Https Doc 14 0c Apps Viewer.googlasasdasdeusercontent

Complex Numbers and their Properties

Institute of Lifelong Learning, University of Delhi pg. 1

Lesson: Complex Numbers and their Properties

Lesson Developer: Vinay Kumar

College: Zakir Husain Delhi College , University of Delhi



Table of Contents

Chapter : Complex Numbers and their Properties

1. Learning Outcomes

2. Introduction

3. Complex Numbers

o 3.1. Graphical representation

o 3.2. Polar form of a complex number

o 3.3. nth roots of unity

o 3.4. Some Geometric Properties of Complex Numbers

3.4.1. Distance between two points

3.4.2. Dividing a line segment into a given ratio

3.4.3. Measure of an angle

o 3.5. Collinearity, Orthogonality and Concyclicity of

Complex numbers

o 3.6. Similar triangles

3.6.1. Condition for Similarity

3.6.2. Equilateral triangles

o 3.7. Some analytical geometry in complex plane

3.7.1. Equation of a line:

3.7.2. Equation of a line determined by a point and a

direction

3.7.3. The foot of a perpendicular from a point to a

line

3.7.4. Distance from a point to a line

3.7.5. Equation of a circle



Exercises

References

1. Learning Outcomes:

After reading this chapter, you will be able to understand

What is complex number?

How can it be graphically represented?

What is the polar form of complex number?

What are the nth roots of a complex number and unity?

How to solve the equations involving complex numbers?

How can we measure the angle between two complex numbers?

What is collinearity , orthogonality and concyclicity of complex

numbers?

How to define similar triangles and equilateral triangles in complex

plane?

How to write equation of a line in complex plane ?

When two lines in complex plane are perpendicular, parallel and

orthogonal?

How to write the equation of circle in complex plane?

2. Introduction:

This unit is according to the syllabus of undergraduate students . As it is

obvious from the name of this chapter that it contains the complex numbers

and some aspects of geometry of complex numbers in complex plane . In

this unit, starting from basic concepts of complex numbers in detail,

important propositions about geometry of complex numbers have also

been discussed.

3. Complex Numbers:



So far we find the solutions of all algebraic equations in real numbers . But

there are some equations whose solution does not lie in the set of real

numbers . As for example if we take the equation 2 1 0x + = ,then it's

solution does not lie in Â (the set of real number).

Therefore ,to overcome these types of situations, the concept of complex

numbers was introduced.

Definition : A number whose square is -1, is called an imaginary number or

a complex quantity and is denoted by i (pronounced as iota). We have

21 1i or i

A complex number is written as z x iy , where xand y are real numbers and

called the real part and imaginary part of the complex number z

respectively. We write

Re( ) Im( )z x and z y

3.1. Graphical representation:

A complex number z has a simple geometric representation. Consider a

rectangular coordinate system. Then every complex number z = x + iy can

be associated with some point P(x,y) in the x-y plane. This plane is called

the z-plane or the complex plane or the Argand diagram. All real numbers (y

= 0) lie on the x-axis or the real axis and all purely imaginary numbers (x =

0) lie on the y-axis or the imaginary axis.

O

Y

O

x

P(z)=(x,y) y

X



Fig 1: Graphical representation of a complex number.

3.1.1. Modulus of a complex number:

Let z x iy be a complex number. The real positive number

2 2z x iy x y= + = + is called the modulus or the absolute value or the

magnitude of a complex number z.

3.1.2. Properties of modulus of complex numbers

Let 1z and 2z be two complex numbers then

(1) 2121 zzzz ,

(2) 2

1

2

1

z

z

z

z .

3.1.3. Equal complex numbers:

Two complex numbers 1z and 2z are equal i.e 1 2z z= ,if and only if 1 2x x

and 1 2y y , we also have 0 0z z= Û = .

3.1.4. Negative of a complex number:

The complex number z x iy- = - - is called the negative of the complex

number z and |-z|=|z|.

3.1.5. Complex conjugate number:

The complex number ( , )z x iy x y is called the complex conjugate or just

the conjugate of a complex number z x iy . Thus is the reflection of z or

the real axis. We also have



(i) ( )z z (ii) 1 2 1 2z z z z

(iii) z z z is a purely imaginary number

(iv) z z (v) 1 2 1 2z z z z ( 1z and 2z be any two complex numbers)

3.2. Polar form of a complex number:

Let z x iy be any complex number represented by the point ( , )p x y in the

complex plane. Let be the angle XOP measured in the counter clockwise

direction(positive direction) and op r .

