Introduction:
Complex numbers are used in many branches of science: especially quantum mechanics
(a branch of Physics) heavily depends upon complex numbers.
In mathematics the need of complex numbers is to solve the polynomial which do not have
the solution in the set of real numbers. e.g., The polynomial x2 − 1 = 0 has the solutions
x = 1, which are the real numbers. But the polynomial x2 + 1 = 0 do not have any solution
in the set of real numbers, since there is no real number, whose square is –1. To overcome
this difficulty, we extended the set of real numbers to the set of complex numbers by
introducing a number i such that i 2 = −1 or i = −1. Remember that i 2 = −1 is the Euler’s
notation for the imaginary unit number.
Complex number:
Since, the complex number is any number that can be written as x + yi where x and y are
real numbers. The real number ‘x’ is called the real part and the real number ‘y’ is called
the imaginary part of x + yi .
For example, the complex number −4 + 5i has the real part x = −4 and the imaginary part y = 5 .
Usually, the complex number x + yi is denoted by z.
Accordingly, z1 = x1 + y1i, z2 = x2 + y2i, z3 = x3 + y3i,...
The set of all complex numbers is denoted by C, that is


C = x + yi x, y are real numbers and i = −1
Note:
Complex numbers may also be defined as ordered pairs of real numbers. Thus, a
complex number z is an ordered pair ( x, y ) of real numbers x and y, written as z = ( x, y )
Basic Algebraic operations on Complex numbers
1.
Addition of Two complex numbers.
Let z1 = x1 + iy1 = ( x1 , y1 ) and z2 = x2 + iy2 = ( x2 , y2 ) then their sum is:
z1 + z2 = ( x1 , y1 ) + ( x2 , y2 ) = x1 + iy1 + x2 + iy2
= ( x1 + x2 ) + i ( y1 + y2 ) = ( x1 + x2 , y1 + y2 )
2.
Subtraction of Two complex numbers.
Let z1 = x1 + iy1 = ( x1 , y1 ) and z2 = x2 + iy2 = ( x2 , y2 ) then their difference is:
z1 − z2 = ( x1 , y1 ) − ( x2 , y2 ) = x1 + iy1 − x2 − iy2
= ( x1 − x2 ) + i ( y1 − y2 ) = ( x1 − x2 , y1 − y2 )
3.
Product of Two complex numbers
Let z1 = x1 + iy1 = ( x1 , y1 ) and z2 = x2 + iy2 = ( x2 , y2 ) then their product is:
z1z2 = ( x1 , y1 )( x2 , y2 ) = ( x1 + iy1 )( x2 + iy2 )
= x1 x2 + ix1 y2 + ix2 y1 + i 2 y1 y2 = x1x2 + i ( x1 y2 + x2 y1 ) − y1 y2
= ( x1x2 − y1 y2 ) + i ( x1 y2 + x2 y1 ) = ( x1x2 − y1 y2 , x1 y2 + x2 y1 )
Unit – 1
1
4.
Division of complex numbers
The division of the two complex numbers is not simple. Since the number in the
denominator has two independent parts. To make the denominator a single term we
rationalize (multiply and divide) the given complex number by the conjugate of the
denominator. After rationalization the denominator will be converted into a single real
number and thus division can be done easily.
If z1 = x1 + iy1 = ( x1 , y1 ) and z2 = x2 + iy2 = ( x2 , y2 ) , are any two complex numbers, then
z1 ( x1 , y1 ) x1 + iy1
=
=
z2 ( x2 , y2 ) x2 + iy2
 x + iy1  x2 − iy2  x1 x2 − ix1 y2 + ix2 y1 − i 2 y1 y2
= 1

