COMPLEX NUMBERS

Consider the quadratic equation;

x 2 � 1 � 0

It has no solutions in the real number system since

x 2 � � 1 or x � � � � � � �1 � j

j � � � � � � � � 1 ie. j2 � 1

Similarly x 2 � 16 � 0 gives x � � � � � � � � � �16

or x � � � � � � � � � �16 � � � � � � � �� 1 � � 4 j

Powers of j

j � � � � � � � � �� 1j 2 � � 1j 3 � j 2. j � � jj 4 � j.j 3 � j � � j � ! j 2 1

j 5 j 4 j j

j 6 j 4. j 2 j 2

j 7 j 4. j 3 j 3

Clearly all the powers of j can be described by j, j 2, j 3, j 4.

e.g. j 2307 " j 2304. j 3

# $ j 4 % 576. & 'j () '

j

Complex.nb 1

General form of complex numbers

Considernow the quadratic equation

x2 � 2 x � 10 � 0�b2 � 4 ac �

x � � 2 � � 4 40� � � � � � � � � � � � � � � �� � � � � � � �� � � � � � � �� � �2

� 2 � � � � � � � � � � � �� 36� � � � � � � � � � � � � � � �� � � � � � � �� � � � � � �2

� � 2� � � � �2 �

6� � � � �2j

� � 1 � 3 j

This leads us t oa general form for complex numbers

z � a � bj a and b are reala � real part of z � Re � z�

b � imaginary part of z � Im � z�

In the previous example;

roots � z

� 1 ! 3 j,

hence a " # 1, b " $ 3

For complex numbers :3 % 2 j, & 4, 2 & 7 j, 16 j are all special cases

a % bj is purely real if b ' 0 , ( ) a * .a + bj is purely imaginary if a , 0 , - . bj / .

Complex.nb 2

Basic Rules of algebra:

Consider : z1 � a1� b1 j,

z2 � a2� b2 j

Sum : z1� z2 � �

a1 � a2 � � �b1 � b2 � j

Difference : z1 � z2 � a1 � a2 � b1 � b2 j

Example :for : z1 � 3 � 4 j and : z2 � � 4 � 7 j

z1 � z2 � 3 � � 4 � � 4 � 7 j � � 1 � 3 jz1 � z2 � 3 � � 4 � � 4 � 7 j � 7 � 11 j

Product : z1. z2 � a1 � b1 j a2 � b2 j � a1 a2 � a1 b2 j � b1 a2 j � b1 b2 j

2

� a1 a2 � b1 b2 � a1 b2 � a2 b1 jExample :

if z1 � 2 � 3 j and z2 � � 1 � j find z1 2 z2 � 1

z1 2 z2 � 1 � 2 z1 z2 � z1

� 2 � 2 � 3 � � 2 � 3 � j � � 2 � 3 j� 2 � 1 � 5 j � � 2 � 3 j

� 4 � 7 j

Complex.nb 3

Complex Conjugate :

For any complex number a � bj there corresponds a complex number

a � bj

obtained by changing the sign of the "imarigary part".

a � bj is the Complex Conjugate of a � bj.

Notation :

If z � a � bj then z� �

a � bj

Clearly z z_ �

2 a

and z z� �

2 bj,

hence a� 1 2

�z � z

_ � �Re � z �

b � 1� � � � �2

�z � z

_ � �Im � z �

Also z z� � � a � bj � a ! bj � a2 ! � j2 b2,

hence z z� � a2 � b2,

Complex.nb 4

Quotient :

To findz1� � � � � � �z2

where z1 � a1 � b1 j,

z2 � a2 � b2 j

z1� � � � � � �z2

�z1� � � � � � �z2

z2� � � � �� � � � � � �z2� � � � � � z1 z2

� � � � �� � � � � � � �� � � � � � �z2 z2

� � � � � ��a1 b1 j � a2 � b2 j � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � �� � � � � � � � ��a2 � b2 j �

a2 � b2 j

��a1 a2 � b1 b2 � �

b1 a2 � a1 b2 j� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � �� � � � � � � � � �a22 � b22 j2

� a1 a2 � b1 b2� � � � � � � � � � � � � � � �� � � � � � � �� � � � � � �a22 � b22

