Wqd10202 Technicalmathii Complex Number

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 )

E2 - 1 - MATHEMATICS UNIT


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

±−=

±−=

−±−=

−±−=

−±−=



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



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



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



(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

+=−−−+−−=−



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



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 −=−



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 −=++−+−−



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

−=+

−=

−=



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

=

•=

+=+



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 +−

+++

=



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

−−=

−−=

+−−+−

=

+−−+−+

=



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+



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



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



(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



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

−−=−+−=



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



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

+=

−−+−−=



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

∠=

−−+∠

×

=−∠

∠×∠



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 ==•

+


Where θ is in radian

e jα


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



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 +−=



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

°=°=

°



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 +−=+



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[



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 )



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 )



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



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



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° )




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

Wqd10202 Technicalmathii Complex Number

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