From ΔOMP, we get

cos sinx r and y r

Thus, the complex number z = x +iy can be written as

(cos sin ) ( , ) 0z x iy r i r r (A)

Fig.2(Polar form of a complex number)

which is known as polar form of complex number.

2 2r x y z , and = 1tan ( / )y x

where is called argument of z and is expressed as arg(z).

O M

y

r

x



The value of which satisfies is called the principal value of or

the principal argument of z.

Arg( ) 2 , 0, 1, 2, . . .Z k k

Arg(z) is called extended argument of the complex number z.

3.2.1. Method to write a complex number in polar form:

Let z x iy

1. At first we find 2 2r z x y and 1tany

x .

2. According to the quadrant in which the given complex number lies, we

find principal value of

or argument of ' 'z ,which is as follows

Let argt z

In first quadrant take t ,In 2nd quadrant take t

In 3rd quadrant take t ,In 4th quadrant take 2t .

3. Polar form of (cos sin )z r t i t

Extended argument of z i.e 2Argz t k where 0, 1, 2,.....k .

Example 1: Find the polar form of the complex numbers.

(a) 2 2 3z i ,(1st quadrant)

(b) 3z i (3rd quadrant)

Solutions:

(a) Given, 2 2 3z i , 2 2 2 22 (2 3) 4r z x y

1 2 3tan

2 3

y

x

z lies in the first quadrant, arg3

z



Polar form of ' 'z is 4(cos sin )3 3

i

Extended argument of ' 'z i.e.,

rg 2 , 0, 1, 2....3

A z k k

(b) Given 3z i (3rd quadrant)

We have r = 3 1 = 2.

t = 1tan | / |y x = 1tan | 1/ 3 |6

z lies in the 3rd quadrant, so 7

arg6 6

z

Polar form of z =7 7

2(cos sin )6 6

i

Extended argument of complex number z =7

2 , 0, 1, 2....6

k k

Value Additions: Note

Two complex numbers 1z , 2z ≠ 0 represented as 1z = 1r (cos 1t + isin 1t ) and

2z = 2r (cos 2t + isin 2t ) are equal iff

1 2r r and 1t - 2t = 2kπ for an integer k.

Example 2: Find the polar representation of the complex number

1 cos sin , (0,2 )z a i a a

Solution: r = |z| = 2 2(1 cos ) sina a = 2 21 cos 2cos sina a a =

2(1 cos )a = 22.2cos ( / 2)a = 2cos( / 2)a



1 1 1

2

2sin cossin 2 2tan ( ) tan ( ) tan (tan )

1 cos 2 22cos

2

a aa a a

aa

.

Cases:

If a(0,π) (0, )2 2

a

also z lies in first quadrant. So,

t = 2

a2cos (cos sin )

2 2 2

a a az i

If a ( , 2 ) ( , )2 2

a and z lies in 4thquadrant. since

( , )

2 2

a =>

2

a .

2 2 ( )2 2

2cos (cos( ) sin( ))2 2 2

a a

a a az i

If a then 0z .

3.2.2. Power of a complex number: De Moivre’s theorem:

For (cos sin )z r i and n N .

We have ,

(cos sin )n nz r n i n (it can easily be proved by induction method).

Value Addition:



Abraham de

Moivre, a French

mathematician

Abraham de Moivre was born 26 May 1667 in Champagne, France. He was a

famous French mathematician best known for Moivre’s formula, which you may be

familiar with if you’ve taken the right college classes. During the later years of his

life, he studied further into probability and mathematics. He later wrote a book on

the latter called The Doctrine of Chances. His work was so extensive that several

papers were published posthumously. As he aged, he became more and more

lethargic, and started sleeping longer hours. He noted that he was sleeping 15

minutes extra every night, and calculated that he would die on the day that the

extra 15 minutes a night accumulated to 24 hours. That day was November 27,

1754, the actual day of his death. He died in London and was buried at St. Martin-

in-the-Fields.

Remarks:

(i) n nn n n nz r and z r z r z z .

(ii) 2 { arg 2 ; }nArg z n k n z k k Z .

(iii It is also true for negative integers.

Example 3:- Evaluate 1000(1 )i .

Solution: Polar form of 1 i = 2(cos sin )4 4

i .

Using De-Moivre's theorem,we have

10001000 1000 500 500 500(1 ) 2 (cos sin ) 2 (cos250 sin 250 ) 2 (1 0) 2 .

4 4i i i i

3.3. nth roots of unity:

First we shall find the nth roots of a complex number. Let 0 (cos sin )z r i

be a complex number with 0, 0,2r .The number 0z has n distinct nth

roots given by

k

2 2z (cos sin ),n k k

r in n

0,1,2,... 1k n



Geometric images of nth roots of a complex number 0z 0 are the vertices

of a regular n-gon inscribed in a circle with center at origin and radius n r

Example: Find the third roots of the number 1z i and represent them in

the complex plane.

