Complex Number
Study Material:
Definition:
A complex number is a
number which can be expressed in the form 
where 
and 
are real numbers and 
is the solution of the equation 
and 
. is called imaginary number.
In a complex number
, is called real part and is called imaginary part. It is also written
as 
. Here 
and 
.
Example:
etc.
The set of complex numbers is denoted
by
. Since
any real number can be written as 
so all real numbers are complex numbers. But
all complex are may not be real numbers. As example 
is not a real number.
Graphical
representation:
- Axis
is called real axis and 
-axis is
called imaginary axis. If 
both are positive then the graphical
representation of 
is as the figure. So here we can see that lies on real axis and lies on imaginary axis.
Power of:
Since
. So
, 
Continuing this process we
will get
, where
an
integer is.
Complex
conjugate of complex number:
Let 
be a complex
number then its complex conjugate is given by 
Modulus of complex number:
Let then the modulus of 
is denoted by 
and defined by a non-negative real number
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Addition
of two complex numbers:
Two complex are added by
separately adding their real and imaginary parts.
Let and 
. Then
Geometric
interpretation:
The sum of two complex
Numbers 
and 
interpreted as point obtain by Building a
parallelogram whose two adjacent sides are and
.
Addition of two complex number
satisfies closer, associative, commutative, existence of identity and existence
of inverse properties.
In general if 
then
Subtraction
of two complex numbers:
Two complex are subtracted by
separately subtracting their real and imaginary parts.
Let and
. Then
Geometric
interpretation:
is the addition
of and 
, whose
geometrical representation is shown in
the figure.
Multiplication
of two complex numbers:
Let and. Then
their multiplication
Geometric
interpretation:
The
geometric interpretation of complex
numbers and is
stretching and rotation of vectors in the plan, shown in the figure.
Division
of two complex numbers:
Let and where
. Then
their division is given by
Example
1: Express 
in
the form of
.
Solution:
.
Example
2: Express 
in the form of.
Solution:
.
Some
important result in complex number: If and be two complex numbers. Then
(i)
(ii) 
provided 
(iii)
(iv)
(v)
provided 
Exercise of NCERT book
Express
each of the complex number given in the exercise 1 to 10 in the form.
1.
Solution:
2.
Solution:
3.
Solution:
.
4.
Solution:
.
5.
Solution:
.
6.
Solution:
7.
Solution:
8.
Solution:
9.
Solution:
10.
Solution:
.
Find
the multiplicative inverse of each of the complex numbers given in the exercise
11 to 13.
11.
Solution:
Let 
Then 
and 
So the multiplicative inverse
of is given by,
.
12.
Solution:
Let 
. Then 
and
So the multiplicative inverse
of is given by,
.
13.
Solution:
Let 
Then 
and
So the multiplicative inverse
of is given by,
.
14.
Express the following expression in the
form
.
Solution:
.
Polar
representation of complex number:
Let P represent a non-zero complex
number
. Let 
be the polar coordinate of the point P. where 
is the distance OP and 
be the angle
which makes OP with the positive -axis.
Then 
and 
i.e., 
.
Then 
and
.
is called modulus of and is denoted by 
and is called argument of and is denoted by
.
Now arg 
, when lies in 1st quadrant
When lies in 2nd quadrant
, when lies in 3rd quadrant
When lies in 4th quadrant.
Exercise of NCERT
Find
the modulus and arguments of each of the complex numbers in exercise 1 to 2.
1.
Solution:
Here 
and 
Then modulus 
And lies in 3rd quadrant. So argument
of is
.
2.
Solution:
Here 
and 
So 
Since lies in 2nd quadrant. So
.
Convert
each of the complex numbers given in the expression 3 to 8 in polar form:
3.
.
Solution:
Let 
. Then 
and 
So 
Since lies in 4th quadrant so
Polar form of is
.
4.
.
Solution:
Let 
. Then and
So 
Since lies in 2nd quadrant so
Polar form of is 
.
5.
.
Solution:
Let 
. Then and
So 
Since lies in 3rd quadrant so
Polar form of is 
.
6.
-3.
Solution:
Let 
then 
and 
So 
Lies in 2nd quadrant.
.
7.
.
Solution:
Here 
. So 
and
.
Here lies in 1st quadrant. So
And
.
So the polar representation
of is
.
8.
.
Solution:
.
Comparing with we have
,
And
Hence 
.
Quadratic equation
The equation of the form
, where
are real and 
is called quadratic equation.
The solution of this equation
is given by
Now if 
then this quadratic equation will have
imaginary roots.
This quadratic equation has at least one root.
Exercise of NCERT
Solve
each of the following equatin:
1.
Solution:
2.
Solution:
3.
Solution:
.
4.
Solution:
.
5.
Solution:
.
6.
Solution:
.
7.
Solution:
.
8.
Solution:
.
9.
Solution:
.
10.
Solution:
.
11.
Convert 
into polar form
Solution:
Let 
.
