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Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866

COMPLEX NUMBERS

There is no real number x which satisfies the polynomial equation 2 1 0x . To permit solutions of this

and similar equations, the set of complex numbers is introduced.

We can consider a complex number as having the form a + bi where a and b are real number and i, which

is called the imaginary unit, has the property that 2 –1i . It is denoted by z i.e. z = a + ib. ‘a’ is called as

real part of z which is denoted by (Re z) and ‘b’ is called as imaginary part of z which is denoted by (Im

z).

Any complex number is:

(i) Purely real, if b = 0 (b) Purely imaginary, if a = 0 (c) Imaginary, if b 0

Note :

(a) The set R of real numbers is a proper subset of the Complex Numbers. Hence the complete

number system is N W I Q R C.

(b) Zero is purely real as well as purely imaginary but not imaginary.

(c) –1i is called the imaginary unit. Also 2 3 4–1; – ; 1 .i i i i etc

(d) a b ab only if atleast one of a or b is non–negative.

(e) is z = a + ib, then a – ib is called complex conjugate of z and written as –z a ib .

Fundamental operations with complex numbers

In performing operations with complex numbers we can proceed as in the algebra of real numbers,

replacing 2 –1i by when it occurs.

(1) Addition ( ) ( ) ( ) ( )a bi c di a bi c di a c b d i

(2) Subtraction ( ) – ( ) – – ( – ) ( – )a bi c di a bi c di a c b d i

(3) 2Multiplication ( ) ( ) ( – ) ( )a bi c di ac adi bci bdi ac bd ad bc i

1.00 The complex number system

1.01

111

1

Algebraic Operations


(4) 2

2 2 2 2 2 2 2 2 2

– – – ( – ) –Division = = = + i

– – –

a bi a bi c bi ac adi bci bdi ac bd bc ad i ac bd bc ad

c di c di c di c d i c d c d c d.

Inequalities in complex numbers are not defined. There is no validity if we say that complex number is

positive or negative. e.g. z > 0, 4 + 2i < 2 + 4i are meaningless.

In real numbers if 2 2 2 2

1 2 1 20 then 0 however in complex numbers, 0 does not imply 0.a b a b z z z z

Illustration 1 : Find multiplicative inverse of 3 + 2i, then

Solution:

Two complex numbers 1 1 1 2 2 2 & z a ib z a ib are equal if and only if their real and imaginary parts are

equal respectively. 1 2 2 1 m 2. . Re( ) and I ( ) I ( )mi e z z z z z .

Illustration 2 : Find the value of x and y for which 2(2 3 ) – (3 – 2 ) 2 – 3 5 where , .i x i y x y i x y R

Solution:

Let z be the multiplicative inverse of 3 2 , then

.(3 2 ) 1

1 3 – 2

3 2 3 2 3 – 2

3 2 –

13 13

3 2 – Ans.

13 13

i

z i

iz

i i i

z i

i

1.02

111

1

Equality In Complex Number

2

2

2

2

( 3 ) – (3 – 2 ) 2 – 3 5

2 – 3 2 – 3

– 0

0, 1 and 3 2 5

5 if 0, and if 1, 1

2

5 0, and 1, 1

2

z i x i y x y i

x y x y

x x

x x y

x y x y

x y x y


Illustration 3 : Find the value of expression 4 3 2– 4 3 – 2 1 when 1x x x x x i is a factor of expression.

Solution:

Illustration 4 : Solve for z if 2 | | 0z z .

Solution:

2 2 2 2

2

2

– 0 and 2 0

0 or 0

when 0 – | | 0

0, 1, –1 0, , –1

when 0 | | 0 0

0 Ans. 0, , –1

x y x y xy

x y

x y y

y z i

y x x x

z z z i z

Illustration 5 : Find square root of 7 + 40i.

Solution:

2

2 4 3 2

2 2

2

4 3 2

1

–1 ( –1) –1

– 2 2 0 Now – 4 3 – 2 1

( – 2 2)( – 3 – 3) – 4 7

when 1 . . – 2 2 0

– 4 3 – 2 1 0 – 4(1 ) 7

– 4 7 – 4

3 – 4 A

x i

x i x

x x x x x x

x x x x x

x i i e x x

x x x x i

i

i ns

2 2 2

Let

( ) 0

z x iy

x iy x y

2

2 2

4 4 2 2 2 2

Let ( ) 9 40

– 9 ........( )

and 20 ..........( )

squing ( ) and adding with 4times the square of ( )

we get – 2 4 81 1600

x iy i

x y i

xy ii

i ii

x y x y x y

2 2 2

2 2

( ) 168

4 ........( )

from ( ) ( ) we can see that & are of same sign

(5 4 ) or (5 4 )

Sq. roots of 40 (5 4 )

(5 4 ) Ans.

x y

x y iii

i iii x y

x iy i i

a i i

i


Illustration 6 : Express the complex number –1 2z i in polar form.