��a2 b1 � a1 b2 j� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � �

a22 � b22

Complex number in general form

NB : to find the quotientz1� � � � � � � � �z2

in general a � bj form we multiply above

and below by the complex conjugate of the denominatorie. by z2

��� � �

Equality

Let z1 � a1 � b1 jand z2 � a2 � b2 j

then z1 � z2 when a1 � a2 and b1 � b2

ie Re � z1 � � Re � z2 �and Im � z1 � � Im � z2 �

Complex Number zero :

0 � 0 j � a 0 and b 0 ! ie. both real part and imaginarypart are zero.

Example :

find the real numbers " and # such that z1 $ z2 % 0,given that z1 & 3 ' ( j and z2 ) * + 5 j.

z1 , z2 - 3 . / 0 1 2 3 5 4 j 5 0 if 3 6 7 8 0 or 7 8 9 3and : 9 5 ; 0 or < = 5

Complex.nb 5

For complex numbers, z1, z2 and z3 :

Commutative property :

z1� z2 � z2 � z1 and z1 z2 � z2 z1

Associative property:�z1 � z2 � � z3 � z1 � z2 � z3 � and

z1 z2 � z3 � z1 z2 z3 �Distributive property :

z1 z2 � z3 � � z1 z2 � z2 z3

Geometrical Representation:A complex number a b j can be represented either as a point � a, b �or as a position vector of a, b � in the x y � plane � complex plane �

or z � plane.

x axis is called real axisy axis is called imaginary axis.

This geographical representation is called Argand DiagramExample :

Modulus of a complex number

Let z � a � bj, the modulus of z is denoted by � z � and

� z � � � � � � � � � � � � � � � � � � � � � � � ��a2 � b2 �

Example : For z � 3 � 4 j

�z

� � ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !32 " # $ 4 % 2

& ' ( ( ( ( ( ( ( ( ( ( ( ( ( (9 ) 16* + ( ( ( ( ( ( (25* 5

,z

, -length of the vector for z and is always . 0

Also z z/ 0 1 a 2 bj 3 4 a 5 bj 67 a2 8 b27 9 z 9 2

Complex.nb 6

Polar form of z

Point�x, y� represents z � x � yj, where x � r cos � , y � r sin �

thus z � r cos � j r sin � r � cos � j sin � � This is called the polar form of z

Here r � � � � � � � � � � � � � � � � � � � � � � ���x2 � y2 � � � z �

�is called the "argument of z" denoted by � � arg z

The � � arg z is not unique since � � 2 k ! k an integer" producesanother value of arg z.

However # $ % & ' ( gives the principle value of arg z .

Clearly r2 ) x2 * y2 and + , tan - 1 ./00 x1 1 1 1 1y

23 44

Also cos 5 6 x7 7 7 7 7r

6 x7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 77 7 7 7 7 7 7 77 7 7 7 7 78 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 99:x2 ; y2 <

sin = > y? ? ? ? ?r @ y? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ? ? ? ? ?A B B B B B B B B B B B B B B B B B B B B BBC

x2 D y2 E

Complex.nb 7

Determination of pricipal

Argument of z : � � � � � �1. z � x � jy in first quadrant � x 0, Y 0

� �arg z� � � tan � 1 ���� x� � � � �

y

�� �� 0 � � � �� � � � � �2

Example : determine the modulus and argument of

z � 3 2 j

r � z ! 3 " 2 j # $ % & & & & & & & & & & & & & & & &32 ' 22( ) * * * * * * * *13

+ ,arg - z . / tan 0 1 1233 24 4 4 4 4

3

56 77 80.588 radians

2. z 9 x : zy in second quadrant ; x < 0, y = 0 >? @

arg A z B C D tan E 1 FGHH xI I I I Iy

JK LL, MN N N N N

2 O P Q R

Example : Determine S z S and arg T zU for z V W 1 X j

r Y Z z Z [ Z \ 1 ] j ^_ ` a a a a a a a a a a a a a a a a a a a a a a a ab b12 c d 12e f g g g g2

Since tan h 1 iiiiiiiiijkll 1m m m m m m m mn 1

op qq rrrrrrrrr s tan t 1 1 u vw w w w w4

thus x y arg z z { | } ~ �� � � � �4 �

3 �� � � � � � � � � �4

rad.