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So
Since lies in 2nd quadrant so
Polar form of is
.
TRIGNOMETRIC OR POLAR FORM:
Let z = x + iy be a complex
number represented by a point P(x, y) in the Argand plane. Then by geometrical
representation we have
OP = and 
ϴ = arg (z)
P(x, y)
O ϴ
M
In POM, we have
Cos ϴ = 
= 
implies x = cos ϴ
sin ϴ
= 
= 
implies
y = sin ϴ
z = (cos ϴ + i sin ϴ)
z = r
(cos ϴ + i sin ϴ) where = r and ϴ = arg (z)
This
form of z is called the polar form of z.
In
general, z = (cos (2n
+ ϴ) + i sin (2n + ϴ)) where = r and ϴ = arg (z)
In the
graph plotted below, complex number in the polar (or trigonometric) form has
been represented. The complex number z = 2.29 + 1.47i
EQUALITY OF COMPLEX
NUMBERS:
Let us
take 2 complex numbers z₁ and z₂ where
z₁
= a₁ + i b₁
And z₂
= a₂ + i b₂.
Then
z₁ and z₂ are said to be equal iff a₁ = a₂ and b₁ = b₂.
ALGREBRA OF COMPLEX NUMBERS
1. ADDITION OF COMPLEX NUMBERS:
The
addition of 2 complex numbers z₁ = a₁ + ib₁ and z₂ =
a₂ + ib₂
z₁ + z₂ = (a₁ +
ib₁) + (a₂ + ib₂)
= (a₁ + a₂) + i (b₁ +
b₂)
Similarly
we can add z₁ , z₂ ,
z₃ , z₄ , ………. , zn as
z₁ + z₂ + ………. + zn = (a₁ + ib₁) + (a₂ + ib₂)
+ …………… + (an + ibn)
= (a₁ + a₂ +
…………… + an) + i
(b₁ + b₂ + …………… + bn)
Example:
Consider 2 complex numbers z₁ = 1 + 3i and z₂ = 5 – 1i. What will
be z₁ + z₂ =?
Solution:
z₁ + z₂ = (1 + 3i) + (5 – 1i)
= (1 + 5) + i (3
– 1)
= 6 + 2i
The graph plotted below is the Victoria representation of
addition of complex numbers.
2.
SUBTRACTION OF COMPLEX NUMBERS:
The
subtraction of 2 complex numbers z₁ = a₁ + ib₁ and z₂ =
a₂ + ib₂
z₁ - z₂ = (a₁ + ib₁) - (a₂ +
ib₂)
= (a₁
- a₂) + i (b₁ - b₂)
Similarly we can add z₁, z₂,
z₃, z₄… zn
z₁ - z₂ - ………. - Zn = (a₁ +
ib₁) - (a₂ + ib₂) - …………… - (an + ibn)
=
(a₁ - a₂ - …………… - an) + i
(b₁ - b₂ - …………… - bn)
Example:
Subtract z₁ = 1 + 3i from z₂ = 4 + 2i.
Solution:
z₁
= 1 + 3i
z₂ = 4 + 2i
z₂
- z₁ = (4 + 2i) - (1 + 3i)
= (4 – 1) + (2 – 3) i
= 3 – i
In the graph plotted below, it is clear that
negative of z₂ = 1 + 3i is taken.
It is so because we subtracting z₂ from z₁. So, we can write
z₂ - z₁ = (4 +
2i) - (1 + 3i)
= (4 + 2i) + (-1 + (-3) i)
= 3 - i
3.
MULTIPLICATION OF
COMPLEX NUMBERS:
The multiplication of 2 complex numbers z₁ = a₁ + ib₁
and z₂ = a₂ + ib₂
Z₁ x z₂ = (a₁ + ib₁)
x (a₂ + ib₂)
= (a₁a₂ - b₁b₂) +
i (a₁b₂ + a₂b₁)
= (
Re(z₁)Re(z₂) – Im(z₁)Im(z₂) ) + i
(Re(z₁)Im(z₂) + Re(z₂)Im(z₁) )
NOTE:
The product z₁ x z₂
can also be obtained if we actually carryout the multiplication (a₁ +
ib₁) x (a₂ + ib₂) as given below:
(a₁ + ib₁) x (a₂ +
ib₂) = a₁a₂ + ia₁b₂ + ia₂b₁ + 
b₁b₂)
= (a₁a₂ - b₁b₂) + i (a₁b₂
+ a₂b₁)
Example: Multiply z₁ = 3 - i and
z₂ = 2 + 2i.
Solution:
We know that
Z₁
x z₂ = (a₁a₂ - b₁b₂) + i (a₁b₂ +
a₂b₁)
Z₁ = 3 - i = a₁ +
ib₁
Z₂ = 2 + 2i = a₂ +
ib₂
Z₁ x z₂ = 8 + 4i
The graph plotted below is the vectorial
representation of the above example.
4.
