Mathematics

What is loge(-1) ?

Not Defined

It’s a complex number

One of its value is i

loge(-1) is defined and is complex no.

Session opener

Session Objectives

Session Objective

1. Complex number - Definition

2. Equality of complex number

3. Algebra of complex number

4. Geometrical representation

5. Conjugate of complex number

6. Properties of modules and arguments

7. Equation involving variables and locus

Solve x2 + 1 = 0

D = –4(<0) No real roots

x -1

Euler Leonhard ( 1707-1783)

-1 i (known as iota )

“i” is the first letter of the latin word ‘imaginarius’

Complex Numbers Intro

0i 1(as usual)1i i2i 13 2i i .i i 4 3i i .i i.i 1

12

1 ii i

i i

22

1i 1

i

33

1 1i i

ii

4

4

1i 1

i

Evaluate:33

17 2i

i

3 3

3163

8 8i .i i i 8i

ii

Solution

Ans: 343i

Integral powers of i(iota)

If p,q,r, s are four consecutive integers, then ip + iq + ir + is =

a)1 b) 2

c) 4 d) None of these

Solution:Note q = p + 1, r = p + 2, s = p + 3

= ip(1 + i –1 – i) = 0

Given expression = ip(1 + i + i2 + i3)

Remember this.

Illustrative Problem

If un+1 = i un + 1, where

u1 = i + 1, then u27 is

a) i b) 1

c) i + 1 d) 0

Solution:u2 = iu1 + 1 = i(i+1) +1 = i2 + i + 1

Hence un = in + in-1 + ….. + i + 1

u3 = iu2 + 1 = i(i2+i+1) +1 = i3 + i2 + i + 1

2827 26

27i 1

u i i ..... i 1 0i 1

Note by previous question:

u27 = 0

Illustrative Problem

z 4 5 4 i 5

If a = 0 ?

If b = 0 ?

If a = 0, b = 0 ?

Complex Numbers - Definition

z = a + i b

Mathematical notation

re(z)= a

im(z)=b

a,bR

Re(z) = 4, Im(z) = 5

z is purely real

z is purely imaginary

z is purely real as well as purely imaginary

If z1 = a1 + ib1 and z2 = a2 + ib2

z1 = z2 if a1 = a2 and b1 = b2

Find x and y if 3

x 5 i2 5y 25

Equality of Complex Numbers

Is 4 + 2i = 2 + i ? No

One of them must be greater than the other??

Order / Inequality (>, <, , ) is not defined for complex numbers

Find x and y if

(2x – 3iy)(-2+i)2 = 5(1-i)

Hint: simplify and compare real and imaginary parts

Solution:

(2x – 3iy)(4+i2-4i) = 5 -5i

(2x – 3iy)(3 – 4i) = 5 –5i

(6x – 12y – i(8x + 9y)) = 5 – 5i

6x – 12y = 5, 8x + 9y = 57 1

x , y10 15

Illustrative Problem

(I) Addition of complex numbers

z1 = a1 + ib1, z2 = a2 + ib2 then

Properties:

z1 + z2 = a1 + a2 + i(b1 + b2)

1) Closure: z1 + z2 is a complex number

2) Commutative: z1 + z2 = z2 + z1

3) Associative: z1 + (z2 + z3) = (z1 + z2) + z3

4) Additive identity 0: z + 0 = 0 + z = z

5) Additive inverse -z: z + (-z) = (-z) + z = 0

Algebra of Complex Numbers – Addition

(II) Subtraction of complex numbers

z1 = a1 + ib1, z2 = a2 + ib2 then

Properties:

z1 - z2 = a1 - a2 + i(b1 - b2)

1) Closure: z1 - z2 is a complex number

Algebra of Complex Numbers - Subtraction

z1 = a1 + ib1, z2 = a2 + ib2 then

Properties:

z1 . z2 = a1a2 – b1b2 + i(a1b2 + a2b1)

1) Closure: z1.z2 is a complex number

2) Commutative: z1.z2 = z2.z1

3) Multiplicative identity 1: z.1 = 1.z = z

4) Multiplicative inverse of z = a + ib (0):