The polar representation of 1z i is 2z (cos sin )4 4

i .

The cube roots of z are

6

6

2 24 42(cos( ) sin( ))

3 3

2 22(cos( ) sin( )), 0,1,2

12 3 12 3

k

k k

z i

k i k k

Or in explicit form

6

0

6

1

6

2

2(cos sin )12 12

3 32(cos sin )

4 4

17 172(cos sin )

12 12

z i

z i

z i

Geometrical images of 0 1 2, ,z z z are the vertices of equilateral triangle.

Fig 3: Geometrical image of the cube roots of (1+i)

Now , we will find nth roots of unity

The roots of equation 1 0nz are called the thn roots of unity.

0M 0z

0z )

1 1M z

2 2( )M z

y

x



Now

2 2

1 cos0 sin0 cos sin , 0,1,2.... 1.n n

k

k kz z i z i k n

n n

Explicitly, we have

2

0 1 2

2 2 4 4cos0 sin0 1, cos sin , cos sinz i z i z i

n n n n

Continuing in the same manner

1

1

2( 1) 2( 1)cos sin .n

n

n nz i

n n

The set of nth roots of unity is 2 3 1{1, , , ,........ }n denoted as nU .

The geometric images of nth roots of unity are the vertices of a regular n-

gon inscribed in a unit circle with one of the vertices at 1.

Note: (i) The cube roots of unity form an equilateral triangle inscribed in the

circle (0,1)C .

2

3 {1, , }U

Fig 4: Geometrical image of the cube roots of unity

(ii)The forth roots of unity form a square inscribed in a circle (0,1)C .

4 {1, 1, , }U i i



Fig 5: Geometrical image of the forth roots of unity

Value Addition: Remark

1. If n divides q, then any root of 1 0nz is a root of 1 0qz .

in other words, n qU U if q is the multiple of n

2. The common roots of 1 0mz and 1 0nz are the roots of 1 0dz ,

where d = gcd(m,n). In other words m n dU U U

Example 4: Two regular polygons are inscribed in the same circle. The first

polygon has 1982 sides and the second has 2973 sides. If the polygons have

any common vertices, how many such vertices there will be?

Solution:- The number of common vertices is given by the number of

common roots of 1982 1 0z and 2973 1 0z i.e 1982 2973 dU U U where d =

gcd(2973,1982) = 991

Example 5: Find the number of ordered pairs (a,b) of real numbers such

that 2002( )a bi a bi .

Solution: Let , ,z a bi z a bi and |z| = 2 2a b

2002 20012002 2002 2002( ) ( 1) 0a bi a bi z z z z z z z z

0z or2001

( 1) 0z 0z or2001

1z 0, 1z z

Now, 0z => (a,b) = (0,0).



When 1z , we have

22002 2003 . 1z z z z z z

Thus, 2003U have 2003 distinct solutions. Total number of ordered pair = 2003

+ 1 = 2004.

Example 6:- Solve the equation 7 4 32 2 0z iz iz .

Solution:- Given equation is

7 4 32 2 0z iz iz

4 3 3

4 3

4 3

( 2 ) ( 2 ) 0

( )( 2 ) 0

, 2

z z i i z i

z i z i

z i z i

Now, we can easily find out the roots of these equations.

Example 7: Solve 4 25( 1)( 1)z z z z .

Solutions: Given equation is

4 25( 1)( 1)z z z z

4 3 2

5 4 3 2

5 4 3 2

5

5

1

5

5 10 10 5 0

5 10 10 5 0

5 10 10 5 1 1 0

( 1) 1 0

( 1) 1

1 ( 1)

z z z z

z z z z z

z z z z z

z

z

z

3.4. Some Geometric Properties of Complex Numbers:

3.4.1. Distance between two points:

Suppose that the complex numbers 1z and 2z have the geometric images 1M

and 2M . Then the distance between the points 1M and 2M is given by

1M 2M = 1 2z z



Let M be the geometric image of complex number z. Then M is between 1M

and 2M if 1M ≠ M , 2M ≠ M , the following relation holds

1 2 1 2 .M M M M M M

The set 1 2 1 2{ : }M M M M M M is called ‘Open Segment’ determined by

points 1M and 2M .

The set 1 2 1 2 1 2[ ] ( ) { , }M M M M M M represent the ‘Closed Segment’.

