Free Study Material On Complex Number am61

COMPLEX NUMBERS 1.00

The complex number system

There is no real number x which satisfies the polynomial equation x2 1 0 . 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 i 2

–1 . 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) (d)

i

–1 is called the imaginary unit. Also i 2 a b

–1; i3

–i; i 4

1 etc.

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 .

1.01 Algebraic Operations 111 Fundamental operations with complex numbers 1 In performing operations with complex numbers we can proceed as in the algebra of real numbers, replacing i 2 by –1 when it occurs. (1)

Addition (a bi) (c di)

(2)

Subtraction (a bi) – (c di)

(3)

Multiplication (a bi) (c di)

a bi c di a bi – c – di

(a c) (b d )i (a – c) (b – d )i

ac adi bci bdi 2

(ac – bd ) (ad bc)i

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(4)

Division

a bi a bi c – bi = c di c di c – di

ac – adi bci – bdi 2 ac bd (bc – ad )i ac bd bc – ad = = 2 + i. c2 – d 2i 2 c2 – d 2 c d 2 c2 d 2

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 a 2 b2

0 then a 0 b however in complex numbers, z12

Illustration 1 :

Find multiplicative inverse of 3 + 2i, then

Solution:

Let z be the multiplicative inverse of 3 2i, then

z22

0 does not imply z1

z2

0.

z.(3 2i) 1 1

z

3 – 2i 3 2i 3 – 2i

3 2i 3 2 – i 13 13

z

3 2 – i 13 13

Ans.

1.02 Equality In Complex Number 111 Two 1 complex numbers z1 a1 ib1 & z2 a2 equal respectively. i.e. z1

z2

ib2 are equal if and only if their real and imaginary parts are

Re( z2 ) and I m ( z1 )

I m ( z2 ) .

Illustration 2 :

Find the value of x and y for which (2 3i) x2 – (3 – 2i) y 2 x – 3 y 5i where x, y R.

Solution:

( z 3i ) x 2 – (3 – 2i ) y 2x – 3y x2 – x x if x x

0

0, 1 and 3x 2 0, y 0, y

2 x – 3 y 5i

2x – 3y

2

2y

5

5 and if x 1, y 1 2 5 and x 1, y 1 2

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Illustration 3 :

Find the value of expression x4 – 4x3 3x2 – 2x 1 when x 1 i is a factor of expression.

Solution:

x 1 i x –1 i

( x –1) 2

x – 2x 2

–1

0 Now x – 4 x3 3 x 2 – 2 x 1

2

4

( x 2 – 2 x 2)( x 2 – 3 x – 3) – 4 x 7 when x 1 i i.e. x 2 – 2 x 2

0

x 4 – 4 x3 3 x 2 – 2 x 1 0 – 4(1 i) 7 – 4 7 – 4i 3 – 4i

Ans

Illustration 4 :

Solve for z if z 2 | z | 0 .

Solution:

Let z

x iy ( x iy ) 2

x2

y2

0

x2 – y 2

x2

y2

0 and 2 xy

x

0 or y

when x

0

0– y

2

| y| 0

0, 1, –1

y when y

Ans.

x

Find square root of 7 + 40i.

Solution:

Let (x iy ) 2

0

0, z

z

Illustration 5 :

x –y

0, i, –1

z

x2 | x | 0

0 0

z

0

i, z

–1

9 40i 9

2

2

and xy

20

........(i) ..........(ii )

squing (i ) and adding with 4times the square of (ii ) we get x 4

y 4 – 2x2 y 2

( x2 x

2

y 2 )2 y

2

4x2 y 2

81 1600

168

4

........(iii)

from (i) (iii) we can see that x & y are of same sign x iy

(5 4i) or

Sq. roots of a 40i (5 4i)

(5 4i) (5 4i)

Ans.

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1.03 Representation Of A Complex Number 111 1 Illustration 6 : Express the complex number z –1 Solution:

2i in polar form.

–1 i 2

z |z|

(–1) 2

2

2

2 1

– tan –1

Arg z

3(cos

z

1 2

i sin )

3

– tan –1 2

( say ) – tan –1 2

where

1.04 Modulus Of A Complex Number 111 1 Illustration 7 : If | z – 5 – 7i | 9 , 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|. |1 z | 1 – z

Solution:

1 z

|1 z 1 – z | (triangle inequality)

1– z

2

minimum value of (|1 z | |1 – z |)

2

Geometrically | z 1| |1 – 2 | | z 1| | z –1| which represents sum of distances of z from 1 and –1 it can be seen easily that minimum ( PA PB ) 1

21/ 4 e

8

AB

2

n

Ans.

