Download Complex number

Mathematics
Session opener
What is loge(-1) ?
Not Defined
It’s a complex
number
loge(-1) is defined and is complex no.
One of its value is i
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
Complex Numbers Intro
Solve x2 + 1 = 0
D = –4(<0)  No real roots
x   -1
-1  i (known as iota )
“i” is the first
letter of the latin
word ‘imaginarius’
Euler Leonhard ( 1707-1783)
Integral powers of i(iota)
i0  1(as usual)
i1  i
i2   1
i3  i2 .i   i
i4  i3 .i   i.i  1
1 i
i1   2   i
i i
1
i2  2   1
i
1 1
i3  3   i
i
i
1
i4  4  1
i
Evaluate:
3
 17  2 3 
i    

 i  

Solution
3
3
8  8
3
 16
i
.i


i


i

8i



i3   i 

Ans: 343i
Illustrative Problem
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
Given expression = ip(1 + i + i2 + i3)
= ip(1 + i –1 – i) = 0
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
u3 = iu2 + 1 = i(i2+i+1) +1 = i3 + i2 + i + 1
Hence un = in + in-1 + ….. + i + 1
i28  1
27 26
u27  i  i  .....  i  1 
0
i 1
Note by previous
question:
u27 = 0
Complex Numbers - Definition
z=a+ib
Mathematical
notation
re(z)
=a
z  4  5  4  i 5
If a = 0 ?
a,bR
im(z)
=b
Re(z) = 4, Im(z) =  5
 z is purely real
If b = 0 ?  z is purely imaginary
If a = 0, b = 0 ?
z is purely real as well as
purely imaginary
Equality of Complex Numbers
If z1 = a1 + ib1 and z2 = a2 + ib2
z1 = z2 if a1 = a2 and b1 = b2
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
3
5
x  5  i2 5y  2
Illustrative Problem
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 = 5
7
1
x
,y 
10
15
Algebra of Complex Numbers
– Addition
(I) Addition of complex numbers
z1 = a1 + ib1, z2 = a2 + ib2 then
z1 + z2 = a1 + a2 + i(b1 + b2)
Properties:
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
- Subtraction
(II) Subtraction of complex numbers
z1 = a1 + ib1, z2 = a2 + ib2 then
z1 - z2 = a1 - a2 + i(b1 - b2)
Properties:
1) Closure: z1 - z2 is a complex number
Algebra of Complex Numbers
- Multiplication
z1 = a1 + ib1, z2 = a2 + ib2 then
z1 . z2 = a1a2 – b1b2 + i(a1b2 + a2b1)
Properties:
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):
z1 
1
1
a  ib
a  ib

.
 2
(remember)
2
a  ib a  ib a  ib a  b
5) Distributivity:
z1(z2 + z3) = z1z2 + z1z3
(z1 + z2)z3 = z1z3 + z2z3
Algebra of Complex Numbers
- Division
z1 = a1 + ib1, z2 = a2 + ib2 then
z1
a  ib1
a  ib1 a2  ib2
 1
 1
.
z2 a2  ib2 a2  ib2 a2  ib2
a1a2  b1b2  i(a2b1  a1b2 )

a22  b22
Illustrative Problem
If one root of the equation
ix2  2(i  1)x  (2  i)  0
is 2 – i then the other root is
(a) 2 + i
(c) i
(b) 2 – i
(d) -i
Solution:
2(1  i)
Sum of the roots is
  2i  2
i
 (2 – i) +  = -2i +2
  = -2i +2-2 + i = -i
Geometrical Representation
Representation of complex numbers
as points on x-y plane is called
Argand Diagram.
Im (z)
Y
P(z)
b
O
Representaion of z = a + ib
a
X
Re (z)
Modulus and Argument
Modulus of z = a + i b
OP  z  a2  b2
Argument of z
Im (z)
Y
Arg(z) = Amp(z)
  tan1
b
a
Argument  (-, ] is called
principal value of argument
P(z)
|z|
b

O
a
X
Re (z)
Principal Value of Argument
Argument  (-, ] is called
principal value of argument
1
Step1 : Find   tan
b
a
 
for   0, 
 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
Illustrative Problem
The complex number which satisfies
the equation z  2 | z  1 | i  0 is
(a) 2 – i
(b) –2 - i
(c) 2 + i
(d) -2 + i
Let z  x  i y
 z  2 | z  1| i  0
 x  i y  2 (x  1)2  y2  i  0
 x  2 (x  1)2  y2  i(1  y)  0
 x  2 (x  1)2  y2  0
andy  1  0
Solution Cont.
 x  2 (x  1)2  y2  0
 y1
 x  2 (x  1)2  1  0
 x2  2(x2  2x  2)
 (x  2)2  0
 x  2
and y  1  0
Illustrative Problem – Principal argument
The principal value of
argument in  2  2 3 i
a)