= 2
x2 − ix2 y2 + ix2 y2 − i 2 y22
 x2 + iy2  x2 − iy2 
x x + i ( x2 y1 − x1 y2 ) + y1 y2 ( x1 x2 + y1 y2 ) + i ( x2 y1 − x1 y2 )
= 1 2
=
x22 + y22
x22 + y22
xx +yy
x y −x y xx + y y x y −x y 
= 1 22 12 2 + i 2 21 12 2 =  1 22 12 2 , 2 21 12 2 
x2 + y2
x2 + y2
x2 + y2 
 x2 + y2
Conjugate of a Complex Number:
The conjugate of the complex number x + yi is x − yi and x − yi is x + yi . We denote the
conjugate of any complex number z as z is obtained by changing the sign of the
imaginary part of z.
Note:
A complex number times its conjugate is always real. i.e., its imaginary part is zero.
Magnitude of Modulus of a Complex Number:
Let z ( x, y ) = x + yi be a complex number. Then absolute value or modulus of z, denoted
by z , is defined by
z = x2 + y 2
Unit – 1
2
1.
2.
3.
EXERCISE 1.1
Simplify and write the complex number as i, − i, − 1 and 1.
(iii) i 9 + i19
(i) −i 40
(ii) i 223
(v) i −1
(vi) ( −1) 2
7
−5
(vii) ( −1) 2
Prove that i107 + i112 + i122 + i153 = 0
Add the following complex numbers.
1 2 1 1
− i, − i
3 3 2 4
4 3  3 4
, 
(iv)  ,
 , 
3 4   4 5
(i) 4 ( 2 + 3i ) , − 3 (1 − 2i )
(iii)
4.
(iv) i 0
(ii)
( 3,1)(1, 3 )
Subtract the following complex numbers.
(i) 2 2 − 5 7i from 5 2 − 9 7i
(iii) ( x, 0 ) from ( 3, − y )
1

 1
(ii)  −7,  from  7, 
3

 3
(iv) 2 x − 3 yi from 4 x − 7 yi
5.
Multiply the following complex numbers:
(i) (8i + 11)( −7 + 5i ) (ii) ( 5i )(1 − 2i )
(iii) ( 9 −12i )(15i + 7 )
6.
Perform the division operation and write the answer in the form a + bi.
4+i
1
6+i
1
(i)
(ii)
(iii)
(iv)
3 + 5i
7 − 3i
−8 + i
i
Prove that the sum as well as product of complex number and its conjugate is a real
number.
Write each expression as a complex number in the form z = a + bi.
2
3
2
(i) (1 − i ) − 2 ( 4 + i )
(ii) (1 − i )
(iii) ( 2i )(8i )
(iv) ( −6i )( −5i )
7.
8.
9.
Find the indicated absolute value of each complex number.
(i) 3 + 4i
(ii) 8 − 5i
10.
If z1 = 3 + 2i and z2 = 4 + 5i, then evaluate.
z1
z2
Simplify and write your answer separately into real and imaginary parts.
(i) z1 + z2
11.
(i)
12.
13.
(1 + 2i )
1 − 3i
(ii) z1 − z2
2
(ii)
(iii) z1 z2
(iv)
1− i
(1 + i )
2
Find the values of “ a ” and “ b ”.
(i) 6a + i ( 4a − b ) = 12 + 3i
(ii) 8a + i ( 5a − b ) = 16 − 7i
(iii) 3a + i ( a − b ) = 6 + 2i
(iv) 4a + i ( 3a − b ) = 8 − 5i
Find the values of “ m ” and “ n ”.
(i) m2 − 2m + 11ni = 10 − n 2i + 14i
(iii) m2 − 4m + 10ni = 8 − n 2i + 6i
(ii) m2 − 9m + 8ni = n 2i + 18i − 7
(iv) m2 − 3m + 7ni = n 2i + 15i − 5
Unit – 1
3
Real and imaginary parts of the Complex Numbers of the Types:
(i)
( x + iy ) ;
(ii)
 x1 + iy1 

 ;
 x2 + iy2 
n = 1, 2
n
n
x2 + iy2  0; n = 1, 2
Properties of Complex Numbers
(i)
Properties of Addition
Addition is commutative i.e. z1 + z2 = z2 + z1
Addition is associative i.e. z1 + ( z2 + z3 ) = ( z1 + z2 ) + z3
Additive identity i.e. 0 + 0i = ( 0,0 )
Additive inverse i.e. if z = a + bi then − z = −a − bi is additive inverse
(ii)
Properties of Multiplication
Multiplication is commutative i.e. z1.z2 = z2 .z1
Multiplication is associative i.e. z1.( z2 .z3 ) = ( z1.z2 ).z3
Multiplication identity i.e. 1 + 0i = (1,0 )
Multiplication inverse i.e. if z = a + bi , z  ( 0,0 ) then z −1 =
a
b
− 2 2 i is
2
a +b a +b
2
Multiplicative inverse
Some properties of the conjugate and modulus of complex numbers
z, z1 , z2  C , then
(a)
z = −z = z = −z
(b)
z=z
(c)
zz = z
(d)
z1 + z2 = z1 + z2
(c)
z1 z2 = z1 z2
(f)
 z1  z1
  = , z2  0
 z2  z2
Unit – 1
2
4
Q.1
(i)
EXERCISE 1.2
Show that for any complex number.
Re ( iz ) = − Im ( z )
Im ( iz ) = Re ( z )
(ii)
Q.2
Use the algebraic properties of complex numbers to prove that
Q.3
Prove that for z  C .
(i)
z is real Iff z = z
(iii)
z is either real or pure imaginary iff ( z ) = ( z )
(iv)
Q.4
Show that z.z is a real number.
If z1 = 2 − 3i and z1 z2 = 16 . Find z2 .
Q.5
If z1 and z 2 are any two complex numbers then prove that
( z1z2 )( z3 z4 ) = ( z1z3 )( z2 z4 ) = z3 ( z1z2 ) z4
z−z
 Im z 
= i