Complex.nb 8

3. z � x � zy in third quadrant � x � 0, y � 0 � :

� �arg � z � tan 1 �� x� � � � �

y

�� �� � � , � � � � � � �� � � � �2

ExampleDetermine the polar form of complex number

z � � � � � � � � �� 6 � � 2 j

r ! " z "# $ % & ' ' ' '6 % & ' ' ' '2 j $

( ) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *+ , -62 . / 0 1 2 3 3 3 32 4 2

5 6 7 7 7 78

Alsotan 8 1 9:;; x< < < < <y => ?? @ tan A 1 BCDDDDD E F G G G G2H H H H H H H HH H H H H HI J K K K K6 LM NNNNNO tan P 1 QRSSSSS 1T T T T T T T TT T TU V V V V

3 WX YYYYYZ [\ \ \ \ \6rad.

Complex.nb 9

Example : Verify the inequailities

�z1 � z2 � � � z1 � � � � z � 2 Triangle inequality�

� 5 � � � � �2

Also � z1 � z2 � � � 2 � 5 j �� � � � � � � � �29

� 5 � � � � � �2 � � z ! 1 " z2 #$ % & & & & & & &29

Polar form of z :

z ' ( ) ) ) )8 * cos +,-- . 5 /0 0 0 0 0 0 0 0 06

12 33 4j sin 5677 8 5 9: : : : : : : : :

6

;< == >

? @ A A A A8 B cos CDEE 5 FG G G G G G G G G6

HI JJ Kj sin LMNN 5 OP P P P P P P P P

6

QR SS T

4. z U x V jy in fourth quadrant W x X Y , y Z 0 [ :\ ]

arg ^ z _ ` a tan b 1 cccccccccdeff xg g g g gy

hi jj kkkkkkkkk lmn n n n n2 o p q 0

Example : find the priciple argument of z r 1 s j.t r arg u z v w x tan y 1 zzzzzzzzz

{|}} x~ ~ ~ ~ ~y

�� �� ���������� � tan � 1 ���������

���� � 1� � � � � � � �1

�� �� ��������� � �tan � 1 1� �� � � � �

4rad.

Complex.nb 10

Exponential form of z :

Recall�refer to "Tables of mathmatical formulas"�

ex � 1 � x � x2� � � � � � � �2 � � x3� � � � � � � �

3 � � x4� � � � � � � �4 � � ...

Let x � j j � � � 1 �

ej � � 1 � j � � � j � � 2� � � � � � � �� � � � � � � �2 � �

�j � � 3� � � � � � � �� � � � � � � �3 � �

�j � � 4� � � � � � � �� � � � � � � �4 � � ...

� 1 �� 2� � � � � � � �2 � �

� 4� � � � � � � �4 � � ... � j !"""" # $

% 3& & & & & & & &3 ' (

% 5& & & & & & & &5 ' ( ... )* ++++

, cos - . jsin /Thus z 0 r 1 cos 2 3 jsin 2 4 has a new form

z 5 r ej 6 the exponential form

Complex.nb 11

Important Points :

1. For the correct value in exponential form � must bein radians not in degrees.

2. Negative � is measured in the clockwise direction.Replacing � by � � in

ej � � cos � � jsin we get e j � � cos � � � � sin � � � � ,

or ej � � cos � � jsin �3. ej � and e � j � are complex cojugates hence

z � rej � and z� re ! j "Consider ej " # cos $ % & ' sin ( ) * and

e + j , - cos . / 0 1 sin 2 3 4Adding and substracting

ej 5 1 e 6 j 5 7 2 j sin 3gives two important results :

cos 8 9 1: : : : :2

;ej < = e > j ? @

sin A B 1C C C C C2

Dej E F e G j H I

Log of complex numbers :Let z J rej K L M 2 k N O P k Q 0, 1, 2, ... R

ln z S ln T rej U V W 2 k X Y Z[ ln r \ j ] ^ _ 2 k ` a ln eb ln r c j d e c 2 k f aThe principal value of ln z is ln z g ln r h j i j k g 0 kExample :

