 2 + 3 x + 5 = 0
23
43
463
4363
44093
)2(2)5)(2(433 2
jx
jx
x
x
x
±−=
±−=
−±−=
−±−=
−±−=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
1.3 POWERS OF j
j 0
0)1( −
1
j 1
1−
j
j 2
)1)(1( −−
-1
j 3
j2j = (-1)j
-j
j 4
j2j2 = (-1)(-1)
1
j 5
j4j = (1)j
j
In general we can bring the power to the nearest multiplication of 4 : j 4 p + 0 = 1 j 4 p + 1 = j j 4 p + 2 = -1 j 4 p + 3 = - j where p ∈ Z 1.4 DOMAIN The domain of the complex number is C where R is an element of C R ∈ C R C
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
1.5 THE ARGAND DIAGRAM A complex number may be represented graphically on rectangular or cartesian axes . The horizontal ( or x ) axis is used to represent the real axis and the vertical ( or y ) axis is used to represent the imaginary axis . Such a diagram is called an ARGAND DIAGRAM . EXAMPLES : Represent Argand points A = 3 + j2 , B = -2 + j4 , C = -3 – j3 , D = 2 – j2
Real Axis
Imaginary Axis
D
C
B
A
2
-2
-3
-3 -2
4
2
3
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
2. ADDITION AND SUBTRACTION - ALGEBRAIC FORM
Two complex numbers are added / subtracted by adding / subtracting separately the two real parts and two imaginary parts . Given two complex number Z = a + j b and W = c + j d 2.1 IDENTITY If two complex numbers are equal , then their real parts are equal and their imaginary parts are equal . Hence , two complex numbers are identical , i.e Z = W if : a = c and b = d
EXAMPLE : Solve the complex equations ;
(a) 36)(2 jjyx −=+ SOLUTION 3622 jyjx −=+ Therefore [ Re ]
32662
=
=
=
x
x
x
[ Im ]
2332
−=
−=
y
y
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
(b) ( )( ) jbajj +=−−+ 3221 SOLUTION
jbaj
jbajjbajj
+=−+=−−++−
+=−−+
74)34()62(
)32)(21(
Therefore : [ Re ] : a = 4 and [ Im ] : b = - 7
2.2 ADDITION & SUBTRACTION
The sum of two complex number , i.e Z + W
)()()()(
dbjcawzjdcjbawz
+++=++++=+
EXAMPLE Given : and 32 jz += 41 jw −−=
jwzjwz
−=+−++−+=+
1)]4(3[)]1(2[
The difference of two complex number , i.e Z - W
)()()()(
dbjcawzjdcjbawz
−+−=−+−+=−
EXAMPLE Given : and 32 jz += 41 jw −−=
73)]4(3[)]1(2[
jwzjwz
+=−−−+−−=−
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
The addition and subtraction of complex numbers may be achieved graphically in the Argand diagram . Represent Example 1 and Example 2 in the Argand diagram .