z+z
 Re z 
(ii)
2
2
z1 + z2 − z1 − z2 = 4 Re ( z1 ) Re ( z2 ) + Im ( z1 ) Im ( z2 )
2
2
z1
+  =  + 2 ; where z1 = 3 + i and z2 = 1 + i .
z2
Q.6
Find the value of  ; If
Q.7
Verify that
Q.8
Write the following equations and inequations in terms of x and y by taking
2 z  Re ( z ) + Im ( z )
z = x + yi
(i)
2z − i = 4
(ii)
z −1 = z + i
(iii)
z − 4i + z + 4i = 10
(iv)
1
Re ( i z ) = 4
2
(v)
 z −1 
Im 
 = −5
 2i 
(vi)
−2  Im ( z + i )  3
Q.9
Find real and imaginary parts of following.
(i)
( 2 + 4i )
(iii)
 7 + 2i 


 3−i 
−1
(ii)
( 3 − −4 )
(iv)
 4 + 2i 


 2 + 5i 
−1
−2
−2
Q.10
3 − 7i
 5 − 4i 
(vi)


2 + 5i
 5 + 4i 
For z1 = −3 + 2i and z2 = 1 − 3i verify the following.
(i)
z1 = − z1 = z1 = − z1
(ii)
 z1  z1
 =
 z2  z2
(iii)
z1 z2 = z1 z2
(iv)
z1 + z2 = z1 + z2
(v)
z1 z2 = z1 z2
(vi)
z1 + z2  z1 + z2
2
(v)
Unit – 1
5
Solution of Equation:
Consider the following equation z + mw = n ________(i)
where, m and n are complex numbers. The equation (i) is called a linear equation in two
complex variables (or unknown) z and w.
Note:
•
A system of equations is consistent if it has at least one solution
• A system of equations which has no solution is called inconsistent
Complex Polynomial:
Factorization of Polynomial of the type z 2 + a 2 as a polynomial of linear factor
Factorization of polynomial of the type az 3 + bz 2 + cz + d = 0 where a  0
Solution by completing square method:
EXERCISE 1.3
1.
2.
Solve the simultaneous linear equations with complex co-efficients.
(i)
z + 2w = 4i ; 2z + 5w = 3
(ii)
2z − w = 3i + 5 ; z + 3w = 12 − 4i
(iii)
2z + ( 3 + i ) w = 13i ;
(iv)
3z + w = 3 + 2i ; 5z − w = 11 + 3i
Find solutions to the following equations.
(i) z 2 + z + 3 = 0
3.
(ii) z 2 − 1 = z
(iii) z 2 − 2 z + i = 0
(iv) z 4 + 4 = 0
(iii) ( z − 1) = −1
(iv) z 4 + z 2 + 1 = 0
Find solutions to the following questions.
(i) z 2 = 1
4.
(3 − i ) z − 2w = 2i
(ii) z 3 = 8
3
Find the roots of the polynomial p ( z ) = z 2 − 2 z + 2 and use this to factor the
polynomial. Verify the factorization by expanding it.
5.
Show that 1 + i, 1 − i and 2 are roots of the polynomial p ( z ) = z 3 − 4 z 2 + 6 z − 4 use
this to factor the polynomial z 2 − 8 (1 − i ) z + 63 −16i = 0.
6.
Solve the given quadratic equation and write the solutions in the form z = a + bi.
(i) z 2 + 2 z + 2 = 0
(ii) 6 z 2 − 5 z + 5 = 0
(iii) 2 z 2 + z + 3 = 0
(iv) 3z 2 + 2 z + 4 = 0
7.
Show that each z1 = −1 + i and z2 = −1 − i does not satisfy the equation z + 2 z + 1 = 0
8.
Determine whether 1 + 2i is a solution of z 2 − 2 z + 5 = 0.
2
Unit – 1
6
Argand Plane:
The plane having a complex number assigned to each of its point is called the complex
plane or the argand plane.
Polar representation of a complex number:
Consider a complex number z = x + iy in cartesian form. Then z = r cos + ir sin  = rcis
 y
