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Mathematics
Session
Complex Numbers
Session Objectives
Session Objective
1. Polar form of a complex number
2. Euler form of a complex number
3. Representation of z1+z2, z1-z2
4. Representation of z1.z2, z1/z2
5. De-Moivre theorem
6. Cube roots of unity with properties
7. Nth root of unity with properties
Representation of complex number in
Polar or Trigonometric form
z = x + iy
z

x
y
x y 
i
2
2
 x2  y2
x

y

2
2
Y
z(x,y)
x2  y2
y

O




x
X
This you
have learnt
in the first
session
Representation of complex number in
Polar or Trigonometric form
x2  y2  r
x
x
  cos  ,
r
x2  y2
y
x2  y2

y
 sin 
r
z = r (cos  + i sin )
where r | z | and   arg  z 
z(x,y)
Y
r
Examples: 1 = cos0 + isin0
-1 = cos  + i sin 
i = cos /2 + i sin /2
-i = cos (-/2) + i sin (-/2)

O x = rcos
y =rsin
X
Eulers form of a complex number
z = x + iy
z = r (cos  + i sin )
z  rei where ei  cos   i sin 
| ei |  cos2   sin2   1
| z | | r || ei |  r
z  r  cos   i sin    rei
Examples:
i0
i
i

2
1  e ,  1  e ,i  e ,  i  e
i

2
Express 1 – i in polar
form, and then in
euler form
i 


 1

1 i  2 

  2  cos 4  isin 4 
2


 2

i

 
  
 2  cos     i sin      2e 4
 4
 4 

Properties of eulers form ei
ei  cos   i sin 
| ei |  cos2   sin2   1
z | z |ei arg z remember this
 cos a  i sin a  cosb  i sinb   eiaeib
i a b
 e    cos  a  b   i sin  a  b 
eia  eib   cos a  i sin a   cosb  i sinb 
ab
ab
ab
 2 cos 
cos

i
sin

2
2 
 2 
a b
ab i 2
 2 cos 
e

 2 
ab i
e  e  2i sin 
e
2


ia
ib
a b
2
Illustrative Problem
The value of ii is ____
a) 2
b) e-/2
c) 
d) 2
Solution:
i = cos(/2) + i sin(/2) = ei/2
ii = (ei/2)i = e-/2
Now find  i
Ans :  i

i i
Illustrative Problem
Find the value of loge(-1).
Solution:
-1 = cos  + i sin  = ei
loge(-1) = logeei = i
General value: i(2n+1), nZ
As cos(2n+1) + isin(2n+1) = -1
Illustrative Problem
If z and w are two non zero complex
numbers such that |zw| = 1, and
Arg(z) – Arg(w) = /2, then z w is
equal to
a) i
b) –i
c) 1
d) –1
Solution
By eulers form z | z | eiArg(z)
iArg zw 
zw | zw | e
| zw | | z || w |  | z || w |  | zw |  1(given)
 

Arg zw  Arg z  Arg  w    Arg  z   Arg  w   
zw  e
i

2
i

2
Representation of z1+z2
z1 = x1 + iy1, z2 = x2 + iy2
z = z1 + z2 = x1 + x2 + i(y1 + y2)
z2(x2,y2)
z(x1+x2,y1+y2)
Note OAz2  z1Bz
z1(x1,y1)
O
A
B
Oz1 + z1z  Oz
ie |z1| + |z2|  |z1 + z2|
Representation of z1-z2
z1 = x1 + iy1, z2 = x2 + iy2
z = z1 - z2 = x1 - x2 + i(y1 - y2)
z2(x2,y2)
z1(x1,y1)
O
-z2(-x2,-y2)
Oz + z1z  Oz1
ie |z1-z2| + |z2|  |z1|
also |z1-z2| + |z1|  |z2|
 |z1-z2|  ||z1| - |z2||
z(x1-x2,y1-y2)
Representation of z1.z2
z1 = r1ei1, z2 = r2ei2
z = z1.z2 = r1r2ei(1+ 2)
| z1z2 | r1r2 | z1 || z2 |
Arg  z1z2   1  2  Arg  z1   Arg  z2 
Y
r1r2ei(1+ 2)
1+ 2
r2ei2
O
2
1
r1ei1
x
Representation of z1.ei and z1.e-i
z1 = r1ei1
z = z1. ei = r1ei(1+ )
| z |  r1 and Arg  z   1  
Y
r1ei(1+ )