12 2

1 1 a ib a ibz . (remember)

a ib a ib a ib a b

Algebra of Complex Numbers - Multiplication

5) Distributivity: z1(z2 + z3) = z1z2 + z1z3

(z1 + z2)z3 = z1z3 + z2z3

z1 = a1 + ib1, z2 = a2 + ib2 then

1 1 1 1 1 2 2

2 2 2 2 2 2 2

z a ib a ib a ib.

z a ib a ib a ib

1 2 1 2 2 1 1 2

2 22 2

a a b b i(a b a b )a b

Algebra of Complex Numbers- Division

2(1 i)Sum of the roots is 2i 2

i

Solution:

Illustrative Problem

If one root of the equation

is 2 – i then the other root is

(a) 2 + i (b) 2 – i (c) i (d) -i

2ix 2(i 1)x (2 i) 0

(2 – i) + = -2i +2

= -2i +2-2 + i = -i

Representation of complex numbers as points on x-y plane is calledArgand Diagram.

Representaion of z = a + ib

O X

Y

Geometrical Representation

Im (z)

Re (z) a

b

P(z)

2 2OP z a b

Modulus of z = a + i b

1 btan

a

Argument of z

Arg(z) = Amp(z)

Modulus and Argument

O X

YIm (z)

Re (z) a

b

P(z)

|z|

Argument (-, ] is called principal value of argument

Argument (-, ] is called principal value of argument

1 bStep1: Find tan for 0,

a 2

Step2: Identify in which quadrant (a,b) lies

( -,+) ( +,+)

( -,-) ( +,-)

1+2i-1+2i

-1-2i 1 -2i

Principal Value of Argument

Step3: Use the adjoining diagram to find out the principal value of argument

Based on value of and quadrant from step 1 and step 2

Principal Value of Argument

Let z x iy

2 2x iy 2 (x 1) y i 0

2 2x 2 (x 1) y 0 andy 1 0

z 2 | z 1| i 0

2 2x 2 (x 1) y i(1 y) 0

The complex number which satisfies the equation

(a) 2 – i (b) –2 - i

(c) 2 + i (d) -2 + i

z 2 | z 1 | i 0 is

Illustrative Problem

Solution Cont.

2 2x 2 (x 1) y 0 and y 1 0

y 1 2x 2 (x 1) 1 0

2 2x 2(x 2x 2) 2(x 2) 0

x 2

Illustrative Problem – Principal argument

The principal value of

4 2a) b) c) d)

3 3 3 3

argument in 2 2 3 i

1 2 3tan

2 3

Step1:

Step2: 3rd quadrant ( -,-)

Solution

23

Step3:

For z = a + ib,

_3 iFor z , z ?

2

Conjugate of a Complex Number

Conjugate of z is z a ib

Q(z)

-b-

Image of z on x – axis

_ 3 iz

2

ba

P (z)Y

X

_

I f z z a ib a ib

b 0, z is puerly real

z lies on x -axis_

What if z z ?

z za Re(z)

2

z zb im(z)

2i

2 2z a ib a b z

1 bArg(z) A rg(a ib) tan Arg(z)

a

Conjugate of a Complex Number

1 2 1 2z .z z . z nnz z

z z

2z.z z

11

2 2

zzz z

1 2 1 2z z z z 1 2 1 2z z z z

1 2 1 2z z z z 1 2 1 2z z z z

2 2 2 2

1 2 1 2 1 2z z z z 2 z z

(Triangle inequality)

22 2Pr oof : z a ib, zz (a ib)(a ib) a b z

Properties of modulus

1 2 1 2Arg z .z Arg z Arg z

11 2

2

zArg Arg z Arg z

z

1 2 n 1 2 nArg z .z ....z Arg z Arg z .... Arg(z )