Principal value of ln l 2 ej m nopo o o4 q 2 k r s t

is ln 2 u j vw w w w w4 x 0.693 y 0.785 j

Complex.nb 12

Multiplication in polar form

Let z1 � r1�cos � 1 � j sin � 1 �

z2 � r2 � cos � 2 � j sin � 2 �then

z1 z2 � r1 r2 � cos � 1 � j sin � 1 � � cos � 2 � j sin � 2 �� r1 r2 cos 1 cos � 2 � sin 1 sin 2� j � sin � 1 cos � 2 � cos � 1 sin � 2 �� r1 r2 � cos � � 1 � � 2 � � jsin � � 1 � � 2 � �

� r1 r2 e � j � 1 ! 2 "Similarly

z1# # # # # # # # #z2 $ r1% % % % % % % % %

r2

&cos ' ( 1 ) * 2 + , jsin - . 1 / . 2 0 1

2 r13 3 3 3 3 3 3 3 3r2

e 4 j 5 1 6 7 2 8Clearly

9z1 z2

9 :r1 r2

: ; <z = 1 >?> z2 @@@@@@@@@A B

z C 1D D D D D D D D D D D D D D D DD D D D Dz2

EEEEEEEEE F r1D D D D D D D D Dr2 F G H

z I 1 HD D D D D D D D D D D D D D D DD D D D D D D DG Hz I 2 H

and arg z1 z2 F J 1 K L 2 M arg z1 K arg z2arg

z1N N N N N N N N Nz2 O P 1 Q R 2 S arg z1 Q arg z2

Example :for z1 T 2 U cos VW W W W W

4 X j sin YZ Z Z Z Z4

[and z2 \ 3 ] cos ^_ _ _ _ _

3 ` j sin ^_ _ _ _ _3

[z1 z2 \ 2 a 3 b cos c de e e e e

4 f de e e e e3 g h j sin i jk k k k k

4 l mk k k k k3 n op 3 q cos rstt 7 uv v v v v v v v v

12

wx yy zj sin {|}} 7 ~� � � � � � � � �

12

�� ��

andz1������� ��� � � �z2 � 2� � � � �

3 � cos � �� � � � � � �12 � �

j sin � � � �

Complex.nb 13

De Moivre ' s Theorem :

If z1 � r1 ej

�1

z2 � r2 ej � 2

zn � rn ej � n

thenz1 z2 zn � r1 r2 rn e

j � � 1 � � 2 � .. � � n r1 r2 rn

�cos � 1 � � 2 � . � � n � �

j sin � � 1 � � 2 � . � � n � �Set z1 � z2 � . .. � zn � z � r ej � � r � cos � � j sin � �then, zn � rn cos ! " j sin ! # n$ rn ejn %

$ rn cos n ! " j sin n ! #Set r $ 1 to give De Moivre ' s Theorem

&cos ' ( j sin ) * n + , cos n - . j sin n / 0

Application :

If 1 z 1 2 1 then z 3 cos 4 5 j sin 4 3 ej 6

17 7 7 7 7z

8 17 7 7 7 7 7 7 7 7 7 7 7 7 7 7 77 7 7 7 7 7 7 7 7 7 7 7 7 7 7 77 7 7 7 7 7 7 7cos 9 : j sin 9

cos 9 ; j sin <= = = = = = = = = = = = = = = == = = = = = = = = = = = = = = == = = = = = = =cos < ; j sin <

> cos < ; j sin <= = = = = = = = = = = = = = = == = = = = = = = = = = = = = = == = = = = = = =cos2 < ? sin2 <

> cos < ; j sin <= = = = = = = = = = = = = = = == = = = = = = = = = = = = = = == = = = = = = =1