3
b) 

3
c)
Solution


4
3
d)
2
3
Step1:


  tan1
2 3 

2
3
Step2: 3rd quadrant ( -,-)
Step3:
   
2
3
Conjugate of a Complex Number
For z = a + ib,

Conjugate of z is z  a  ib
Image of z on x – axis
For z 
_
3 i _
, z  ?
2
Y
_
z 
3 i
2
If z  z  a  ib  a  ib
 b  0, z is puerly real
_
What if z   z ?
z lies on x
-axis
P (z)

-
a
b
-b
Q(z)
X
Conjugate of a Complex Number
a  Re(z) 
z  z
2
b  im(z) 
z z
2i
z  a  ib 
a2  b2  z
A rg(z)  A rg(a  ib)   tan1
b
  A rg(z)
a
Properties of modulus
z1 .z2  z1 . z2
n
z z
n
zz
z.z  z
2
2
2
Pr oof :z  a  ib, zz  (a  ib)(a  ib)  a  b  z
2
z1
z1

z2 z2
z1  z2  z1  z2
z1  z2  z1  z2
z1  z2  z1  z2
z1  z2  z1  z2
2
2

2
z1  z2  z1  z2  2 z1  z2
2

(Triangle inequality)
Properties of Argument
Arg  z1.z2   Arg  z1   Arg  z2 
 z1 
Arg    Arg  z1   Arg  z2 
 z2 
Arg  z1.z2....zn   Arg  z1   Arg  z2   ....  Arg(zn )

1
Arg z   Arg  z  , Arg     Arg  z 
z
Arg(purely real) = 0 or  or 2n and vice versa



or

or
2n

1


Arg(purely imaginary) =
2
2
2
and vice
versa
Conjugate Properties
zz
Pr oof :z  a  ib, z  a  ib  a  ib  z
z1  z2  z1  z2
z1.z2  z1 .z2
 z1   z1 
     provided z2  0
 z2   z2 
Illustrative Problem
1
Conjugate of
is
2i
(a)
(c)
2i
5
1
2i
(b)
(d)
5
2i
2i
5
Solution:
z 
1 2  i 2  i



2  i 2  i
5
2i
2i
 z (
) 
5
5
Square Root of a Complex
Number
Find
a  ib
Let x  iy  a  ib
x2 – y2 + 2ixy = a + ib
x2 – y2 = a
2
2
 x y 
x
(Squaring)
• Find x2 , take positive
value of x
2xy = b
2
y
2

2
 4x2 y2
Other root will be – (x+iy)
• 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
Illustrative Problem
Find 8  15i
Solution
Let x  iy  a  ib
x2 – y2 + 2ixy = 8 –15i
x2 – y2 = 8
2xy = -15 x,y are of opposite sign
 x2  y2  64  225  17
2
 2x
 8  17  x  
5
2
Illustrative Problem
Find 8  15i
x
Solution
2
25
3
 y2  17  y  
2
2
2
15
5
3
As xy  
, for x 
y2
2
2
For x 
5
5
,
 One of the square root 
5
i
2
5
3
Other square root  
i
2
2
3
2
Equation involving complex
variables and locus
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
Equation in complex variable (z) defines
defines shapes as shapes as circle , parabola etc.
circle , parabola
Distance between Distance between P(z1) and Q(z2) = |z1-z2|
P(x1,y1) and
Q(x2,y2) = PQ
Illustrative Problem
If z is a complex number then
|z+1| = 2|z-1| represents
(a)
Circle
(b)
Hyperbola
(c)
Ellipse
(d)
Straight Line
Solution
Let z  x  iy then
(x  1)2  y2  2 (x  1)2  y2
 3x2  3y2  10x  3  0
Illustrative Problem
 z 5  
If arg 
 , then the locus of z

 z 5 2
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
Solution
Let z = x + iy, then
 x  iy  5  
arg 


 x  iy  5  2
 x2  y2  25  i10y  

arg 
2
2

 2
x

5

y





As argument is
complex number
2
is purely imaginary
 x2 + y2 = 25, circle with center (0,0) and radius 5
Illustrative Problem
If z is a complex no. such that | z  4 | 3
the maximum value of | z  1| is
(a) 2
(b) 6
(c) 0
(d) -6
Solution
| z  1  3| |z  1|  3
Locus of z
 |z  1|  3  | z  1  3|
 |z  1|  3  3
As | z+4|  3
 |z  1|  6
Least value = ?
-7
-1
Class Exercise
Class Exercise - 1
The modulus and principal argument
of –1 – i 3 are respectively

(a) 4 and
3
(c) 4 and –

3
2
(b) 2 and
3
2
(d) 2 and –
3
Solution:
The complex number lies in the third quadrant and
principal argument  satisfying     
is given by  – .
Solution contd..
arg(z) =   tan1
 3


1
3

2
is the principal argument.