3.4.2. Dividing a line segment into a given ratio:

Let 1M ( 1z ) and 2M ( 2z ) be two distinct points. A point ( )M z on the line 1M 2M

divides the segment 1M 2M into ratio k \ {1}. if

1 2

1 2

1 2

( )

1

MM kMM

z z k z z

z kzz

k

Note: for k = -1, z = 1 2

2

z z [Midpoint]

3.4.3. Measure of an angle:

A triangle is oriented if an ordering of its vertices is specified. It is positively

oriented if the vertices are oriented counterclockwise otherwise negatively

oriented.

Value Addition: Note

The measure of directly oriented angle 1 2M OM equals 1

2

argz

z.



M1(1+i) M2(-1+i)

Example 8: (a) Find the 1 2M OM and 2 1M OM where 1M & 2M are the

geometric images of 1z and 2z where 1 21 , 1z i z i

Solution: 1 21 , 1z i z i

2

1

1 ( 1 )(1 )

1 2

z i i ii

z i

1 2M OM arg2

i

, 2 1M OM arg( ) 2 3 / 22

i

.

Fig: 6

(b) 21 2

1

1, 1

zz i z i

z i

1 2 2 1

3arg( ) arg

2 2M OM i and M OM i

Value Addition: Note:

Let us consider three district points 1M ( 1z ), 2M ( 2z ), 3M ( 3z ). The measure of

oriented angle 2 1 3M M M is 3 1

2 1

arg( )z z

z z

.

Example 9: Suppose that 1 2 34 3 , 4 7 , 8 7z i z i z i . Find 3 1 2M M M and

2 1 3M M M



Solution:

2 1

3 1

3 1 2

2 1 3

4 (1 ) 1

4 4 2 2

1arg

2 4

2 7arg arg(1 )

1 4

z z i i i i

z z i

iM M M

M M M ii

Value Addition: Note:

Angle between two lines: Let us consider four distinct points iM ( iz )

1,2,3,4i . Angles between two lines 1 3M M and 2 4M M equals 3 1

4 1

arg( )z z

z z

or

4 2

3 1

arg( )z z

z z

.

3.4.4. Rotation of a Point:

Let us consider an angle α and a complex number cos sini

Let (cos sin )z r t i t be complex number and M be its geometric image.

. (cos( ) sin( ))z r t i t

Also, .r z and arg( . ) argz t z

It means that the geometric image M’ ( .z ) is the rotation of M with respect

to the origin by the angle .

M(z)

0

M’(z)



3.5. Collinearity, Orthogonality and Concyclicity of Complex

numbers:

Consider ( )i iM z i=1,2,3,4

3.5.1. Collinear:

Points 1 2 3, ,M M M are collinear iff *3 1

2 1

z z

z z

where, * - the set of non-zero

real numbers.

3.5.2. Orthogonal:

Lines 1 2M M and 3 4M M are orthogonal iff *1 2

3 4

z zi

z z

.


If 2 4M M then 1 2M M 3 2M M iff *1 2

3 2

z zi

z z

.

3.5.3. Concyclic:

The distinct points 1 1 2 2 3 3 4 4( ), ( ), ( ), ( )M z M z M z M z are concyclic (or collinear) iff

*3 2 3 4

1 2 1 4

:z z z z

kz z z z


1) The points 1 2 3 4, , ,M M M M are collinear iff



* *3 2 3 4

1 2 1 4

z z z zand

z z z z

2) The points 1 2 3 4, , ,M M M M are concyclic iff

*3 2 3 4

1 2 1 4

:z z z z

kz z z z

But 3 2 3 4

1 2 1 4

,z z z z

z z z z

Example10: If 1 2 3 4( ), ( 1 2 ), ( 2 ) (1 2 )M z i M i M i andM i be geometric images

Check whether 1 2M M 3 4M M or not?

Solution: 1 2

3 4

,z z

i hencez z

1 2M M 3 4M M .

Example11: Let 1 2 3, ,z z z be the co-ordinates of vertices of A,B,C of a

triangle. If 1 1 2w z z and 2 3 1w z z . Prove that 90A iff Re( 1 2w w ) = 0.

Solution: 90A => BA CA

2 1 1 1 1 21 22

3 1 2 2 2

( ) 0 ( ) 0 ( ) 0z z w w w w

i i e e e w wz z w w w

Example12: 1 2 3 4(1), ( ), ( 1) ( )M M i M and M i are concyclic

as *1 1: 1 ,

1 1

i ik

i i

Also 1 1

1 1

i iand

i i



Example 13: 1 2 3 4(2 ), (3 2 ), ( 1 2 ) ( 2 3 )M i M i M i and M i are collinear

as *4 4 1: 1

1 4 4

i ik

i i

3.6. Similar triangles:

Let 1 1 2 2 3 3 1 1 2 2 3 3( ), ( ), ( ), ( ), ( ) ( )A a A a A a B b B b and B b be six points in the complex plane.