1Arg z Arg z , Arg Arg z

z

Arg(purely real) = 0 or or 2n and vice versa

Arg(purely imaginary) = or or 2n 12 2 2

and vice versa

Properties of Argument

z z

1 2 1 2z z z z

1 2 1 2z .z z .z

1 12

2 2

z zprovided z 0

z z

Pr oof : z a ib, z a ib a ib z

Conjugate Properties

1Conjugate ofi s

2 i

(a) (b)2 i5 5

2 i

(c) (d)1

2 i2 i

5

Illustrative Problem

1 2 i 2 iz

2 i 2 i 5

2 i 2 iz ( )

5 5

Solution:

Find a ib

Let x iy a ib

x2 – y2 + 2ixy = a + ib

x2 – y2 = a 2xy = b

22 2 2 2 2 2x y x y 4x y

• Find x2 , take positive value of x

• Find y2, take value of y which satisfies 2xy = b

Note if b > 0 x,y are of same sign, else if b < 0 x,y are of opposite sign

• Square root will be x +iy

Square Root of a Complex Number

(Squaring)

Other root will be – (x+iy)

Find 8 15i

Let x iy a ib

x2 – y2 + 2ixy = 8 –15i

x2 – y2 = 8 2xy = -15 x,y are of opposite sign

2 2x y 64 225 17

Illustrative Problem

Solution

2 52x 8 17 x

2

Find 8 15i

5 3One of the square root i

2 2

Illustrative Problem

Solution

25 25 3For x , y 17 y

22 2

15 5 3As xy , for x y -

2 2 2

5x

2

5 3Other square root i

2 2

Cartesian System – 2D

Argand Diagram

Point ( x,y) Complex No.( z)

Locus of point Locus of complex no. ( point in argand diagram)

Equation in x,y defines shapes as circle , parabola

Equation in complex variable (z) defines shapes as circle , parabola etc.

Distance between P(x1,y1) and Q(x2,y2) = PQ

Distance between P(z1) and Q(z2) = |z1-z2|

Equation involving complex variables and locus

Illustrative Problem

Let z x iy then

2 2 2 2(x 1) y 2 (x 1) y

2 23x 3y 10x 3 0

If z is a complex number then |z+1| = 2|z-1| represents

(a) Circle (b) Hyperbola

(c) Ellipse (d) Straight Line

Solution

If , then the locus of z

is given by

a) Circle with centre on y-axis and

radius 5

b) Circle with centre at the origin and radius 5

c) A straight line

d) None of these

z 5arg

z 5 2

Illustrative Problem

Let z = x + iy, then

x iy 5arg

x iy 5 2

2 2

2 2

x y 25 i10yarg

2x 5 y

As argument is complex number

is purely imaginary2

x2 + y2 = 25, circle with center (0,0) and radius 5

Solution

Illustrative Problem

| z 1 3| |z 1| 3

| z 1| 3 | z 1 3|

| z 1| 3 3 As | z+4| 3

| z 1| 6 Least value = ?

-7 -1

Locus of z

(a) 2 (b) 6 (c) 0 (d) -6

I f z is a complex no. such

the maximum value of | z 1| isthat | z 4 | 3

Solution

Class Exercise

Class Exercise - 1

The modulus and principal argument

of –1 – i are respectively3

(a) 4 and

3

2(b) 2 and

3

(c) 4 and –

32

(d) 2 and –3

Solution:

The complex number lies in the third quadrant andprincipal argument satisfying is given by – .

Solution contd..

arg(z) =

1 3

tan1 3

23 3

is the principal argument.

The modulus is = 221 3 2

Hence, answer is (d).

Class Exercise - 2

If , then x2 + y2 is equal to

22a 1x iy

2a i

42

2

a 1(a)

4a 1

42

2

a 1(b)

4a 1

22

2

a 1(c)

4a 1

(d) None of these

Solution

22a 1x iy

2a i

Taking modulus of both sides,

22a 1x iy

2a i

222 2

2

a 1x y

4a 1

422 2

2

a 1x y

4a 1

Hence, answer is (a).

Class Exercise - 3

If |z – 4| > |z – 2|, then

(a) Re z < 3 (b) Re z < 2(c) Re z > 2 (d) Re z > 3

Solution:

If z = x + iy, then |z – 4| > |z – 2|

2 22 2x 4 y x 2 y

|x – 4| > |x – 2|

x < 3 satisfies the above inequality.