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Illustration 9 :

z–

2 z

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

Solution:

z–

2 z

1

z –

Let z

r

2 r

1

r–

2 z

2 r

and

r–

r

(1, 2)

r

R

1

(1, 2)

z

–

2 z

..........(i ) –1 r –

2 1 r

..........(ii )

from (i) and (ii) r r

2 2

2 r

r

2 1 r

r

z–

(1, 2)

Ans.

1.05 Argument Of A Complex Number 111 1 Illustration 10: Solve for z, which satisfy Arg ( z – 3 – 2i)

Solution:

6

and Arg ( z – 3 – 4i)

2 . 3

From the figure, it is clear that there is no z, which satisfy both ray.

Illustration 11: Sketch the region given by (i)

Solution:

(i)

Arg ( z –1– i)

/3

(ii)

z

5 & Arg ( z – i –1)

/3.

(ii)

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1.06 111 1

Conjugate Of A Complex Number

Illustration 12:

If

Solution:

Re

z –1 is purely imaginary, then prove that z z 1 z –1 z 1

1.

0

z –1 z 1

z –1 z 1

0

z –1 z 1

z –1 0 z 1

z z – z z –1 z z – z z –1 0 z

z

Hence Proved.

1

1.07 Rotation Theorem 111 z –1 1 Illustration 13: If Arg z 1

Solution:

arg

z –1 z 1

arg

2

zz 1

then interrupter the locus.

3

3

1– z –1 – z

Here arg

1

1– z –1– z

3

represents the angle between lines

ing –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 arg

1– z –1– z

will be equal to –

out radius and centre of this circle. centre

Illustration 14:

0,

1 3

2 . Now try to geometrically find 3 Radius

2 3

Ans.

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.

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Solution:

Let affix of C and D are z3 Considering

DAB

90

z4 respectively

0

AD

AB

z4 – (2 3i) (3 4i) – 2 3i AD AB z4 – (2 3i ) (1 i )i

we get

2 3i i –1

z4

Illustration 16:

z3

3 4i i –1 z 5i

192

192

194

.

194

–

2

2

Ans.

If 1,

,

are cube roots of unity prove.

(i )

(1 –

2

(ii )

(1 –

2 5

)(1 )

(iii ) (1 – )(1 –

Solution:

1 zi

z3

1.08 Cube Root Of Unity 111 1 Illustration 15: Find the value of

1

i 2

z3 – (3 4i ) ( z 3i ) – (3 – 4i) i e– CB AB 2 3 4i – (1 i )(–i )

and

Solution:

e

(iv)

(1 –

(i )

(1 –

2

(–2 )(–2

)(1

)

–

)(1 –

)(1 –

2

2

(1

2 2

–

4

)

4

2

32

2 5

)(1 –

2

–

4

8

)(1 –

)

9 4

8

)..........to 2n factors

22 n

)

)

4

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th

1.09 n Roots Of Unity 111 1 Illustration 17: Find the roots of the equation Solution:

z6

–64

6

z 6 .e

z

i (2 n 1)

i (2 n 1)

x

z6

64 0 where real part is positive.

z

6

z

ze

z

2e 6 , 2e 2 , ze 2 , ze

i

i

i

i

5 6

e

i

7 6

, ze

i

3 2

i

11 6

i

roots with +ve real part are i –

2e

6

Find the value

sin

k 1

6

Solution:

sin

k 1 6

e

i

11 2

Ans.

6

Illustration 18:

e6

, ze

2 k 2 k . – cos 7 7

6 2 k 2 k – cos 7 7 k 1

sin

k 0

2 k 6 2 k – cos 1 7 7 k 0

6

(Sum of imaginary part of seven seventh roots of unity) k 0 6

–

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

k 0

0–0 1 1 i

1.10

Ans.