is called Polar form of the complex number where, r = x 2 + y 2 and  = arg ( z ) = tan −1  
x
Principal Argument:
The principal argument of the complex number lies in the interval −    
Properties of Complex Number in Polar Form:
1
( cos − i sin  )
r
2. If z1 = r1 ( cos1 + i sin 1 ) and z2 = r2 ( cos2 + i sin 2 ) then
1. If z = r ( cos + i sin  ) , then z −1 =
z1 + z2 = ( r1 cos1 + r2 cos2 ) + i ( r1 sin 1 + r2 sin 2 )
3. If z1 = r1 ( cos1 + i sin 1 ) and z2 = r2 ( cos2 + i sin 2 ) then
z1 − z2 = ( r1 cos1 − r2 cos2 ) + i ( r1 sin 1 − r2 sin 2 )
4. If z1 = r1 ( cos1 + i sin 1 ) and z2 = r2 ( cos2 + i sin 2 ) then
z1.z2 = r1r2 cos (1 + 2 ) + i sin (1 + 2 )
5. If z1 = r1 ( cos1 + i sin 1 ) and z2 = r2 ( cos2 + i sin 2 ) then
z1 r1
= cos (1 − 2 ) + i sin (1 − 2 )
z2 r2 
De Moivre’s Theorem:
De Moivre’s Theorem is a fundamental concept in mathematics that offers a method for
raising complex numbers to a power using trigonometry
Such that z n = r n ( cos n + i sin n ) = ( r n , n )
Note:
De Moivre’s Theorem states that this is true for any rational number “n”
Euler’s Theorem:
The Euler form of complex numbers is a way to represent complex numbers using
trigonometric functions. It’s a useful and compact representation that involves the
exponential function and imaginary unit. The Euler form is expressed as: z = r.ei . This
form is closely related to De Moivre’s theorem, for any complex number z = r.ei , raising
it to a power n yields: z n = r n .ein
Application of Complex Numbers in Real World:
Step into the real world of complex numbers, where we’ll see how they help solve practical
problems in areas like electronics, waves, and fluid dynamics. This section connects
classroom theory with real-life applications, showing how complex number are key in
understanding and solving everyday challenges in various fields.
Unit – 1
7
1.
EXERCISE 1.4
Write the number in polar form with argument between − and  .
(i) −3 + 3i
2.
(ii) 2 3 + 2i

i
2
(ii) e

5
2 i
(ii) 7cis
4
(
(iii) 1 − i 3
)
5
(iv) (1 − i )
8
(iii) e

i
3
(iv) e − i
(iii) 8cis

(iv) 10cis0.41
2
Perform the indicated operations and give the results in polar form.
(
(i) (1 + i ) 1 − 3i
(iii) (1 + i )
6.
)
Express each of the following complex numbers in rectangular form.
(i) 3cis
5.
(
20
Write the number in the form of a + bi
(i) e
4.
(iv) 8i
Find the indicated power using De Moivre’s Theorem.
(i) (1 + i )
3.
(iii) 3 + 4i
(ii) 1 − 3i
(
)
)
(ii) − 3 − i ( −1 + i )
4
(iv)
−1 − i
−1 + i
Perform the indicated operations in each express the results in the form a + ib.
6 ( cos 51 + i sin 51 )
1

(i)  4 ( cos 29 + i sin 29 )   ( cos16 + i sin16 )  (ii)
2 ( cos 21 + i sin 21 )
2

(iii)
7.
( cis180 )( 6cis99 )
3 ( cos 39 + i sin 39 )
Calculate the position of a particle from mean position when amplitude is 0.004mm
and angle is
(i)
8.