Triangle 1 2 3A A A and 1 2 3B B B are similar if the angle at kA is equal to the angle

at kB , {1,2,3}k .

3.6.1. Condition for Similarity:

1. The triangle 1 2 3A A A and 1 2 3B B B are similar, having the same

orientation iff 2 1 2 1

3 1 3 1

a a b b

a a b b

2. Condition (1) is equivalent to 1 2 3

1 2 3

1 1 1

0a a a

b b b

.

3. The triangle 1 2 3A A A and 1 2 3B B B are similar, having the opposite

orientation iff 2 1 2 1

3 1 3 1

a a b b

a a b b

.

3.6.2 Equilateral triangles:

Proposition (I): Suppose 1 2 3,z z and z are coordinates of the vertices of the

triangle 1 2 3A A A .Then following statements are equivalent,

(a) 1 2 3A A A is an equilateral triangle

(b) 2 2 2

1 2 3 1 2 2 3 3 1.z z z z z z z z z

(c) 3 22 1

3 1 1 2

.z zz z

z z z z

(d) 1 2 3

1 2 3

1 1 10. ( ).

3

z z zwhere z

z z z z z z



(e) 2 2

1 2 3 1 2 3

2 2( )( ) 0, ( cos sin ).

3 3z z z z z z where i

(f) 1 2 3

2 3 1

1 1 1

0.z z z

z z z

Proof : (a)Let 1 2 3A A A is an equilateral triangle

3

2 1 3 2

33 1 1 2

( ) ....( )

( ) ....( )

i

i

z z e z z i

z z e z z ii

(since2 1z z and

3 1z z can be obtained from 3 2z z and

1 2z z by rotation of

an angle 3

respectively)

Dividing (i) by (ii) we have

3 22 1

3 1 1 2

.z zz z

z z z z

(c) Proved.

(f) 1 2 3

2 3 1

1 1 1

z z z

z z z

= 1 2 1 3 1

2 3 2 1 2

1 0 0

z z z z z

z z z z z

After solving this determinant we get

3 22 1

3 1 1 2

.z zz z

z z z z

(c)Proved.

2 2 2

2 1 2 1 1 2 3 3 2 1 3 1 2 ,z z z z z z z z z z z z z

2 2 2

1 2 3 2 1 3 2 1 3.z z z z z z z z z

(b) Proved.



(e)L.H.S =

2 2

1 2 3 1 2 3( )( )z z z z z z = 2 2 2 2 2 2

1 2 3 2 1 3 2 1 3( ) ( ) ( ),z z z z z z z z z

= 2 2 2

1 2 3 2 1 3 2 1 3( 1) ( 1) ( 1),z z z z z z z z z

= 2 2 2

1 2 3 2 1 3 2 1 3,z z z z z z z z z

L.H.S = 0 2 2 2

1 2 3 2 1 3 2 1 3.z z z z z z z z z

(b) (converse is also true) Proved.

At last we shall prove (d) (b)

1 2 3

2 3 1 3 1 2

1 2 3

2

1 2 3 1 2 2 3 3 1

1 1 10,

( )( ) ( )( ) ( )( )0,

( )( )( )

3 (2 2 2 ) 0,

z z z z z z

z z z z z z z z z z z z

z z z z z z

z z z z z z z z z z z

2

1 2 2 3 3 1 1 2 33 .2.3 0( 3 ),z z z z z z z z z z z z z

2

1 2 2 3 3 13 0,z z z z z z z

2

1 2 2 3 3 1

2

1 2 31 2 2 3 3 1

2 2 2

1 2 3 1 2 2 3 3 1 1 2 2 3 3 1

2 2 2

1 2 3 1 2 2 3 3 1

3 ,

( )3 ,

9

2( ) 3 3 3 ,

.

z z z z z z z

z z zz z z z z z

z z z z z z z z z z z z z z z

z z z z z z z z z

Proposition 2: Suppose 1, 2, 3,z z z are co-ordinates of the vertices of the

positively oriented triangle 1 2 3A A A .Then following statements are equivalent,

(a) 1 2 3A A A is an equilateral triangle,

(b) 3 1 2 1( )z z z z where cos sin3 3

i

,

(c) 2 1 3 1( )z z z z where 5 5

cos sin3 3

i

,



(d) 2

1 2 3 0z z z where 2 2

cos sin3 3

i

,

Proof : (a) (b)

1 2 3A A A is an equilateral triangle and positively oriented iff 3A is

obtained from 2A by rotating about 1A through an angle of

3

.