IMAGINARYAXIS
3
2 -1
-4w
z
Addition
jwz −=+ 1 REAL
AXIS Subtraction
73 jwz +=−
IMAGINARY AXIS
7
3
z
z
w -4
-1
3
2
REAL AXIS
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
2.3 SCALAR MULTIPLICATION If Z = a + j b and k ∈ R , where k is a scalar ; then k Z ,
jkbkakzjbakkz
+=+= )(
EXAMPLE 3 : Given Z1 = 2 + j4 and Z2 = 3 - j Determine :
(a) 4Z1 = 168)42(4 jj +=+
(b) 5 Z2 = 515)3(5 jj −=−
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
3. MULTIPLICATION AND DIVISION - ALGEBRAIC
FORM 3.1 MULTIPLICATION Multiplication of complex numbers is achieved by assuming all quantities involved are real and using j 2 = -1 to simplify : Given two complex numbers : Z = a + jb and W = c + jd The product of two complex number , i.e Z . W ))(( jdcjbawz ++=• by using F O I L method
)()(
2
bcadjbdacwzbdjbcjadacwzbdjjbcjadacwzjjbdjbcjadacwz
++−=•−++=•+++=•
+++=•
Z . W = )()( bcadjbdac ++− EXAMPLE : multiply the following complex number (a) ( 3 + j2 )( 4 - j5 ) = 1222)5)(2()4)(12()5)(3()4)(3( jjjjj −=−++−+ (b) ( -2 + 5j )( -5 + 2j ) = 29)5)(5()2)(5()2)(2()5)(2( jjjjjj −=++−+−−
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
3.2 COMPLEX CONJUGATE The complex conjugate of a complex number is obtained by changing the sign of the imaginary part . Hence the complex conjugate of :
• Z = a + j b is Z = a - j b
• W = c - j d is W = c + j d EXAMPLE : Let Z = 2 + j5
1. The complex conjugate of Z , is 52 jz −=
2. Calculate Z Z. : 22 52 +=zz = 4 9+ = 13 CONCLUSION : The product of the complex number and its conjugate Z Z. is always a real number. EXAMPLE : Let Q = 1 + j2 and R = 3 + j4 1. Calculate RQ + Solution
64)43()21( jjjRQ +=+++=+ Therefore
64 jRQ −=+ or 64
43
21
jRQ
jR
jQ
−=+
−=
−=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
2. Calculate QR Solution
105
)64()83(
)43)(21(
jQR
jQR
jjQR
−−=
−−+−=
−−=
or
105
105)64()83(
)43)(21(
jQR
jQRjQRjjQR
−−=
+−=++−=
++=
3. Calculate 2Q Solution
43
441
)21(
2
22
22
jQ
jjQ
jQ
−−=
+−=
−=
or
43
43441
)21(
2
2
22
22
jQ
jQjjQ
jQ
−−=
+−=
++=
+=
From the previous examples , we can conclude that the : 3.3 PROPERTIES OF the COMPLEX CONJUGATES
( )nn zz
wzzw
wzwz
=
•=
+=+
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
The geometric interpretation of the complex conjugate ( shown below ) Z is the reflection of Z in the real axis . Im Z a j b= +
Z a jb= −
j
-j
O
Re
3.4 DIVISION Division of complex numbers is achieved by multiplying both numerator and denominator by the complex conjugate of the denominator . Given two complex numbers : Z = a + jb and W = c + jd The quotient of two complex number , i.e ; Z / W
22
)()())(())((
)()(
dcadbcjbdac
wz
jdcjdcjdcjbaz
jdcjba
wz
w
+−++
=
−+−+
++
=
=
2222
)()(dcadbcj
dcbdacz
w +−
+++
=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
EXAMPLE : evaluate the following 4352jj
+−
2523
2514
252314
169)815()206(
)4()3()]4)(2()3)(5[()]4)(5()3)(2[(
22
jwz
jwz
jwz
jwz
−−=
−−=
+−−+−
=
+−−+−+
=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
4. THE TRIGONOMETRIC FORM AND THE POLAR
FORM OF A COMPLEX NUMBER 4.1 INTRODUCTION
Let a complex number Z = a + jb as shown in the Argand Diagram below. Let the distance OZ be r and the angle OZ makes with the positive real axis be θ.