Solution:

Illustration 7 : If | – 5 – 7 | 9z i , then find the greatest and least values of |z – 2 – 3i.|

Solution: We have 9 = |z – (5 + 7i)| = distance between z and 5 + 7i. Thus locus of z is the circle

of radius 9 and centre at 5 + 7i. For such a z (on the circle), we have to find its greatest

`and least distance as from 2 + 3i, which obviously 14 and 4.

Illustration 8 : Find the minimum value of |1 + z| + |1 – z|.

Solution:

1.03

111

1

Representation Of A Complex Number

22

–1 –1

–1

–1 2

| | (–1) 2 1 2 3

2Arg – tan – tan 2 ( )

1

3(cos sin ) where – tan 2

z i

z

z say

z i

1.04

111

1

Modulus Of A Complex Number

|1 | 1– |1 1– | (triangle inequality)

1 1– 2

minimum value of (|1 | |1– |) 2

Geometrically | 1 | |1– 2 | | 1 | | –1| which represents sum of distances of from 1 and –1

it can be seen easily t

z z z z

z z

z z

z z z z

1 1/4 8

hat minimum ( ) 2

2 Ans.n

PA PB AB

e


Illustration 9 : 2

– 1zz

then find the maximum and minimum value of |z|.

Solution:

Illustration 10: Solve for z, which satisfy

2Arg ( – 3 – 2 ) and Arg ( – 3– 4 ) .

6 3z i z i

Solution: From the figure, it is clear that there is no z, which satisfy

both ray.

Illustration 11: Sketch the region given by

(i) ( –1– ) / 3Arg z i (ii) 5& ( – –1) / 3z Arg z i .

Solution: (i) (ii)

2 2 2 2– 1 – – –

2

Let

2 2 – 1

2 1 ..........( )

2 2 and – 1 –1 – 1

(1, 2) .

z z z zz z z

z r

r rr r

r r R ir

r rr r

r .........( )

from (i) and (ii) (1, 2)

(1, 2) Ans.

ii

r

r

1.05

111

1

Argument Of A Complex Number


Illustration 12: If –1

1

z

z is purely imaginary, then prove that 1z .

Solution:

Illustration 13: If –1

1 3

zArg

z then interrupter the locus.

Solution:

Here 1–

arg–1–

z

z represents the angle between lines

joining –1 and z and 1 + z. As this angle is constant, the locus of z will be a of a circle

segment.(angle in a segment is count). It can be seen that locus is not the complete side

as in the major are 1–

arg–1–

z

z will be equal to

2–

3. Now try to geometrically find

out radius and centre of this circle. 1 2

centre 0, Radius Ans.3 3

Illustration 14: If A(z + 3i) and B(3 + 4i) are two vertices of a square ABCD (take in anticlock wise

order) then find C and D.

1.06

111

1

Conjugate Of A Complex Number

2

–1Re 0

1

–1 –1 –1 –1 0 0

1 1 1 1

– –1 – –1 0

1 1

1 Hence Proved.

z

z

z z z z

z z z z

zz z z zz z z

zz z

z

1.07

111

1

Rotation Theorem

–1arg

1 3

1–arg

–1– 3

z

z

z

z


Solution: 3 4

0

Let affix of C and D are respectively

Considering 90

z z

DAB AD AB

4

4

4

3

3

3

– (2 3 ) (3 4 ) – 2 3we get

2

– (2 3 ) (1 )

2 3 –1 1

– (3 4 ) ( 3 ) – (3 – 4 ) and –

2

3 4 – (1 )(– )

3 4 –1 5

z i i i ie

AD AB

z i i i

z i i zi

z i z i i ie

CB AB

z i i i

z i i z i

Illustration 15: Find the value of 192 194 .

Solution:

Illustration 16: 2If 1, , are cube roots of unity prove.

2 2

2 5 2 5

2 4 8

2 2 4 4 8 2

( ) (1– )(1 – ) 4

( ) (1– ) (1 – ) 32

( ) (1– )(1– )(1– )(1– ) 9

( ) (1– )(1– )(1– )..........to 2n factors 2 n

i

ii

iii

iv

Solution:

1.08

111

1

Cube Root Of Unity

192 194

21 – Ans.

2 2

2

( ) (1– )(1 – )

(–2 )(–2 )

4

i


Illustration 17: Find the roots of the equation 6 64 0z where real part is positive.

Solution:

Illustration 18: Find the value 6

1

2 2sin – cos

7 7k

k k.

Solution:

Illustration 19:

1.09

111

1

nth

Roots Of Unity

6

6 6 (2 1)

(2 1)6

5 7 3 11

6 6 62 2 2 2

11

6 6

–6

–64

.

2 , 2 , , , ,

roots with +ve real part are

2 Ans.

i n

i n

i i ii i i i

ii

i

z

z z e x z

z ze

z e e ze ze e ze ze

e e

e

6 6

1 1

6 6

0 0

6

0

6

0

2 2sin – cos

7 7

2 2 sin – cos 1

7 7

(Sum of imaginary part of seven seventh roots of unity)

– (Sum of real part of seven seventh roots of unity) 1

0 – 0

k k

k k

k

k

k k

k k

1 1

i Ans.