4
(ii)

3
(iii)

6
When particle is at position of x = 2 + 3i from its mean position and xmax = 1 + 4i is
position at maximum distance from mean position as it can be seen under
microscope at this point
(i)
Calculate the angle at time t = 0 and find the position of the particle.
(ii)
If xmax = 1 + 4i calculate the frequency when t = 2
9.
Find the impedance z for the following values.
(i)
E = ( −50 + 100i ) Volts I = ( −6 − 2i ) amps
(ii)
E = (100 + 10i ) Volts I = ( −8 + 3i ) amps
Unit – 1
8
(i)
REVIEW EXERCISE
If x 2 = −9 then x = ________
(a) 3
(ii)
(b) –3
The real part of complex number z = 7i is:
(a) 0
(iii)
(b) 7
(vi)
(b) 10
2
6+i
(b)
2
6−i
(d) 3
(d) 3
(b) 25
(c) –25
(d) 25i
(b) –1
(c) i
(d) −i
(c) −7 − 4i
(d) 7 + 4i
i = ___________
The conjugate of 7 + 4i = _______
(b) 7 − 4i
If we replace i by −i in z = x + iy then another complex number obtained in known
as:
(a) Prime factor of z
(c) Additive inverse
(b) Reciprocal of z
(d) Complex conjugate of z
If z1 = 3 + i and z2 = 1 + 4i then Re ( z1 − z2 ) = _______
(a) –3
(xii)
1
2
(d) 3 − i
10
(a) −7 + 4i
(xi)
1
2
(c) −3 − i
−1250
= _________
2
(a) 1
(x)
(d) 8
The multiplicative identity of complex number is:
(a) 0
(b) 1
(c) 2
The additive identity of a complex number is:
(a) 0
(b) 1
(c) 2
(a) −25i
(ix)
(c) 20
1
2
(vii)
(viii)
(d) 1
The additive inverse of 3 + i is _________
(a)
(v)
(c) –7
The imaginary part of complex number z = 8 + 10i
(a) 0
(iv)
1
3
(d) − i
(c) 3i
(b) 2
(c) 3
(d) 2
(b) ( x + y )( x − y )
(c) ( x + yi )( x − y )
(d) ( x + y )( x − yi )
(a) A circle
(b) Real axis
(c) Imaginary axis
The conjugate of the complex number sin x − i cos 2 x is
(d) None of these
(b) cos x − i sin 2 x
(d) − sin x + i cos 2 x
x + y = ________
2
2
(a) ( x + yi )( x − yi )
(xiii) If z 2 + 1 = z 2 − 1 then z lies on:
(xiv)
(a) sin x + i cos 2 x
Unit – 1
(c) − sin x − i cos 2 x
9
(xv)
If z = −1 − i, then arg z is
(a)

4
8
(xvi)
 1+ i   1− i 

 +

 2  2

4
3
4
5

(c) −
(b) 4
(c) 2
(d) 8
(c) i
(d) −i
(b)
(d) −
8
(a) 1
1+ i 

 1− i 
(xvii) If n is positive integer then 
(a) 1
4 n +1
=
(b) –1
(xviii) If i = −1, then i + i + i + ... to 1000 terms is equal to
2
2
(a) 1
(xix)
If
(b) –1
5 ( −8 + 6i )
(1 + i )
2
(c) i
(d) 0
(c) ( −15, 20 )
(d) ( −15, −20 )
= a + ib then ( a, b ) equals
(b) ( 20,15)
(a) (15,20 )
2.
3
Solve the following and write the result in standard from.
(i) −8i ( 2i − 7 )
(ii) ( −4 − 8i )( 3 + i )
(v) (8 − 4i )( −3 + 9i )
(vi)
2 + 3i
2+i
(iii) ( 7 − 5i )( −2 − 3i )
(vii)
3 − 4i
4 + 3i
(iv)
2i
1+ i
(viii) −
6i
3 + 2i
3.
Evaluate: x 2 − 2 x + 2 for x = 1 + i .
4.
Evaluate: x 2 − 7 x − 5 for x = 1 − 2i
5.
Represent the following complex numbers on a graph in standard from.



(i) 5  cos + i sin 

6.
9
(ii) 9 ( cos58 + i sin 58 )
9
Perform the operation on the following & represent result in Trigonometric Form.


 

 
(i)  2  cos + i sin   6  cos + i sin  
 
7.
4
4   
12  
12
(ii)
12 ( cos52 + i sin 52 )
3 ( cos110 + i sin110 )
Use DeMoivre’s theorem to find the find indicated power of the complex number
(i) ( 3 − 2i )
Unit – 1
8
 

 
10
(ii)  2  cos
+ i sin
 
5

10  
10