Imaginary axis Z r jb θ Real axis O a A
From the trigonometry of right angle triangle :
a = r cos θ and b = r sin θ
Hence :
Z = a + jb = θcosr + θsinjr = )sin(cos θθ jr +
4.2 TRIGONOMETRIC FORM AND POLAR FORM Z = r ( cos θ + j sin θ ) known as the TRIGONOMETRIC FORM is usually abbreviated to Z = [ r , θ ] or Z = r ∠ θ which is known as the POLAR FORM of a complex number . 4.3 MODULUS / MAGNITUDE r is called the modulus or magnitude of Z and is written as mod Z or Z r is determined by using Pythagoras Theorem on triangle OAZ :
mod Z = Z = r = a b2 2+
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
4.4 ARGUMENT / AMPLITUDE θ is called the argument or amplitude of Z and is written as arg Z . By trigonometry on triangle OAZ :
arg Z = θ = arctan ab
algebraic form Z = + j b a ↓ ↓ ↓ Z = r cos θ + j r sin θ trigonometric form Z = )sin(cos θθ jr +
polar form Z = [ ],θr or r ∠ θ EXAMPLE 1 : Determine the modulus and argument of the complex number Z = 2 + j3 and express Z (i) in trigonometric form and (ii) in polar form Solution Find r and θ,
139432 22 =+=+=r
°== 3.5623arctanθ
(i) trigonometric form
)3.56sin3.56(cos13 °+°= jz (ii) Polar form
°∠= 3.5613z
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
EXAMPLE 2 Express the following complex numbers in (i) in trigonometric form and (ii) in polar form Represent each complex numbers on the Argand diagram SOLUTION
θ
Im
3
z
j4
(i) 43 jz +=
o1.5334tantan
52516943
11
22
===
==+=+=
−−
ab
r
θ
Re
Therefore; )1.53sin1.53(cos5 °+°= jz
θ'θ Re
Im
-3
j4
(ii) 43 jz +−=
°−=−
=
==+=+−=
− 1.533
4tan'
5251694)3(
1
2
θ
r
Therefore;
°=°−°= 9.1261.53180θ
)9.126sin9.126(cos5 °+°= jz
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
(iii) 43 jz −−=
θ
'θ
Im
-j4
-3
1.5334tan'
525169)4()3(
1
22
=−−
=
==+=−+−=
−θ
r
Re
Therefore the actual θ
°=°+°= 1.2331.53180θ
)1.233sin1.233(cos5 °+°= jz (iv) 43 jz −=
1.5334tan'
525169)4()3(
1
22
−=−
=
==+=−+=
−θ
r
Therefore ; ))1.53sin()1.53(cos(5 °−+°− j=z
θ
'θ
Im
Re
3
z - j4
or
°=°−°= 9.3061.53360θ
))9.306sin()9.306(cos(5 °+° j=z
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
EXAMPLE 3 : Convert the following complex numbers into a + jb form , correct to 4 significant figures .
(a) Z = 4 ∠ 30°
000.2464.3)30sin30(cos4
jzjz
+=+=
(b) Z = 7 ∠ -145°
015.4734.5)145sin145(cos7
jzjz
−−=−+−=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
5. MULTIPLICATION AND DIVISION – TRIGONOMETRIC /
POLAR FORM 5.1 TRIGONOMETRIC FORM Given two complex numbers :
Z = r ( cos θ + j sin θ ) and W = p ( cos α + j sin α ) 5.1.1 CONJUGATE The conjugate of Z is Z = r ( cos θ - j sin θ ) The conjugate of W is W = p ( cos α - j sin α ) 5.1.2 MULTIPLICATION The product of two complex numbers , i.e Z.W Z.W = )sin(cos)sin(cos ααθθ jpjr +•+ = )sin)(cossin(cos ααθθ jjrp ++ = rp )sinsincossincossincos(cos 2 αθαθθααθ jjj +++ = )cossincossinsinsincos(cos αθθααθαθ jjrp ++− = )]cossincos(sin)sinsincos[(cos αθθααθαθ ++− jrp Apply the trigonometry-sum identities
)]sin()[cos( αθαθ +++=• jrpwz EXAMPLE Given Z = 2 ( cos 30° + j sin 30° ) and W = 5 ( cos [-45°] + j sin [-45°] ) Evaluate in trigonometric form Z .W Solution Therefore
)]15sin()15[cos(10)]45(30sin)45(30)[cos5)(2(
−+−=•−++−+=•
jwzjwz
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
5.1.3 DIVISION
The quotient of two complex numbers , i.e ZW
ZW
= )sin(cos)sin(cosααθθ
jpjr
++
We have to multiply the numerator and the denominator with the conjugate of the denominator
ZW
= )sin(cos)sin(cos
)sin(cos)sin(cos
αααα
ααθθ
jj
jpj
−r −
•++
= )sin(cos
)sinsincossincossincos(cos22
2
αααθαθθααθ
+−+−