3 1 2 1( )z z z z .where cos sin3 3

i

.

(a)(c)

1 2 3A A A is an equilateral triangle and positively oriented

iff 3A is obtained from 2A by rotating about 1A through an angle of 5

3

.

3 1 2 1( )z z z z .where 5 5

cos sin3 3

i

.

(a) (d)

1 2 3A A A is an equilateral triangle.

3 2 2 1

3 2 2 1

3 2 1

2

3 2 1

2

3 2 1

( )

0

(1 ) 0

( ) 0

0.

z z z z

z z z z

z z z

z z z

z z z

where2 2

cos sin3 3

i


negatively oriented triangle 1 2 3A A A .Then following statements are equivalent,

(a) 1 2 3A A A is an equilateral triangle,

(b) 3 1 2 1( )z z z z where cos sin3 3

i

,

(c) 2 1 3 1( )z z z z where 5 5

cos sin3 3

i

,

(d) 2

1 2 3 0z z z where 2 2

cos sin3 3

i

,



Proof: 1 2 3A A A is an equilateral triangle and negatively oriented iff 1 3 2A A A is

an equilateral triangle andpositively oriented. By using the proposition 2 we

can prove it.


equilateral triangle 1 2 3A A A .Consider the statements :

(1) 1 2 3A A A is an equilateral triangle,

(2) 1 2 2 3 3 1,z z z z z z

(3) 2 2

1 2 3 2 1 3, .z z z and z z z

Then (2)(1),(3)(1) and (2) (3).

Proof: (2)(1) Taking the modulus of the terms in the given relation

1 2 2 3 3 1

1 2 2 3 3 1

,

,

z z z z z z

z z z z z z

1 2 3

2 2

1

2

1 1

2 2 2

1 2 3

1 2 3

, ( )

,

,

, , ,

z z z r say

z r

z z r

r r rz similarly z z

z z z

Substituting in the given relation (2) we have,

1 2 2 3 2 3 3 1

2 2 2 2

1 2 2 3

2 3 3 1

31 2 2

2 3 3 1

31 2

2 3 1

,

, ,

, ,

,

z z z z z z z z

r r r rz z z z

z z z z

zz z z

z z z z

zz z

z z z

2 2 2

1 2 3 2 1 3 3 1 2

2 2 2

1 2 3 2 3 1 3 1 2

, ,

,

z z z z z z and z z z

z z z z z z z z z



By proposition 1 we can say that 1 2 3A A A is an equilateral triangle. Thus we

have shown that (2) (3) ,(2) (1).Also the argument for (2) (3) are

reversible. So (2) (3) .Proved.

Example14: Suppose 1, 2, 3,z z z are co-ordinates of the vertices of the triangle

1 2 3A A A .If 1 2 3z z z

and 1 2 3 0z z z .Prove that triangle 1 2 3A A A is an

equilateral.

Solution: Given

2 3

1 2 3

1 2 3

1z z z 0

z z z 0

z z z 0 .....(*)

Also let 1 2 3z z z r (say)

22 22 2 2

1 2 3

2 2 2

1 1 2 2 3 3

z r z r z r ,

z z r z z r z z r ,

2 2 2

1 2 3

1 2 3

r r rz z z

z z z

Substituting in (*) we have

2 2 2

1 2 3

1 2 3

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

0,

1 1 10,

0

0

r r r

z z z

z z z

z z z z z z

z z z

z z z z z z

2 2 2 2

1 2 3 1 2 3 1 2 2 3 3 1

2 2 2

1 2 3

( ) 2( )

0

z z z z z z z z z z z z

z z z

Thus 2 2 2

1 2 3 1 2 2 3 3 1 0z z z z z z z z z

Hence by proposition 1, 1 2 3A A A is an equilateral triangle.

3.7. Some analytical geometry in complex plane:

3.7.1. Equation of a line:



Equation of a line in the complex plane is 0 z z where is a non-

zero complex number and Î Â .

The equation of line in Cartesian plane 0Ax By C+ + = where A,B,CÎ Â and

2 2 0A B+ ¹ .Set 2 2

z z z zz x iy then x y

i

+ -= + = =

Substituting the value of x and y in we have

( ) ( ) 02 2

( ) ( ) 02 2

z z z zA B C

i

z z z zA Bi C

+ -+ + =

+ -Þ - + =

( ) ( ) 02 2

A Bi A Biz z C

+ -Þ + + =

,2

A Bilet C

-= =

Then 0 z z

Also angular coefficient

Ai

B

.