O
1 
r1ei1
r1ei(1- )
What about
z1e-i
x
Representation of z1/z2
z1 = r1ei1, z2 = r2ei2
z = z1/z2 = r1/r2ei(1- 2)
z1 r1 z1
z
 
z 2 r2 z 2
 z1 
Arg  z   Arg    1  2  Arg  z1   Arg  z 2 
 z2 
2
r1ei1
1
r2ei2
r1/r2ei(1- 2)
1- 2
Illustrative Problem
If z1 and z2 be two roots of the
equation z2 + az + b = 0, z being
complex. Further, assume that the
origin, z1 and z2 form an
equilateral triangle, then
a) a2 = 3b
b) a2 = 4b
c) a2 = b
d) a2 = 2b
Solution
z2 = z1ei /3
/3
z1

Sum z1  z2  z1  z1ei / 3  z1 ei0  ei / 3
 i 6
 2z1 cos e  a
6
i
Pr oduct z1z2  z12e

3
.......I
b
i
Squaring I we get 3z12e

3
 a2
Hence a2 = 3b

De Moivre’s Theorem
1) n  Z,
 cos   i sin  
n
 cosn  i sinn
 cos   i sin    cosn  i sinn
n
 cos a  i sin a
m
 cos a  i sin a
n
 cos b  i sinb 
m
 cosb  i sinb   cos ma  nb   i sin ma  nb 
n
 cos ma  nb   i sin ma  nb 
De Moivre’s Theorem
2) n  Q,
cos n + i sin n is one of the
values of (cos  + i sin )n
n  p /q
 cos   i sin  
p/q
  cosp  i sinp 
1/q
  cos  p  2k   i sin  p  2k  
1/ q
2k  p
2k  p
 cos
 i sin
,k  1,2,..., q  1
q
q
Particular case
 cos   i sin  
1/n
 cos
2k  
2k  
 i sin
,k  0,1,2,...,n  1
n
n
Illustrative Problem
7
2
2
 cos5  i sin 5   cos   i sin  
7
7 

Simplify
3
1
2
2


 cos 4  i sin 4  4  cos   i sin  
3
3 

2
5
Solution:
 cos   i sin  
2
5 5

 cos   i sin 
  cos   i sin  
 cos3  isin3
4

7
2


7
cos


i
sin







1
2 3


4
3
cos


i
sin







2  2 1 2
  cos   i sin  
3
Illustrative Problem
Pr ove that : 1  i   1  i   2
n
n
n
1
2
 n 
cos  
 4 
Solution:



1  i  2  cos  i sin 
4
4




1  i  2  cos  isin 
4
4

1  i 
n
1  i 
n
n
n 

 2  cos
 i sin 
4
4 

n
2
 1  i 
n
1  i 
n
n
n 

 2  cos
 i sin 
4
4 

n
1
n
n
2
 2 2cos
 2 cos
4
4
n
2
n
2
Cube roots of unity
1
3
x  1 or x3  1


x3  1  0 or  x  1 x2  x  1  0
1  i 3 1  i 3
,
2
2
2
or x  1, ,  or 1, 2 , 
 x  1,
2
i
1  i 3 1
3
2
2


i
 cos
 i sin
e3
2
2
2
3
3
2
i
1  i 3 1
3
2
2
 

i
 cos
 i sin
e 3
2
2
2
3
3
2
Find using (cos0 + isin0)1/3
2
 1  i 3   1  i 3 
Note : 
 

2
2

 

2
 1  i 3   1  i 3 
and 
 

2
2

 