pjjjr
= )1(
)]sin)(sin1()cossincossin(cos[cosp
jr αθαθθααθ −−+−+
= pj )]cossincos(sin)sinsincos[(cosr θααθαθαθ −++
)]sin()[cos( αθαθ −+−= jpr
wz
Apply the trigonometry-sum identities
)]sin()[cos( αθαθ −+−= jpr
wz
EXAMPLE Given Z = 2 ( cos 30° + j sin 30° ) and W = 5 ( cos [-45°] + j sin [-45°] )
Evaluate in trigonometric form ZW
Solution
]75sin75[cos
52
))]45(30sin())45(30[cos(52
jwz
jwz
+=
−−+−−=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
5.2 POLAR FORM Given two complex numbers : Z = r ∠ θ and W = p ∠ α 5.2.1 MULTIPLICATION The product of two complex numbers , i.e Z.W
Z.W = r ∠ θ x p ∠ α or αθαθ +∠=+=• rprpwz ],[ EXAMPLE : Determine in polar form
(1) 8∠ 25° × 4∠ 60°
°∠=°+°∠= 85326025)4)(8(
(2) 3∠ 16° × 5∠ -44° × 4∠ 60°
°∠=
°+°−°∠=3260
604416)4)(5)(3(
5.2.2 DIVISION
The quotient of two complex numbers , i.e ZW
ZW
= ],[],[
αθpr
or )](,[ αθ −=pr
wz
EXAMPLE : Evaluate in polar form
(1) 16 752 15∠∠
o
o °∠=°−°∠= 608)1575(2
16
(2) πππππ
ππ
121320
32461210
36
212
410
∠=
−−+∠
×
=−∠
∠×∠
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
6. THE EXPONENTIAL FORM OF A COMPLEX NUMBER 6.1 EXPONENTIAL FORM
The exponential form of complex number : z r e j= θ
where : e jjθ θ θ= +c o s s in
Given two complex numbers : Z re j= θ and W p=
6.1.1 CONJUGATE
The conjugate of Z re j= θ is Z re j= − θ
The conjugate of is W pe j= α W pe j= − α
6.1.2 MULTIPLICATION The product of two complex number , ie Z.W
Z . W = ))(( αθ jj pere
z )( αθαθ ++ ==• jjj rperpew
Example
Given two complex numbers : 210πj
ez = and 35πj
ew = Therefore
65
)32
(50)5)(10(
πππjjeewz ==•
+
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Where θ is in radian
e jα
UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
6.1.3 DIVISION The quotient of two complex numbers , ie Z / W
Z / W = α
θ
j
j
pere
)( αθαθ −− == jjj e
pre
prz
w
Example
Given two complex numbers : 210πj
ez = and 35πj
ew = Therefore
6)
32(
25
10 πππjjee
wz
==−
6.1.4 EXAMPLES EXAMPLE 1 : Change ( 3 - j4 ) into (a) polar form (b) exponential form Solution Find r and θ ; 525169)4(3 22 ==+=−+=r
°−=−
= 1.5334arctanθ
Actual θ = 180 °=°−° 9.1261.53
(a) Polar form : °∠ 9.1265
(b) Exponential form: (Convert 126.1 into 2.21 radian) 21.25 je
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
EXAMPLE 2 : Convert 7.2 e j 1.5 into algebraic form Solution Find a and b
18.794.85sin2.7sin51.094.85cos2.7cos
=°===°==
θθ
rbra
Therefore algebraic form : 0.51+j7.18 EXAMPLE 3 : Express Z = 2e 1 + j π/3 in algebraic form Solution
603 22 jjeeeez ×=×=
π
71.460sin2sin72.260cos2cos
=°===°==
erbera
θθ
Algebraic form: 71.472.2 jz += EXAMPLE 4 : Change 6 e 2 - j3 into the algebraic form Solution
326 jeez −×=
( mode radian ) 26.63sin6sin
89.433cos6cos2
2
===
−===
erbera
θ
θ
Algebraic form: 26.689.43 jz +−=
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
6.2 DE MOIVRE’S THEOREM By repeating the multiplication of the same complex number , we get : TRIGONOMETRIC EXPONENTIAL POLAR FORM FORM FORM Z = r ( cos θ + j sin θ ) Z = r e j θ Z = r ∠ θ Z2 = r2 ( cos 2θ + j sin 2θ ) Z2 = r2 e j 2θ Z2 = r2 ∠ 2θ Z3 = r3 ( cos 3θ + j sin 3θ ) Z3 = r3 e j 3θ Z3 = r3 ∠ 3θ In general we can write the above results , named after the French mathematician ,Abraham De Moivre , as De Moivre’s Theorem Zn = rn ( cos nθ + j sin nθ ) Zn = rn e j nθ Zn = rn ∠ nθ The theorem is true for all positive , negative or fractional values of n . The theorem is used to determine powers and roots of complex numbers . 6.2.1 POWERS OF COMPLEX NUMBERS EXAMPLE 1: Determine the following complex numbers in polar form .