Proposition 5: Let us consider two lines as

1 1 1 1 2 2 2 2: 0 : 0d z z and d z z

Then the lines 1 2d and d are

(a) parallel iff 1 2

1 2

.

=

(b) perpendicular iff 1 2

1 2

0.

+ =

(c) concurrent iff 1 2

1 2

.

¹



Solution:(a) 1 2d and d are parallel if

1 2

1 1 2 2

1 1 2 2

1 1 2 2 1 1 2 2

1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

1 2 1 2

1 2

1 2

,

( )( ) ( )( ),

,

2 2 ,

.

m m

i i

=

+ +Û =

- -

Û + - = - +

Û - + - = + - +

Û =

Û =

(b) 1 2d and d are perpendicular if 1 2 1m m = -

1 1 2 2

1 1 2 2

1 1 2 2 1 1 2 2

1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

( )( ) 1

( )( ) ( )( ),

,

i i

+ +Û = -

- -

Û + + = - -

Û + + + = - - +

1 2 1 2

1 2

1 2

0,

0.

Û + =

Û + =

(c) 1 2d and d are concurrent iff 1 2m m¹ and by (a) 1 2

1 2

.

¹ Proved.

Value Addition: Note

(a) Equation of a line determined by two points 1 1 2 2( ) ( )P z and P z is

1 1

2 2

3 3

1

1 0.

1

z z

z z

z z

æ öç ÷ç ÷ç ÷ç ÷=ç ÷ç ÷ç ÷ç ÷ç ÷è ø

Hint: Equation of a line determined by two points 1 1, 1 2 2, 2( ) ( )P x y and P x y in



Cartesian plane is given as

1 1

2 2

1

1 0.

1

x y

x y

x y

æ öç ÷ç ÷ç ÷=ç ÷ç ÷ç ÷ç ÷è ø

take1 1 1 2 2 2, .z x iy z x iy and z x iy= + = + = +

(b) The area of triangle 1 2 3A A A whose vertices have coordinates 1, 2, 3,z z z is equal

to the absolute value of the number

1 1

2 2

3 3

1

14

1

z zi

z z

z z

æ öç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷è ø

Hint: In Cartesian coordinates the area of the triangle with vertices

1, 1 2, 2 2, 2( ), ( ), ( )x y x y x y is equal to the absolute value of the determinant

1 1

2 2

11

1 .2

1

x y

x y

x y

æ öç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷è ø

3.7.2. Equation of a line determined by a point and a direction:

Proposition 6: Let : 0 d z z be a line and let 0 0( )P z be a point. The

equation of a line parallel to the line d and passing through the point 0P is

0 0( ).z z z z

- = - -

Proof: Hint (Equation of line parallel to the line d passing through the point

0 0 0( , )P x y is

0 0( ).y y i x x

+- = -

- (Using Cartesian coordinates)



Put 0 0 0 00 0, , ,

2 2 2 2

z z z zz z z zy y x x

i i

- +- += = = = and simplify it).


equation of a line perpendicular to the line d and passing through the point

0P is

0 0( ).z z z z

- = -

Proof: Hint (Equation of line perpendicular to the line d passing through the

point 0 0 0( , )P x y is

0 0

1( ).y y x x

i

-- = - -

+ (Using Cartesian coordinates)

Put 0 0 0 00 0, , ,

2 2 2 2

z z z zz z z zy y x x

i i

- +- += = = = and simplify it).

3.7.3. The foot of a perpendicular from a point to a line:


foot of the perpendicular from 0P to the line d has the coordinate

0 0 .2

z zz

Proof: Hint: The point z is the solution of the equations

0 0

0

( ) ( )

z z

z z z z



3.7.4. Distance from a point to a line:

Proposition 9: The distance from a point 0 0( )P z to the line

: 0 d z z where is a complex number, is equal to

0.

2

z zD

+ +=

Proof:(Hint) By using the previous proposition we have

0 00

2

z zD z

+ += -

(It can be simplified to above form easily).

3.7.5. Equation of a circle:

Consider a fixed complex number 0z and let z be any complex number which

moves in such a way that its distance from 0z is always equals to ' 'r .Thus

z lies on a circle whose centre is 0z and radius ' 'r . It’s equation is

0z z r- = .

Squaring both sides we have

2 2

0

2

0 0

2

0 0

2

0 0 0 0

( )( ) 0,

( )( ) 0,

( ) 0,

z z r

z z z z r

z z z z r

zz zz z z z z r

- =

Þ - - - =

Þ - - - =

Þ - - + - =

Taking 2

0 0 0,z z z r = - - = we have

0zz z z + + + = (Centre is - and radius is2

- .