Hence, answer is (a).

Class Exercise - 4

For x1, x2 , y1, y2 R, if 0< x1 < x2, y1 = y2andz1 = x1 + iy1, z2 = x2 + iy2

and z3 = , then

z1, z2 and z3 satisfy

(a) |z1| < |z3| < |z2| (b) |z1| > |z3| > |z2| (c) |z1| < |z2| < |z3| (d) |z1| = |z2| = |z3|

1 2z z

2

Solution

y1 = y2 = y (Say)

2 21 1z x y

2 22 2z x y

1 23

x xz iy

2

221 2

3x x

z y2

1 21 2

x xx x

2

(As arithmetic mean of numbers)

= |z1| < |z3| < |z2|

Hence, answer is (a).

Class Exercise - 5

If

then value of z13 + z2

3 – 3z1z2 is

(a) 1 (b) –1 (c) 3 (d) –3

1 21 3i 1 3i

z and z ,2 2

Solution:

We find z1 + z2 = –1. Therefore,

3 3 3 31 2 1 2 1 2 1 2 1 2z z 3 z z z z 3z z (z z )

3 31 2(z z ) ( 1) 1.

Hence, answer is (b).

Class Exercise - 6

If one root of the equationix2 – 2(1 + i) x + (2 – i) = 0 is2 – i, then the other root is

(a) 2 + i (b) 2 – i (c) i (d) –i

Solution:

Sum of the roots = = –2i + 2 1 i

2 2i 1 ii

One root is 2 – i.

Another root = –2i + 2 – (2 – i) = –2i + 2 – 2 + i = –i

Hence, answer is (d).

Class Exercise - 7

If z = x + iy and w = , then |w| = 1,in the complex plane

(a) z lies on unit circle(b) z lies on imaginary axis (c) z lies on real axis(d) None of these

1 izz i

Solution:

1 izw 1 1

z i

1 iz z i

Putting z = x + iy, we get

1 i x iy x iy i

1 y ix x i (y 1)

Solution contd..

2 2 2 21 y x x (y 1)

1 + y2 + x2 + 2y = x2 + y2 – 2y + 1

4y = 0

y = 0 equation of real axis

Hence, answer is (a).

Class Exercise - 8

The points of z satisfying arg

lies on

(a) an arc of a circle (b) line joining (1, 0), (–1, 0)(c) pair of lines (d) line joining (0, i) , (0, –i)

z 1

z 1 4

Solution:

If we put z = x + iy, we get

x 1 iyz 1

z 1 x 1 iy

By simplifying, we get

2 2

2 2

x 1 y i 2yz 1

z 1 x 1 y

Solution contd..

Equation of a circle.

Note: But all the points put togetherwould form only a part of the circle.

Hence, answer is (a).

Class Exercise - 9

The number of solutions of Z2 + 3 = 0 is

(a) 2 (b) 3 (c) 4 (d) 5

z

Solution:

Let z = x + iy(x + iy)2 + 3(x – iy) = 0x2 – y2 + 2ixy + 3x –3iy = 0x2 – y2 + 3x + i(2xy – 3y) = 0x2 – y2 + 3x = 0, 2xy – 3y = 0Consider y(2x – 3) = 0

Case 1: y = 0, then x2 + 3x = 0, i.e. x = 0 or –3i.e. two solutions given by 0, –3

Solution contd..

Case 2: x = , then – y2 + = 0 3

2

9

4

9

2

i.e. two solutions given by 3 3 3 i

2

So in all four solutions.

Hence, answer is (c).

Class Exercise - 10

Find the square root of –5 + 12i.

Solution:

Let x iy 5 12i

Squaring, x2 – y2 + 2ixy = –5 + 12i

x2 – y2 = –52xy = 12xy = 6, Both x and y are of same sign.

22 2 2 2 2 2x y x y 4x y 25 144 13

2x2 = 8 x = ±2, y = ±3

2 + 3i and –2 – 3i are the values.

Thank you