> e @ j A

Hence zn B ejn A B cos n C D j sin n Cand z @ n B e @ jn A B cos n C E j sin n C

Adding and subtracting gives

2 cos n C B zn D 1F F F F F F Fzn

and 2 j sin n C B zn E 1F F F F F F Fzn

Complex.nb 14

Example : Evaluate

�i � �

cos �� � � � �6 � j sin �� � � � �

6 � 3ii � 1 � j 25,

� iii 1 � 1 � j 8 .Solution : By De Moivre ' s Theorem

�i � � cos �� � � � �

6 � j sin �� � � � �6 � 3 �

cos 3 � �� � � � �6 � � j sin 3 � �� � � � �

6 ! cos � �� � � � �2 � j sin � �� � � � �

2 ! 0 � j! j

"ii # Let z $ 1 % j, then r $ & z & ' ( ) ) ) )2

and * + arg z + tan , 1 - ./ / / / /4

Thus z25 0 1 2 3 3 3 32 4 cos 56 6 6 6 64 7 j sin 89 9 9 9 9

4 : : 25; < = > > > >2 ? 25 @ABB

cos25 CD D D D D D D DD D D D4 E j sin 25 FG G G G G G G GG G G G

4

HI JJK 2

25L L L L L L L2 M cos M 6 N O PQ Q Q Q Q

4 R S j sin T 6 U S UV V V V V4 R RW 212 X Y Y Y Y2 Z cos [\ \ \ \ \

4 ] j sin ^_ _ _ _ _4 `

a 212 b c c c c2 defffff 1g g g g g g g gg g gh i i i i2 j j klmmmmm 1n n n n n n n nn n no p p p p

2

qr sssss tu sssssv 212 w 1 x j yz

iii y Set z { 1 | j thus from } ii ~ r � � z � � � � � � �2and � � arg z � tanh � 1 � �� � � � �

4

Thus1� � � � � � �z8 � z � 8 � � � � � � �2 � cos �� � � � �

4 � j sin �� � � � �4 � � � 8

Complex.nb 15

� � � � � � �2 � � 8 ����

cos �� � 8 � � � � � � � � �4

�� �� �j sin ���� � 8 �� � � � � � � � �

4

�� �� �� ��

� 1 24 ! cos 2 " # j sin 2 $ %

& 1' ' ' ' ' ' '16

Powers of cos ( and sin ( :

Example : cos6 ) * + ,-.. 1/ / / / /2

0122z 3 14 4 4 4 4

z

56 77 8 6

Binomial expansion gives

cos6 9 : 1; ; ; ; ; ; ;26 < z6 = 6 z4 = 15 z2 = 20 = 15 z > 2 ? 6 z @ 4 A z @ 6 B

C 1D D D D D D D64

E FGHHz6 I 1J J J J J J J

z6

KL MM N6 OPQQ z4 R 1S S S S S S S

z4

TU VV W15 XYZZ z2 [ 1\ \ \ \ \ \ \

z2

]^ __ [ 20 `a 1b b b b b b b

64 c 2 cos 6 d e 12 cos 4 f e 30 cos 2 f e 20 gh 1i i i i i i i

32cos6 j k 3i i i i i i i

16cos 4 j k 15i i i i i i i

32cos 2 j k 5i i i i i i i

16

Complex.nb 16

Example : Express cos 6 � and sin 6 � in terms of cos � and sin �Example : Sketch z � 1� � � � �

2

� � � � � �3 � j � on an Argand Diagram

and evaluate z 24z

� � 3� � � � �4 � 1� � � � �

4 � � � � � �1 � 1

arg z � � �� � � � �6

Polar form of z is

z � 1 � cos � � �� � � � �6 � � j sin � !" " " " "

6 # #$ cos !" " " " "

6 j sin !" " " " "

6

$ e % j &' ' ' '6

Hence z ( 24 ) * e + j ,- - - -6 . / 24 0

ej 4 1 2 cos 4 3 4 j sin 4 5 6 1

Complex.nb 17

Roots of complex numbers:

Recallde Moivre' s theorem;

For z � r �cos � � j sin � �zn � rn � cos n � j sin n �

fFor any value of n � integer or fraction, positive or negative This is a very important result.

When n is a fraction weare finding roots of complex numbers

Consider wn � z � integer�The n different solutions w0, w1, w 2, , ..., wn � 1 of this equation

are the ' nth roots of z' denoted by

� � � � �z

nor z

1� � � �n

Let z � r � cos � � j sin � �and w � � � cos � � j sin !

