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COMPLEX NUMBER
P.M./Complex/p.1
Define i 2 = -1 .
A complex number z can be expressed in the form z = x + iy where x, y  R and
x - real part of z (Re z) ;
y - imaginary part of z (Im z) :
imaginary axis
i - purely imaginary unit .
The complex number z = x + iy is represented by the point
(x, y) on the complex plane / Gaussian plane /Argand
× (x , y)
diagram.
0
Note :
real axis

1.
For a real number x (x = x + i 0), it is represented by a point lying on the real axis in complex plane.
2.
A complex number z is purely imaginary iff Re z = 0 , and the point representing z lies on the
imaginary axis.
3.
Let P be the point representing the complex number z = x + iy on the complex plane.
y
(i) modulus of z :
imaginary axis
x2 + y2
z =
(ii) argument or amplitude of z :
x (x, y)
z
arg z = the angle  counted from the positive real axis

to the vector OP
0
(anti-clockwise : +ve ; clockwise : -ve)
= tan -1
☼
4.
y
x
Principle range of  is (-π, π].
z = x + iy - standard form
= r (cos  + i sin  ) or r cis 
= r  e i
where r = z
- polar form
- exponential form
where ei = cos  + i sin  - Euler's Formula
ex =
(

xr
x2
x3
 r ! = 1 + x + 2 ! + 3 ! + ...
r=0
 eix = 1 + ix +
∴
(ix ) 2
(ix ) 3
+
+ ...
2!
3!

cos x + i sin x = 1 


x2
x4

+
cos x = 1 2!
4!



x2
x4
+
- ...  + i
2!
4!



- ...  and sin x =



x 


x 


x3
x5
+
- ... 
3!
5!



x3
x5
+
- ...  . )
3!
5!


x
real axis
Example
P.M./Complex/p.2
-1 = _____________________________________________________
ln (-1) = __________________________________________________
∴ ln (-8) = __________________________________________________
Express the following complex numbers in the polar form :
(a)
2i
(b)
(1 + i)i
(c)
1 + i
2 - i
(d) (4 + i)(3 - i)
Solution
(a) ______________________________________________________________
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(b) ______________________________________________________________
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(c) ______________________________________________________________
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(d) ______________________________________________________________
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(e) ______________________________________________________________
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Five Arithmetic Operations
(a) Addition/ Subtraction :
z1 = x1 + i y1
and z 2 = x 2 + i y 2
where x1 , y1 , x 2 , y 2 R
 z1  z2 = (x1  x2 ) + i(y1  y2 )
(b) Multiplication :
z1  z2 = (x1 + iy 1)(x2 + iy 2 )
(x1x2 - y1y2 ) + i(x1y2 + y1x2 )
i
i
z1  z2 = r1e 1  r2e 2
i(  )
= r1r2e 1 2
=

z1z2 = r1r2
and arg z 1z2 = arg z1 + arg z 2
n and arg z n = n arg z
Note : z1n = z1
1
1

z1z2 = z1 z2
for any integer n.
(e)
i
Division :
P.M./Complex/p.3
z1
x + iy 1 x2 - iy 2
= 1

z2
x2 + iy 2 x2 - iy 2

=
(x1x2 + y1y2 ) + i(x2 y1 - y2 x1 )
x22 + y22
=
(x1x2 + y1y2 )
(x y - y2 x1 )
+ i 2 1
x22 + y22
x22 + y22
z1
1
=
(x1x2 + y1y2 )2 + (x2 y1 - y2 x1 )2
2
2
z2
x2 + y2
=
=
1
1
2
z
2
=
(x1x2 )2 + (y1y2 )2 + (x2 y1)2 + (y2 x1)2
x2 2 + y2 2
(x 2 + y 2)(x 2 + y 2 )
2
2
1
1
z1
z2
i 

r e 1
z 

arg  1  = arg  1
z 
i 

 2
r e 2
 2

 r i ( -  ) 
= arg  1 e 1 2  = arg z1 - arg z 2
r

 2

(c) Conjugation :
z = x + iy

conjugate of z , z = x - iy
Properties :
x2 + (-y)2 = z
(i)
z =
(ii)
arg z = tan -1
(iii)
z = (x - iy) = x + iy = z
(iv)
z = 0