Logarithm Of A Complex Quantity

Illustration 19:

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(

) )

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Solution:

Let z1

cos

z3

i sin , z2

cos z1

z2

z3

1 z1

cos

1 z1

1 z2

Now z

2 1

(cos

cos

cos ) i(sin

sin

0

1 z1

Also

0 – i.0

sin ) (1)

i sin ) –1

(cos

– i sin , 1 z3

cos

– i sin

1 – cos – sin z3

(cos

cos

cos ) – i(sin

sin

z

z32

( z1

z2

1 z1

1 z2

1 z3

0 – 2 z1 z2 z3 .0

(2)

or (cos 2

z3 ) 2 – 2( z1 z2

z3 z1 )

z 2 z3

0, u sin g (1)and (2)

i sin ) 2

(cos

(cos

i sin 2 ) 2

cos 2

i sin ) 2

i sin ) 2

(cos

i sin 2

cos 2

i sin 2

Equation real and imaginary parts on both sides, cos 2 (ii ) z13

sin )

0 2 2

0 – 2 z1 z2 z3

or

i sin ,

i sin .

0 i.0 (i )

cos

z23

z33

( z1

z2 )3 – 3 z1 z2 ( z1

0 0 i.0 cos 2

cos 2

0 and sin 2

sin 2

z2 ) z33

(– z3 )3 – 3z1 z2 (– z3 ) z33 , using (1) 3z1 z2 z3 (cos 3(cos

i sin )3

(cos

i sin )(cos

i sin )3 (cos i sin )(cos

i sin )3 i sin )

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sin 2

0

or cos 3

i sin 3

cos 3

3{cos(

i sin 3

cos 3

) i sin(

i sin 3

)}

Equation imaginary parts on both sides, sin 3

sin 3

sin 3

3sin(

)

Alternative method Let C S

cos

sin

cos

sin

sin i

C iS

e

C – iS

e–i –i

From (1)

(e

ei 2

ei 2

ei (2

)

cos

e

0

i

e

e–i )

2

i

0

e–i

(e

–i

)

2

(1) 0

(e

–i

(2) )

2

i

(e )(e ) (ei )(ei ) (ei )(ei )

ei ei ei (e –2

ei 2

ei 2

0

ei 2

e–i

i

ei )

0( from 2)

Comparing the real and imaginary parts we cos 2

cos 2

cos 2 – sin 2

sin 2

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

ei 3

ei 3

3ei (

sin 2

0

3ei ei ei

)

Comparing the real and imaginary parts we obtain the results.

Illustration 20:

If z1 and z2 are two complex numbers and c > 0, then prove that 2

z1 + z2

Solution:

(I C) z1

2

2

(I C –1 ) z2 .

We have to prove that: z1 i.e.

z2

2

z1

orz1 z2

(1 C ) z1 3

z2

2

z2 z2

c z1 –

z2 z2

2

C –1 z2

c z1

1 c

(1 C –1 ) z2

z1 z2

(using Re ( z1 z2 ) or

2

2

(1 C ) z1 2

2

(1 C –1 ) z2

or c z1

2

3

1 2 z2 – z1 z2 – z2 z2 c

0

z1 z2 ) 2

z2

0 which is always true.

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Illustration 21:

If ,

/ 3], i 1, 2, 3, 4, 5 and z 4 cos

[ / 6,

cos 2 .z 3 cos 3 .z 2

or cos 1 .z 4

cos 1 .z 4



i [ / 6, / 3]

2 3

cos

i

3 | z |4 2

3 | z |4

cos 2 .z 3

3

z cos

4

cos

5

2 3

5

cos 4 .z cos

cos 3 .z 2

5

cos 4 .z

2 3 cos

5

3 2 3 | z |3 2

3 | z |2 2

| z |3

| z |2

|z|

3 | z | | z |2

| z |3

| z |4

3

cos 4 .z cos

cos 2 .z 3 cos 3 .z 2

2 3

1 2

3 |z| 2

3 2

| z |5 ..........

|z| 3–e| z | | z | 1– | z | 4| z| 3

|z|

3 4

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 z =

Solution:

z 3 cos

Given that cos 1 .z 4

Illustration 21:

2

3 . 4

2 3, then show that | z |

Solution:

z 3 cos

1

a –1 b –1 – c –1 – d –1 . a –1b –1 – c –1d –1

Since point P, A, B are collinear

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z z 1 a a 1

0

z a – b – z (a – b)

ab – ab

0(i )

b b 1 Similarlym, since points P, C, D are collinear z a –b c –d – z c –d a –b 

zz

r2

k ( say )

a

cd – cd a – b – ab – ab c – d (ii ) k ,b a

k ,c b

k etc. c

From equation (ii ) we get z z

k k k k – (c – d ) – z – ( a – b) a b c d

ck kd ak bk – ( a – b) – – (c – d ) d c b a

a –1 b –1 – c –1 – d –1 a –1b –1 – c –1d –1

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