Equation wn " zgives # n $ cos n % & j sin n ' ( ) r * cos + , j sin - .

Equalityof two complex numbers in polar form means that;

/i 0 their modulus are equal

i.e. 1 n 2 r or 3 2 r 14 4 4 4n

5ii 6 their arguments, arg wn and arg z may differ by a

multiple of 2 7 , say 2 k7 ,k 8 0, 1, 2. ..

Thus n 9 : ; < 2 k =or > ? @ A 2 k =B B B B B B B BB B B B B B B BB B B B

n, k C 0, 1, 2, ..., n D 1

Eall other choicesof k duplicate the values of F G

Thus nth roots of z, w0, w1, w 2,, ..., wn H 1 are given by

wk I r1J J J J J Jn K cos LMNN O P 2 k QR R R R R R R RR R R R R R R RR R R R

n

ST UU Vj sin WXYY Z [ 2 k \] ] ] ] ] ] ] ]] ] ] ] ] ] ] ]] ] ] ]

n

^_ `` a

b r1c c c c c cn exp j d e f 2 k gh h h h h h h hh h h h h h h hh h h hh h h h

n i k j 0, 1, 2. .., n k 1

Complex.nb 18

Example : Find all the cube roots of � 8

Set z � � 8,then r � � � 8 � � 8,

arg � � 8 � Hence � 8 � 8 cos � � j sin � �and � � 8 � 1� � � �

3 � 81� � � �3 expj � � 2 k �� � � � � � � �� � � � � � � �

3

� � � � � �83 �cos � !! " # 2 k $% % % % % % % %% % % % % % % %% % % %

3

&' (( )j sin *+,, - . 2 k /0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0

3

12 33 4k 5 0, 1, 2 6 since n 7 3 8

There are 3 cube roots of w3 9 z 9 : 8Thus the 3 cube roots of z ; < 8 are

w0 ; 2 = cos >? ? ? ? ?3 @ j sin AB B B B B

3 CD 2 EFGGGGG 1H H H H H

2 I j J K K K K3L L L L L L L LL L L2

MN OOOOOP 1 Q R S S S S3 j

w1 T 2 U cos V W j sin V XY 2 Z [ 1 \ 0 ]^ [ 2w2 ^ 2 _`aa cos 5 bc c c c c c c c c

3 d j sin 5 ef f f f f f f f f3

gh iij 2 klmmmmm 1n n n n n

2 o j p q q q q3r r r r r r r rr r r2

st uuuuuv 1 w x y y y y3 j

Complex.nb 19

w0, w1, w2 are equally spaced2 �� � � � � � � � �3

radians�120 ° � apart on circle of radius 2

� � � � � � �8

3 �,

centered at 0, 0

For wn � z roots are spaced by2 � n

rad and lie on a circle of radius � � � � �rn

Example : Find the values of j2� � � �3

There are two ways depending upon how you express j2� � � �3

�i � j

2� � � �3 � � j2 � 1� � � �

3 � � � 1 � 1� � � �3

i.e find 3 cube roots of � 1�ii � j

2 3 ! " j 1# # # #

3 $ 2i.e find the 3 cube roots of j

and square them .

Here it appears sensible to use method % i &'i & polar form of ( 1 :

r ) * + 1 , - 1 and . / arg 0 1 1 2 3 4wk 5 6 7 1 8 19 9 9 9

3

: ; < < < <13 =>?? cos @A?? B C 2 k DE E E E E E E EE E E E E E E EE E E E3

FG HH Ij sin JKLL M N 2 k OP P P P P P P PP P P P P P P PP P P P

3

QR SS Tk U 0, 1, 2

w0 U cos V WX X X X X3 Y Z j sin [ \] ] ] ] ]

3 ^ _ 1` ` ` `2̀ a

b c c c c3d d d d d d d dd d d2

j

w1 e cos f g j sin f e h 1w2 i cos jkll 5 mn n n n n n n n n

3

op qq rj sin stuu 5 vw w w w w w w w w

3

xy zz { 1| | | | |2 }

~ � � � �3� � � � � � � �� � �2

j

Exercise : Solve the equation

w3� � � �2 � j

Complex.nb 20