   
3
3
2
The modulus is = 1
Hence, answer is (d).

2
3
 2
Class Exercise - 2
a

If
2

1
2
 x  iy, then x2 + y2 is equal to
2a  i
a

(a)

2
1
4
4a  1
a

(c)

1
2

1
4
4a2  1
2
2
a

(b)
2
2
4a  1
(d) None of these
Solution
a
2

2
1
 x  iy
2a  i
Taking modulus of both sides,
a
2

1
2
 x  iy
2a  i
a  1
2
 x y
2
a
2
2


1
2
4a2  1
4
4a2  1
Hence, answer is (a).

x2  y2
Class Exercise - 3
If |z – 4| > |z – 2|, then
(a) Re z < 3
(c) Re z > 2
(b) Re z < 2
(d) Re z > 3
Solution:
If z = x + iy, then |z – 4| > |z – 2|
 x  4   y2 
2
 x  2  y2
2
|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 = y2and
z1 = x1 + iy1, z2 = x2 + iy2
z1  z2
and z3 =
, then
2
z1, z2 and z3 satisfy
(a) |z1| < |z3| < |z2|
(c) |z1| < |z2| < |z3|
(b) |z1| > |z3| > |z2|
(d) |z1| = |z2| = |z3|
Solution
y1 = y2 = y (Say)
z1 
z2 
z3 
z3 
x1 
x12  y2
2
x2

y
2
x1  x2
2
iy
= |z1| < |z3| < |z2|
2
Hence, answer is (a).
 x1  x 2 
2

y


2


x1  x2
2
 x2 (As arithmetic mean of numbers)
Class Exercise - 5
If z1 
1  3i
1  3i
and z2 
,
2
2
then value of z13 + z23 – 3z1z2 is
(a) 1
(b) –1
(c) 3
(d) –3
Solution:
We find z1 + z2 = –1. Therefore,
z13  z23  3 z1z2  z13  z23  3z1z2 (z1  z2 )
(z1  z2 )3  (  1)3   1.
Hence, answer is (b).
Class Exercise - 6
If one root of the equation
ix2 – 2(1 + i) x + (2 – i) = 0 is
2 – i, then the other root is
(a) 2 + i
(b) 2 – i
(c) i
(d) –i
Solution:
Sum of the roots = 2
1  i
i
 2i 1  i = –2i + 2
One root is 2 – i.
Another root = –2i + 2 – (2 – i)= –2i + 2 – 2 + i = –i
Hence, answer is (d).
Class Exercise - 7
1  iz
If z = x + iy and w =
, then |w| = 1,
in the complex plane z  i
(a) z lies
(b) z lies
(c) z lies
(d) None
on unit circle
on imaginary axis
on real axis
of these
Solution:
w 1 
1  iz
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..
1  y 2
 x2 
x2  (y  1)2
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  z  1   


lies on
(a) an arc of a circle
(c) pair of lines
 z  1
4
(b) line joining (1, 0), (–1, 0)
(d) line joining (0, i) , (0, –i)
Solution:
z  1  x  1  iy

If we put z = x + iy, we get
z  1  x  1  iy
z 1

By simplifying, we get
z 1


x 2  1  y 2  i  2y 
 x  12  y 2
Solution contd..
Equation of a circle.
Note: But all the points put together
would form only a part of the circle.
Hence, answer is (a).
Class Exercise - 9
The number of solutions of Z2 + 3 z = 0 is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
Let z = x + iy
(x + iy)2 + 3(x – iy) = 0
x2 – y2 + 2ixy + 3x –3iy = 0
x2 – y2 + 3x + i(2xy – 3y) = 0
x2 – y2 + 3x = 0, 2xy – 3y = 0
Consider y(2x – 3) = 0
Case 1: y = 0, then x2 + 3x = 0, i.e. x = 0 or –3
i.e. two solutions given by 0, –3
Solution contd..
9
9
3
2
Case 2: x = , then – y +
=0
4
2
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 = –5
2xy = 12
xy = 6, Both x and y are of same sign.
x y
2
2


x y
2
2

2
 4x2 y 2 
2x2 = 8  x = ±2, y = ±3
2 + 3i and –2 – 3i are the values.
25  144  13
Thank you