-y
= - arg z
x
(-y) = 0  x = 0

x = y = 0
∴
(v)
(vi)
(vii)
z = 0 iff z = 0
 y = -y
z = z  x + iy = x - iy
∴ z is self-conjugate , i.e. z = z iff z is real .
z  z = (x + iy )(x - iy ) = x2 + y2 = z 2
z  z = z  z
1
2
1
2
(viii) z  z = z  z
1 2
1 2
(ix)
(x)
z 
z
 1 = 1
z 
z
 2
2
 z + z = 2 Re z

 z - z = 2i Im z


n
 n
 z1  = z1



y = 0
z +z
1 2
Proof :
2
 z1 + z2
z1 + z2
Triangular Inequality :


= z +z z +z
1 2 1 2
=
P.M./Complex/p.4

z1 + z2  z1 + z2 
= z z +z z +z z +z z
11 1 2
2 1 2 2
∴
2


+ 2 Re z z + z
1 2
2
z
1

z
1
=
z
1
=
z1 2 + 2 z1 z2 + z2 2
2
2
+2 z z + z
1 2
2
 
2
=
( z z = z z )
1 2
1 2
2
+2 z z + z
1 2
2
2
=
 z + z 
 1
2
2
 z1 + z2
z1 + z2
OR
imaginary axis
z1 + z 2
z2
z1
0
z1 - z 2
Claim :
real axis
 z1 + z 2
Proof
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Example
1.
Let z1 , z2 , ... , zn be arbitrary complex numbers .
 z1z2 + z1z2 .
(a)
Prove that z1 z1 + z2 z2
(b)
Using (a), or otherwise, show that
z1
2
+ z2
2
+ ... + z n
2
 Re z1 z 2 + z 2 z 3 + ... + z n-1 z n + z n z1 
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2.
(a)
Suppose u, v are 2 non-zero complex numbers such that u + v + 1 = 0.
Show that u = v = 1 iff
1
1
+
+ 1 = 0.
u
v
Hence, or otherwise, show that u = v = 1 iff u2 + v2 + 1 = 0 .
(b)
Let A, B and C be 3 distinct non-collinear points on the complex plane representing the
complex numbers z1 , z2 , z3 respectively. By using the result of (a) to show that △ABC
is an equilateral △ iff z12 + z2 2 + z32 = z1z3 + z3z2 + z2 z1 .
Solution :
(a)
' Only if '
assume u = v = 1 , then _____________________________________________
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1
1
' If ' assume
+
+ 1 = 0
u
v
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(u + v + 1)2 = 0  _____________________________________________
1
1
+
+ 1 = 0
u
v

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
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
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
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(b)
P.M./Complex/p.6
imaginary axis
z1 = x1 + iy 1 and z 2 = x2 + iy 2
z1 - z2 = (x1 - x2 ) + i(y1 - y2 )
z1 - z2 =
C
(x1 - x2 )2 + (y1 - y2 )2
B
= the distance between the points
A
representing z1 and z 2
3.
0
real axis
∴
A, B and C form an equilateral △
iff
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iff
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iff
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iff
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iff
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In the Argand diagram, PQR is an equilateral △ whose circumcentre is at the origin. If P represents
the complex number 4 + i, find the complex numbers represented by Q and R.
Solution :
imaginary axis
By definition,
∠POQ = ∠QOR = ∠ROP = __________
4 + i = 17 e i
the complex number represented by Q
= ____________________________________________
P(4,1)
Q
0
real
axis
= ____________________________________________
= ____________________________________________
= ____________________________________________
R
the complex number represented by R
= ____________________________________________
= ____________________________________________
= ____________________________________________
Classwork (99 II Q.5)
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Geometrical Applications of Complex Numbers
1. z1 = x1 + iy 1 and z 2 = x2 + iy 2
P.M./Complex/p.7
imaginary axis
 z1 + z2 = (x1 + x2 ) + i(y1 + y2 )
z1 - z2 = z1 + (-z2 )
z2
arg  z2 - z1 
z1
arg ( z2 - z1)
arg z1 - z2 
= the angle the vector from z1 to z 2 makes with
real
axis
+ ve real axis
arg ( z1 - z2 )
= the angle the vector from z2 to z1 makes with
+ ve real axis
Note : (i)
(ii)
2
 -ve
arg ( z1 - z2 ) = 
 +ve
if counted clockwisely
if counted anti-clockwisely
arg ( z1 - z2 ) = arg ( z2 - z1) + 
 = arg (z1 - z2 )
imaginary
A(z1)
axis
 = arg (z3 - z2 )
C(z3)
∠ABC =  - 
B(z 2)
= arg (z1 - z2 ) - arg (z3 - z2 )
z -z 
= arg  1 2 
z - z 
 3 2
Note :
 =
Case 1 :
 = 

