19. Complex Numbers-1

7/28/2019 19. Complex Numbers-1

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Mathematics


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What is loge(-1) ?

Not Defined

Its a complexnumber

One of its value is i

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

Session opener


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Session Objectives


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Session Objective

1. Complex number - Definition

2. Equality of complex number

3. Algebra of complex number

4. Geometrical representation

5. Conjugate of complex number6. Properties of modules and

arguments

7. Equation involving variables andlocus


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Solve x2

+ 1 = 0

D = 4(


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0

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

i i .i i.i 1

1

2

1 ii i

i i

2

2

1i 1

i

331 1i iii

4

4

1i 1

i

Evaluate:33

17 2ii

3 3316

3

8 8i .i i i 8i

ii

Solution

Ans: 343i

Integral powers of i(iota)


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If p,q,r, s are four consecutiveintegers, 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


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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 previousquestion:

u27 = 0



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z 4 5 4 i 5

If a = 0 ?

If b = 0 ?

If a = 0, b = 0 ?

Complex Numbers - Definition

z = a + i b

Mathematicalnotation

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 aspurely imaginary


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Find x and y if(2x 3iy)(-2+i)2 = 5(1-i)

Hint: simplify and compare realand 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



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(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


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

Properties:

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

1) Closure: z1.z2 is a complex number2) 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


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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 2

2 2

a a b b i(a b a b )

a b

Algebra of Complex Numbers- Division


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

i

Solution:


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


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Representation of complex numbersas 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)


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2 2OP z a b

Modulus of z = a + i b

1 btana

Argument of z

Arg(z) = Amp(z)

Modulus and Argument

O X

YIm (z)

Re (z)a

b

P(z)

|z|

Argument (-, ] is calledprincipal value of argument


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Argument (-, ] is calledprincipal 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


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Step3: Use the adjoining diagramto find out the principal value ofargument

Based on value of and quadrantfrom step 1 and step 2

Principal Value of Argument


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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



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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


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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 3tan2 3

Step1:

Step2: 3rd quadrant ( -,-)

Solution

2

3

Step3:


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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

b

a

P (z)Y

X

_

If z z a ib a ib

b 0, z is puerly real

z lies on x-axis_

What if z z ?


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z za Re(z)2

z zb im(z)

2i

2 2z a ib a b z

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

a

Conjugate of a Complex Number


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1 2 1 2z .z z . z nn

z zz z

2z.z z

11

2 2

zz

z 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 21 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


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

1 1 22

zArg Arg z Arg z

z

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

1

Arg z Arg z , Arg Arg zz

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


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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


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1Conjugate of is2 i

(a) (b)2 i

5

5

2 i

(c) (d)1

2 i

2 i

5


1 2 i 2 iz

2 i 2 i 5

2 i 2 iz ( )

5 5

Solution:


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Find a ib

Let x iy a ib

x2 y2 + 2ixy = a + ib

x2 y2 = a 2xy = b

2

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

Find x2

, take positivevalue of x

Find y2, take value of ywhich satisfies 2xy = b

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

Square root will be x +iy

Square Root of a ComplexNumber

(Squaring)

Other root will be (x+iy)


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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


Solution

2 52x 8 17 x2


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Find 8 15i

5 3

One of the square root i2 2


Solution

25 25 3

For x , y 17 y22 2

15 5 3As xy , for x y -

2 2 2

5x

2

5 3Other square root i

2 2


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CartesianSystem 2D

Argand Diagram

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

Locus of point Locus of complex

no. ( point inargand diagram)

Equation in x,ydefines shapes ascircle , parabola

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

Distance betweenP(x1,y1) andQ(x2,y2) = PQ

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

Equation involving complexvariables and locus


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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


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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 originand radius 5

c) A straight line

d) None of these

z 5argz 5 2



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Let z = x + iy, then

x i y 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


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| 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

If z is a complex no. such

the maximum value of | z 1| is

that | z 4 | 3

Solution


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Class Exercise


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Class Exercise - 1

The modulus and principal argument

of1 i are respectively3

(a) 4 and

3

2(b) 2 and

3

(c) 4 and 3

2(d) 2 and 3

Solution:

The complex number lies in the third quadrant andprincipal argument satisfyingis given by .


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Solution contd..

arg(z) =

1 3tan1 3

2

3 3

is the principal argument.

The modulus is =221 3 2

Hence, answer is (d).


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Solution

2

2a 1

x iy2a i

Taking modulus of both sides,

2

2

a 1 x iy2a i

22

2 2

2

a 1 x y

4a 1

42

2 22

a 1x y

4a 1

Hence, answer is (a).


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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.



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Class Exercise - 4

For x1, x2 , y1, y2R, 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


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Solution

y1 = y2 = y (Say)

2 21 1z x y

2 22 2

z x y

1 23

x xz iy

2

221 2

3

x x

z y2

1 21 2

x xx x

2

(As arithmetic mean of numbers)

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



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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).


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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).


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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 iz

z 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)


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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



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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


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Solution contd..

Equation of a circle.

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



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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 = 0

Consider 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


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Solution contd..

Case 2: x = , then y2 + = 032

94

92

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

2

So in all four solutions.

Hence, answer is (c).


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Thank you

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19. Complex Numbers-1