i. [ 2∠35° ]5
Solution Solve for
]175,32[
)]35)(5(,2[]35,2[
5
5
°=°=
°
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
ii. ( -2 + j3 ) 6
Solution Convert the algebraic form into trigonometric form
Find r and θ
13
94
)3()2( 22
=
+=
+−=
r
r
r
°−=
−=
56
)23arctan(
θ
θ
Therefore:
)744sin744(cos2197
)]1246sin()1246(cos()13[
)]124sin124(cos13[6
6
°+°=°×+°×=
°+°
jj
j
EXAMPLE 2: Determine the value of ( 2 + j3 )3 , expressing the result in both polar and algebraic form . Solution Convert into polar form 32 j+
°==
=
31.5623arctan
13
θ
r
Therefore
]93.168,13[
]31.563,13[]31.56,13[
23
23
3
°=
°×=°
Algebraic form 946)32( 3 jj +−=+
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
EXAMPLE 3 : Determine the value of ( -7 + j5 )4 , expressing the result in algebraic form . Solution Convert the algebraic form into trigonometric form Find r and θ
74
2549
)5()7( 22
=
+=
+−=
r
r
r
°−=
−=
54.35
)75arctan(
θ
θ
Therefore:
)24.213sin24.213(cos405224
)]54.356sin()54.356(cos()74[
)]54.35sin54.35(cos74[6
6
°−+°−=°−×+°−×=
°−+°−
jj
j
Polar form: [ ]85.577,5476 ° Algebraic form: − 33604324 j− 6.2.2 ROOTS OF COMPLEX NUMBERS The square root of a complex number is determined by letting n = ½ in De Moivre’s Theorem , i.e
[ ] [ ] / /r r r r∠ = ∠ = ∠ = ∠θ θ θ θ1 2 1 2 12 2
There are two square roots of a real number , equal in size but opposite in sign . EXAMPLE : [ ]4 60∠ =o Solution
°∠=°
∠=°∠ 3022
604]604[
or °∠=°+°∠=°∠ 2102301802604[
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
7. EXERCISES CARTESIAN COMPLEX NUMBERS 1. Solve the following quadratic equation and write down :
a : the real part and b : the imaginary part
(a) 2x2 + 5x + 7 = 0 a : ______ b : ______ (b) 2x ( x + 2 ) = - 9 a : ______ b : ______ (c) 9x2 - 2x + 28 = 3 - 2x a : ______ b : ______ (d) 2x2 - 4x + 5 = 0 a : ______ b : ______ (e) x2 + 2x + 2 = 0 a : ______ b : ______ 2. Show on the Argand Diagram the following complex number
(a) Z = 7 (b) W = -4 + j (c) R = -3 -j4 (d) Q = -j4 (e) V = 5 + j12
3. Evaluate : (a) j 24 (b) j 45 (c) j 105
(d) j 86 (e) -4 / j 23 ADDITION AND SUBTRACTION – ALGEBRAIC FORM 4. Calculate the following complex number :
(a) ( 7 + j5 ) + ( -18 + j9 )
(b) ( -6 - j9 ) + ( 5 + j3 )
(c) ( 5 + j ) + ( 5 - j9 )
(d) ( 12 + j4 ) + ( j3 )
(e) ( 7 - j6 ) + ( -6 -j5 )
(f) ( j16 ) - ( 6 - j5 )
(g) ( 10 - j5 ) - ( 2 + j5 )
(h) ( 6 + j8 ) - ( 7 - j4 )
(i) ( 8 + j5 ) - ( 9 )
(j) ( 25 + j8 ) + ( 6 - j5 ) - ( 5 + j )
(k) ( 8 + j2 ) + ( -9 - j )
(l) ( 1 - j ) - ( 2 + j2 ) + ( 3 + j7 )