Example15: Find the centre and radius of the circle (1 ) (1 ) 7 0zz i z i z+ - + + - = .

Solution: Given equation can be rewritten as

(1 ) (1 ) 7 0zz i z i z+ + + + - =

Here 1 , 7i = + = - , so centre is at - i.e. 1 i- - and radius is 2

-

i.e., 2

1 7 2 7 3i+ + = + = .

Summary:

In this unit we covered the following points.

1. The complex number z = x +iy can be written as

(cos sin ) ( , ) 0z x iy r i r r . which is known as polar form of complex

number.

2 2r x y z , and = 1tan ( / )y x

where is called argument of z and is expressed as arg(z).

The value of which satisfies is called the principal value of or

the principal argument of z.

2. For (cos sin )z r i and n N ,We have (cos sin )n nz r n i n

3. Geometric images of nth roots of a complex number 0z 0 are the

vertices of a regular n-gon inscribed in a circle with center at origin and

radius n r . The cube roots of unity form an equilateral triangle inscribed in

the circle (0,1)C .



4. The points 1 2 3, ,M M M are collinear iff *3 1

2 1

z z

z z

where, * - the set of non-

zero real numbers.

The lines 1 2M M and 3 4M M are orthogonal iff *1 2

3 4

z zi

z z

.

The distinct points 1 1 2 2 3 3 4 4( ), ( ), ( ), ( )M z M z M z M z are concyclic (or collinear) iff

*3 2 3 4

1 2 1 4

:z z z z

kz z z z

5. The triangle 1 2 3A A A and 1 2 3B B B are similar, having the same orientation iff

1 2 3

1 2 3

1 1 1

0a a a

b b b

. The triangle 1 2 3A A A and 1 2 3B B B are similar, having the

opposite orientation iff 2 1 2 1

3 1 3 1

a a b b

a a b b

.

6. If 1 2 3A A A is an equilateral triangle

2 2 2

1 2 3 1 2 2 3 3 1.z z z z z z z z z

Or 3 22 1

3 1 1 2

.z zz z

z z z z

Or 1 2 3

1 2 3

1 1 10. ( ).

3

z z zwhere z

z z z z z z

Or 2 2

1 2 3 1 2 3

2 2( )( ) 0, ( cos sin ).

3 3z z z z z z where i

Or 1 2 3

2 3 1

1 1 1

0.z z z

z z z



7. Two lines 1 1 1 1 2 2 2 2: 0 : 0d z z and d z z are

Parallel iff 1 2

1 2

=

, perpendicular iff 1 2

1 2

0

+ =

and concurrent iff 1 2

1 2

.

¹

8. Equation of a line determined by two points 1 1 2 2( ) ( )P z and P z is

1 1

2 2

3 3

1

1 0.

1

z z

z z

z z

æ öç ÷ç ÷ç ÷ç ÷=ç ÷ç ÷ç ÷ç ÷ç ÷è ø

9. The area of triangle 1 2 3A A A whose vertices have coordinates 1, 2, 3,z z z is equal

to the absolute value of the number

1 1

2 2

3 3

1

14

1

z zi

z z

z z

æ öç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷è ø

10. Let : 0 d z z be a line and let 0 0( )P z be a point. The equation of a

line parallel to the line d and passing through the point 0P is

0 0( ).z z z z

- = - -

11. Let : 0 d z z be a line and let 0 0( )P z be a point. The foot of the

perpendicular from 0P to the line d has the coordinate

0 0 .2

z zz

12. The equation of circle in complex plane 0zz z z + + + = (Centre is

- and radius is2

- .

Exercises:



(1): Find the cube roots of the complex number 2 2i+ .

(2): Find the forth roots of the complex number 3 i+ .Also find its fifth,

sixth, seventh roots.

(3): Solve the equations

(a)3 125 0z - = (b) 3 64 0z i+ = (c) 6 3 1 0z iz i+ + - =

(4): Find argz and z for (1 3) (1 3)n nz i i= + + - .

(5): Compute1 1

, 3n

nz if z

z z+ + = .

References:

1. TituAndreescu and DorinAndrica, Complex Numbers from

A to …..Z,

2. Jain Iyenger Jain : Advanced Engineering Mathematics

3. Murray Spiegal and Seymour Lipschutz ,Schaum's outline

of complex variables

4. Liang-shin hann, Complex numbers and geometry

Https Doc 14 0c Apps Viewer.googlasasdasdeusercontent

Documents

Transcript of Https Doc 14 0c Apps Viewer.googlasasdasdeusercontent