DIVISION OF COMPLEX NUMBERS:
The division of a complex number z₁ by a non
- zero complex number z₂ is defined as the multiplication of z₁ by
the inverse of z₂ and is denoted by
.
Thus, 
= z₁ x (
) =
z₁ x ( 
)
The division of 2 complex numbers z₁ =
a₁ + ib₁ and z₂ = a₂ + ib₂
= 
To further simplify it
we will rationalize it by multiplying numerator and denominator by a₂ -
ib₂. By doing this we will get denominator as real number and we can
easily separate the real and imaginary part.
=
= x 
= 
x 
= 
= 
+ i 
The graph plotted below is an example of
vectorial representation of a division of 2 complex numbers z₁ = 24.46 + 8.02i
and z₂ = 2.01 – 3.41i.
CONJUGATE OF A COMPLEX NUMBER
Let z = a +
ib be a complex number. Then the conjugate of z is denoted by 
and is equal to a – ib.
z = a + ib → 
a – ib
It follows
from the definition that conjugate of a complex number is obtained by replacing
i by –i.
The graph
plotted below is the vectorial representation of a conjugate of complex number.
Theorem: Let z₁, z₂ and z₃ be the
complex numbers. Then,
1.
= z
Proof:
Let
z = a + ib
= a – ib
= 
= 
= z
Hence proved.
2. Z + = 2 Re (z)
Proof:
Let z = a + ib
= a – ib
z + = (a + ib) + (a – ib)
= 2a = 2 Re (z)
Hence proved.
3.
z - = 2 Im ( z )
Proof:
Let z = a + ib
= a – ib
z - = (a + ib) - (a –
ib)
= 2ib = 2 i Im (z)
Hence proved.
4.
If z =
, then z is purely real.
Proof:
Let
z = a + ib
= a – ib
z =
i.e. (a + ib) = (a – ib)
2 x ib = 0
2 x Im (z) = 0
Im (z) = 0
Thus, z is
purely real
Hence proved.
5. z + = 0 implies z is purely imaginary.
Proof:
Let
z = a + ib
= a – ib
z
+ = (a + ib) + (a – ib)
= 2a
= 2 Re (z) = 0
Implies
a = 0
Therefore,
z is purely imaginary.
Hence proved.
6. z = 
Proof:
z = (a + ib) x (a – ib)
= 
=
Hence proved.
MODULUS OF A COMPLEX NUMBER:
The modulus
of a complex number z = a + ib is
denoted by and is defined as
= 
= 
Clearly, 
0 for all z є Ȼ
The graph
plotted below shows the modulus of a complex number.
Theorem: If z₁, z₂,
z₃ є Ȼ, then
i.
= 0
z = 0 i.e. Re( z ) = Im( z ) = 0
Proof:
= 0
= 0
= 0
a = 0 and b = 0
Re
(z) = Im (z) = 0
Hence proved
.
ii.
= 
= 
Proof:
Let z = a +
ib. Then, = a – ib and –z = - a – ib
=
= 
=
= 
=
Hence, = =
Hence proved.
iii.
- 
Re
(z) 
; - Im
(z)
Proof:
- a and
- b
- Re
(z) and
- Im
(z)
Hence proved.
iv.
z x = 
Proof:
Let z =
a + ib and 
= a – ib
Then, z x = (a + ib) (a – ib)
= 
+ 
= =
RECIPROCAL OF COMPLEX NUMBER:
Let z = a +
ib be a complex number. Then the reciprocal of z is denoted by 
, where 
.
Multiplying
both numerator and denominator by the conjugate 
of z
= 
x 
= 
= 
= 
- i 
Thus, the multiplicative inverse of a non –
zero complex number z is same as it’s reciprocal and is given by
+ i 
= 
.
SQUARE
ROOTS OF A COMPLEX NUMBER:
Consider a
complex number z = a + ib such that 
= x + iy. Here both x and y are real
numbers. Then,
= x + iy
a + ib =
=(
) + 2 i
xy
Equating
the real and imaginary part
a
= ()
b = 2 xy
Now,
(
) ² = 
+ 4
() ² = + 
() =
On further
solving the equation we can conclude,
= 
{ 
+ i 
} , if Im (z) > 0
= { - i } , if Im (z) < 0
Here is a
vectorial representation of the square root of one of the complex number.
Question: Find
the square root of 7 – 24i.
Solution:
Let 
= x + iy.
7 – 4i =
7 – 4i = () + 2 i
xy
Equating the real and imaginary part
7 = () --------- (1)
-24 = 2 xy
-------- (2)
Now,
(
) ²= + 4
() ²= 49
() = 25 -------- (3)
On solving the values for x and y from the equations (1) and
(2), we get
= 16
implies x = 
4
= 9
implies y = 
--------Since
in (2),
2 xy is
negative, so we will take the values of x and y of opposite sign.
i.e.
either x = 4 and y = -3 or
x = -4 and y = 3
Hence, = 
.