(m) ( 3 + j7 ) + ( 2 - j )
(n) ( 4 + j 3 ) - ( 8 + j4 )
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
5. Given Z = 5 + j6 and W = 6 + j , calculate :
(a) 10Z
(b) 15W
(c) 2Z + 3W
(d) 4W + Z
(e) ½ Z - ¼ W
(f) Z - W
(g) -Z + W
(h) 5Z - 2W
(i) 9Z + 9W
(j) ¾ Z + ½ W
(k) 1/3W - 1/4Z
(l) W + ½ Z
(m) 4Z – W
(n) 2Z + 4W
(o) 5W + 3Z
6. Let Z = ( 4 + j7 ) and W = ( 3 – j2 ) (a) Represent Z and W as two vectors on in the Argand Diagram ( on the
same diagram ) (b) Represent Z + W on the Argand Diagram ( geometrically ) without
calculating 7. Let V = ( 4 + j6 ) and G = ( 2 – j5 ) (a) Represent V and G as two vectors on in the Argand Diagram ( on the
same diagram ) (b) Represent V - G on the Argand Diagram ( geometrically ) without
calculating MULTIPLICATION AND DIVISION– ALGEBRAIC FORM 8. Find the following product and express the answer in the algebraic form .
(a) ( 8 - j7 )( 8 + j7 )
(b) ( -6 - j8 )( -6 + j8 )
(c) ( 2 - j8 )( - j5 )
(d) ( 5 + j7 )2
(e) ( 3 - j5 )( 9 + j5 )
(f) ( 2 + j6 )( 6 - j )
(g) ( 8 - j3 )( 5 + j8 )
(h) ( 6 - j8 )( - j4 )
(i) ( 3 + j8 )( 5 + j9 )
(j) ( 3 - j )( 6 + j2 )
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UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
9. Find the following quotient and express the answer in the algebraic form .
(a) (( )− +− −4 93 6jj
)
(b) (3 54
)− jj
(c) 79j
(d) (( )3 26 5
)++jj
(e) ( )( )− −− −
12 3
jj
10. The total impedance of an ac circuit containing two impedance Z1 and Z2 in
parallel is given by
Z Z ZZ ZT = +
1 2
1 2
(a) Find ZT when Z1 = 1 + j kΩ and Z2 = 1 - j2 kΩ (b) Find ZT when Z1 = 3 + j5 kΩ and Z2 = 5 - j4 kΩ 11. Find the conjugate of Z and the multiplication of Z . Z
(a) Z = 4 + j5
(b) Z = -3 - j6
(c) Z = 4 - j8
(d) Z = 6 - j3
(e) Z = 8 + j4
12. Find Z W+ and W.Z
(a) Z = 5 + j6
W = 3 - j2
(b) Z = 4 + j6
W = -4 + j4
(c) Z = 6 - j5
W = -2 - j6
(d) Z = -6 + j7
W = 6 - j5
(e) Z = 7 - j
W = -j3
13. Represent these complex numbers and their conjugate in the Argand
Diagram (a) Z = 2 + j5 (b) W = 4 - j7 (c) V = -5 + j4 (d) Y = -6 -j8 (e) Q = j8
14. If Z1 = 1 - j3 , Z2 = -2 + j5 , Z3 = -3 - j4 ; determine in a + jb form :
(a) Z1Z2 (b) ZZ
1
3
(c) Z ZZ Z
1 2
1 2+ (d) Z1Z2Z3
E2 - 30 - MATHEMATICS UNIT
UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
TRIGONOMETRIC AND POLAR FORM OF A COMPLEX NUMBER 15 . Given : Z = 7 + j5
a. Draw the projection of the complex number on the Argand Diagram b. Find the modulus c. Find the argument d. Write down the trigonometric form of Z e. Write down the polar form of Z