Properties of cube roots of unity
1    2  0
1..2  3  1
|  | | 2 |  1
1
2
2


,



,


2

4 2
3 3

1,, 2 are the vertices of
equilateral triangle and lie on unit
circle |z| = 1
Why
so?
O
2
1
Illustrative Problem
If  is a complex number such that
2++1 = 0, then 31 is
a) 1
b) 0
c) 2
d) 
Solution
2    1  0

1  i 3
  or 2
2
let   , 31  31    
or    ,    
2
31

2 31
 62  2  
Nth roots of unity
1
n
x  1   cos 0  i sin0 
1
n
2k
2k
 cos
 isin
,k  0,1,...,n  1
n
n
k  0, x  1  ei0
2
i
2
2
k  1, x  cos
 i sin
e n 
n
n
4
i
4
4
k  2, x  cos
 i sin
 e n  2
n
n
.
.
k  n  1, x  cos
2 n  1 
n
 i sin
2 n  1 
n
i
e
2n1
n
  n1
Properties of Nth roots of unity
a) 1      .....  
2
b) 1.. 2...... n1  
n1
n  1

0
 1
n1n
2
  1
n1
1
n is odd
  1 n is even
c) Roots are in G.P
d) Roots are the vertices of n
sided regular polygon lying on
unit circle |z| = 1
3
2
4/n

1 2/n
n-1
Illustrative Problem
Find fourth roots of unity.
Solution:
1
4
x 1
  cos 0  isin 0 
1
4
2k
2k
 isin
, k  0,1, 2,3
4
4
k  0, x  cos 0  isin 0 1
 cos


k  1, x  cos  isin  i
2
2
k  2, x  cos   isin   1
3
3
k  3, x  cos  i sin
 i
2
2
i
-1
1
-i
Illustrative Problem
 1
Find
3
4
in eulers form.
Solution:
 1
3
4
  cos   i sin  
3
4
  cos3  isin 3 
1
4
2k  3
2k  3 

  cos
 i sin
 , k  0,1, 2,3
4
4


e
i
3
4
,e
i
5
4
,e
i
7
4
,e
i
9
4
e
i

4
Class Exercise
Class Exercise - 1
Express each of the following
complex numbers in polar form
and hence in eulers form.
(a)  6  2 i
(b) –3 i
Solution
1.
(a)
r | 6
  tan1
2i | 
2
+
 2 

 6 6
i
7
7 

 6  2i  2 2  cos  isin   2 2 e
6
6

62 2
7
6
(–6, –2 )
 7
=
6
6
Solution Cont.
b) –3i
r  3i  0  9  3
3 

0
2
3
i
3
3 

–3i  3  cos
 isin   3e 2
2
2 

  tan
1
3
2
–3i
Class Exercise - 2
If z1 and z2 are non-zero complex
numbers such that |z1 + z2| = |z1| + |z2|
then arg(z1) – arg (z2) is equal to


(a) –
(b) 
(c) 0
(d)
2
2
Solution
| z  z |  | z |  | z | (triangle inequality)
1 2
1
2
| z |  | z | | z  z |
1
2
1 2
z1 and z2 are in the same line
z1 and z2 have same argument or
their difference is multiple of 2
arg (z1) – arg (z2) = 0 or 2n in general
Class Exercise - 3
Find the area of the triangle on the
argand diagram formed by the
complex numbers z, iz and z + iz.
Solution
We have to find the area of PQR. Note
that OPQR is a square as OP = |z| =
|iz| = OR and all angles are 90°
1
Area of PQR  area of square OPQR
2
1
1
 OP2  | z |2
2
2
Y
Q (Z + iZ)
R (iZ)
P (Z)
O
X
Class Exercise - 4
50
3 i 3 
 325 (x  iy), where
If  2  2 


x and y are real, then the ordered pair
(x, y) is given by ___.
 3 1
a) 
,
 2 2 


 1
3
c) 
,
 2 2 


1 3
b)  ,
 2 2 


 3 1 
d) 
,

 2
2

Solution
50
50
50


3


3

i
3
3

i
50
 325 
  i
  ( 3) 

 2 
2
2
2






50
50






25
3 i
 325 
   3  cos  isin 
 2 2
6
6





 
25
25 



25
25

3
cos
8



isin
8


 3  cos
 isin



3 
3  
3
3 






 