1.10 Logarithm Of A Complex Quantity

If cos cos cos 0 and also sin sin sin 0, then prove that

(i) cos2 +cos2 +cos2 = sin2 +sin2 +sin2 = 0

(ii) sin3 +sin3 +sin3 = 3sin( )

(iii) cos3 +cos3 +cos3 = 3cos( )


Solution:

1 2

3

1 2 3

1

Let cos sin , cos sin ,

cos sin .

(cos cos cos ) (sin sin sin )

0 .0 0 (1)

1( ) Also (

z i z i

z i

z z z i

i

iz

–1

1 3

1 2 3

2 2 2 2

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

cos sin ) cos – sin

1 1 cos – sin , – cos – sin

1 1 1 (cos cos cos ) – (sin sin sin ) (2)

0 – .0 0

Now ( ) – 2( )

i i

iz z

iz z z

i

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

1 2 3

3 1 2

1 2 3

2 2 2

2

1 1 10 – 2

0 – 2 .0 0, sin (1) (2)

(cos sin ) (cos sin ) (cos sin ) 0

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

Equation real and imaginary parts on bot

z z zz z z

z z z u g and

or i i i

or i i cos i i

3 3 3 3 3

1 2 3 1 2 1 2 1 2 3

3 3

3 1 2 3 3

1 2 3

3

h sides, cos 2 cos 2 cos 2 0 and sin 2 sin 2 sin 2 0

( ) ( ) – 3 ( )

(– ) – 3 (– ) , using (1)

3

(cos sin )

ii z z z z z z z z z z

z z z z z

z z z

i 3 3(cos sin ) (cos sin )

3(cos sin )(cos sin )(cos sin )

i i

i i i


cos3 sin 3 cos3 sin 3 cos3 sin 3

3{cos( ) sin( )}

Equation imaginary parts on both sides, sin 3 sin 3 sin 3 3sin( )

Alternative method

Let cos cos cos 0

sin sin sin

or i i i

i

C

S

– – –

– 2 – 2 – 2

2 2 2

0

0 (1)

– 0 (2)

From (1) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )

i i i

i i i

i i i i i i i i i

i i i i i

C iS e e e

C iS e e e

e e e e e e e e e

e e e e e e –2 –

(2 ) 2 2

3 3 3

3 3 3 ( )

( )

0( 2)

Comparing the real and imaginary parts we

cos 2 cos 2 cos 2 – sin 2 sin 2 sin 2 0

Also from (1) ( ) ( ) ( ) 3

3

i i i

i i i

i i i i i i

i i i i

e e e

e e e from

e e e e e e

e e e e

Comparing the real and imaginary parts we obtain the results.

Illustration 20: If 1 2 and z z are two complex numbers and c > 0, then prove that

2 2 2–1

1 2 1 2 + (I C) (I C )z z z z .

Solution:

2 2 2–1

1 2 1 2

3 2 2 3–1

1 2 1 2 2 2 1 2

2 2 2 2–1

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

1 2 1 2

2

1 2

We have to prove that:

(1 ) (1 )

i.e. (1 ) (1 )

1 or – – 0

(using Re ( ) )

1– 0 wh

z z C z C z

z z z z z z C z C z

orz z z z c z C z c z z z z z zc

z z z z

or c z zc

ich is always true.


Illustration 21: 4 3 3

1 2 3 4 5If , [ / 6, / 3], 1, 2, 3, 4, 5 and cos cos cos cos cosi z z z z

32 3, then show that | | .

4z

Solution:

4 3 2

4 3 2

2 3 4 5

3 3 3 3 3 2 3 | | | | | | | |

2 2 2 2 2

3 | | | | | | | |

3 | | | | | | | | | | ..........

| | 3 3 – | | | |

1– | |

3 4 | | 3 | |

4

z z z z

z z z z

z z z z z

ze z z

z

z z

Illustration 21: Two different non parallel lines cut the circle |z| = r in point a, b, c, d respectively.

Prove that these lines meet in the point z given by –1 –1 –1 –1

–1 –1 –1 –1

– –z =

–

a b c d

a b c d.

Solution: Since point P, A, B are collinear

4 3 2

1 2 3 4 5

4 3 2

1 2 3 4 5

4 3 2

1 2 3 4 5

Given that

cos . cos . cos . cos . cos 2 3

or cos . cos . cos . cos . cos 2 3

2 3 cos . cos . cos . cos . cos

[ / 6, / 3]

1 3 cos

2 2i

z z z z

z z z z

z z z z

i


2

1

1 0 – – ( – ) – 0( )

1

Similarlym, since points P, C, D are collinear

– – – – – – – – – – ( )

( ) , , .

From equati

zz

a a z a b z a b ab ab i

b b

z a b c d z c d a b cd cd a b ab ab c d ii

k k kzz r k say a b c etc

a b c

–1 –1 –1 –1

–1 –1 –1 –1

on ( ) we get

– ( – ) – – ( – ) – ( – ) – – ( – )

– –

–

ii

k k k k ck kd ak bkz c d z a b a b c d

a b c d d c b a

a b c dz

a b c d