16. Given : Z = -3 - j4
a. Draw the projection of the complex number on the Argand Diagram b. Find the modulus c. Find the argument d. Write down the trigonometric form of Z e. Write down the polar form of Z
17. Complete the following table
ALGEBRAIC
FORM
MODULUS
Z
ARGUMENT
θ
TRIGONOMETRIC
FORM
POLAR FORM
Z = -5 + j2
Z = 5 - j5
Z = [ 5 , 35° ]
Z = 4 + j3
Z = [ 4 , 55° ]
Z = 5(cos
120+jsin120° )
Z = [ 3 , 110° ]
Z=3√2(cos 310° + jsin 310°)
Z = 5 - j5√3
Z = -2 + 0j
E2 - 31 - MATHEMATICS UNIT
UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
ADDITION / SUBTRACTION / MULTIPLICATION / DIVISION –
TRIGONOMETRIC FORM , POLAR FORM
18. Evaluate in polar form 2∠30° + 5∠-45° - 4∠120° NOTE : Addition and subtraction in polar form is not possible directly . Each complex number has to be converted into cartesian form first 19. Express the given cartesian complex numbers in polar form , leaving
answers in surd form .
(a) 2 + j3 (b) -4 (c) -6 + j (d) -j3 (e) (-2 + j )3 (f) j3 ( 1 - j )
20. Convert the given polar complex numbers into algebraic form giving
answers correct to 4 significant figures .
(a) 5∠30° (b) 3∠60° (c) 7∠45° (d) 6∠125° (e) 4∠π (f) 3.5∠-120°
21. Evaluate in polar form .
(a) 3∠20° × 15∠45° (b) 2.4∠65° × 4.4∠-21° (c) 6.4∠27° ÷ 2∠-15° (d) 5∠30° × 4∠80° ÷ 10∠-40°
(e) 46
38
∠ + ∠π π (f) 2∠120° + 5.2∠58° - 1.6∠-40°
22. Find the product , and then the quotient , WZ.WZ ( in trigonometric form )
(a) Z = 2√3(cos 260° + j sin 260°) and W = 4√3(cos 320° + j sin 320°)
(b) Z = √3(cos 120° + j sin 120°) and W = 2√3(cos 310° + j sin 310°)
(c) Z = 3 ( cos 110° + j sin 110° ) and W = ( cos 28° + j sin 28° )
(d) Z = 5 ( cos 20° + j sin 20° ) and W = 4 ( cos 55° + j sin 55° )
E2 - 32 - MATHEMATICS UNIT
UNIVERSITI KUALA LUMPUR COMPLEX NUMBER – E2
E2 - 33 - MATHEMATICS UNIT
23. Determine the two square roots of the complex number ( 5 + j12 ) in polar and Cartesian forms and show the roots on an Argand Diagram
24. Find the roots of ( 5 + j3 )1/2 in algebraic form , correct to 4 significant figures EXPONENTIAL FORM OF A COMPLEX NUMBER 25. Change the following complex number to exponential form
(a) z = 1 + j (b) w = - 1 (c) Q =-π - jπ
26. Change the following complex number to the algebraic form
(a) z = e πI (b) z = 2 e πi/6 (c) z = e 1 + πi/3
27. Use De Moivre’s Theorem to find the indicated powers . Express the results
in a + jb (a) z = ( 1 + j )
, z20
(b) z = ( - 1 + j )
, z10
(c) z =2(cos 15° + jsin15°)
, z5
(d) z =2(cos 50° + jsin50°)
, z4