 325 cos  8    isin  8    
3
3 






 325  cos  isin 
3
3

= 325 (x + iy)
x  cos
 1

3 2
y  sin

3

3
2
Class Exercise - 5
If cos   cos   cos   0  sin   sin   sin 
then prove that
cos3  cos3  cos3  3cos(     )
sin3  sin3  sin3  3sin       
Solution
Let x  cos   isin 
y  cos   isin 
z  cos   isin 
x + y + z = (cos   cos   cos  )  i(sin   sin   sin  )
= 0 + i0 = 0
x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = 0
x3 + y3 + z3 = 3xyz
Solution Cont.
x3 + y3 + z3 = 3xyz
 (cos   isin )3  (cos   isin )3  (cos   isin  )3
 3 cos(     )  isin(     )
comparing real part cos3  cos3  cos3   3cos(     )
comparing imaginary part sin3  sin3  sin3   3 sin       
Class Exercise - 6
1
 2cos  and
x
yn
y
If x 
1
 2cos , then
y
xm
is equal to ___.

yn xm
a)2cos     
b)2cos(m  n)
c)2cos(m – n)
d) 0
Solution
x
1
 2cos 
x
x2  2cos  x  1  0
2cos   4cos2   4
x
2

2cos   2isin 
2
Take any one of the values say
x  cos   isin 
 cos   isin 
Solution Cont.
x  cos   isin 
Similarly y  cos   isin 
xm yn (cos   isin )m (cos   isin )n




yn xm (cos   isin )n (cos   isin )m
 cos  m – n   isin m – n 
 cos  n  m   isin  n  m 
 2cos  m  n 
Class Exercise - 7
The value of the expression
1(2  )(2  2 )  ...  (n  1)(n  )(n  2 ),
where  is an imaginary cube root
of unity is ___.
Solution
n
2
 (r  1)(r  )(r   )
r2
n
  (r  1)(r 2  (  2 )r  3 )
r2
n
n 3
2
  (r  1)(r  r  1)   (r  1)
r2
r2
n 3
n
n 3
n
  r   1   r –  1
r 1
r 1
r2
r2

2
n2  n  1
4
n
Class Exercise - 8
If are the cube roots of p, p < 0, then
for any x, y, z, x  y  z is equal to
x  y  z
(a) 1
(b) 
(c) 2 (d) None of these
Solution
1
x  p3 , p  0
1 1
1
 p 3 , p 3 , p 3 2
1
say   p 3 ,   ,    2
x  y  z x  y  z2

x  y  z x  y2  z
 2

  2
2  z
x


y



x  y2  z
Class Exercise - 9
6 
2k
2k 
The value of   sin
– icos

7
7


k 1
(a) –1
(b) 0
(c) i
(d) –i
is
Solution:
6 
2k
2k 
– icos
  sin

7
7


k 1
6 
2k
2k 
 –i   cos
 isin

7
7


k 1
...(i)
roots of x7 – 1 = 0 are
cos
2k
2k
 isin , k = 0, 1, …, 6
7
7
6 
2k
2k 
 isin
0
  cos

7
7 
k  0
Solution Cont.
6 
2k
2k 
 isin
0
  cos

7
7 
k  0
6 
2k
2k 
1    cos
 isin
0

7
7 
k  1
6 
2k
2k 
 isin
 –1 ...(ii)
  cos

7
7 
k  1
From (i) and (ii), we get
6 
2k
2k 
– icos
 (–i)(–1)  i
  sin

7
7 
k  1
Class Exercise - 10
If 1, , 2,..., n  1 are the roots of
the equation xn – 1 = 0, then the
argument of 2 is
2
8
(a)
(b) 4 (c) 6 (d)
n
n
n
n
Solution:
Y
2

2
n
O

2
n
X
1
As nth root of unity are the
vertices of n sided regular
polygon with each side
making an angle of 2/n at
the centre, 2 makes an
angle of 4/n with x axis and
hence, arg(2) = 4/n
Thank you