z - z 
arg  2 4 
z -z 
 1 4

Case 2 :

z - z 
and  = arg  2 3 
z -z 
 1 3
imaginary
axis
z4
z3
z - z 
z - z 
arg  2 4  - arg  2 3  = 0
z -z 
z -z 
 1 3
 1 4
z - z 
 arg  2 4 
z -z 
 1 4
z - z 
 2 4
z -z 
 1 4
real
axis
0
z -z 
 1 3 =0
z - z 
 2 3
z2
z1
0
real
axis
z -z 
 1 3  is real and positive
z - z 
 2 3
 + (-) = 
z - z 
arg  2 4 
z -z 
 1 4
imaginary
axis
z4
z -z 
 1 3 =π
z - z 
 2 3
z - z  z - z 
4   1 3  is real and negative
  2
 z - z  z - z 
 1 4   2 3
z2
-
z3
z1
0
real
axis
Note : For the distinct points representing z1 , z2 , z3 , z4 to form a cyclic quadrilateral,
z - z  z - z 
 2 4  1 3
 z - z  z - z 
 1 4   2 3
must be real and non-zero.
3.
z is a variable complex number moving on the complex plane in such a way that P.M./Complex/p.8
 z + 1 
.
arg 
=
 z - 1
3
What is the locus of z ?
Solution :
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4.
Circle with centre representing z0 and radius r :
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5.
Perpendicular bisector of two fixed points representing z1 and z 2 :
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6.
Ellipse with two fixed points representing z1 and z 2 as two foci :
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7.
Hyperbola with two fixed points representing z1 and z 2 as two foci :
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Example
1.
Find the locus of the variable point P which represents the complex number z such that
2 z - 2 = z - 6i .
Solution :
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2.
Find the locus of the point P representing the complex number z such that
z = a +
b(1 + it )
1 - it
P.M./Complex/p.9
where a, b are real constants and t is a real variable.
Solution :
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Classwork
3.
(01)
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Transformation
Suppose two complex variables w and z are connected by a functional relation. Then if the point P
representing z describes a curve in an Argand diagram (called the z-plane), the point representing w
will describe a related curve in another Argand diagram (called the w-plane). Such a relation from a
figure on the z-plane to a figure on the w-plane is called a transformation.
Example
1.
Let z be a complex variable with z = 1. If w = 3z +
1
, find the locus of the points representing
z
w.
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P.M./Complex/p.10
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2.
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(90 I Q.10)
1
Let f : C\0  C be defined by f (z) = z +
.
z
Let z = r(cos  + i sin ) and f(z) = u + iv where r > 0 and  , u , v  R,
(a)
express u and v in terms of r and  .
(b)
Find and sketch the image of each of the following circles under f :
(i)
z = 1
;
(ii)
z = a
where 0 < a < 1 .
(a) ________________________________________________________________________
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(b) (i) ____________________________________________________________________
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(ii) ________________________________________________________
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3.
(89 I Q.12)
The function f : C\-1  C\-i is defined by f(z) =
(b)
i(1 - z)
.
1+z
Find and sketch the image, under f, of each of the following :
(i) the upper half of the imaginary axis (including the origin) ;
(ii) the positive real axis .
Solution :
(i) ___________________________________________________________
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P.M./Complex/p.11
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(ii) ___________________________________________________________
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W.34 Q.6, 7
H.W. : W.34 Q.1 – 5
De Moivre's Theorem
If z = r(cos  + i sin  ) where r = z , for any rational number n, then
zn = r n(cos  + i sin  )n = r n(cos n + i sin n ) .
Proof :
Case 1 : Let n be a positive integer.
When n = 1, the statement is obviously true.
Assume zk = r k (cos k + i sin k ) .
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Case 2 : Let n be a positive rational number, i.e. n =
p and q. Let z
p
q
p
q
= r (cos  + i sin  )
p
for some positive relatively prime integers
q
where - <    .
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Case 3 : Let n be a negative rational number, then (-n) is a positive rational number. P.M./Complex/p.12
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∴ The De Moivre's Theorem is true for all rational numbers n.
Classwork
1.
5
3
+ i sin 3  cos  - i sin  
Prove that
= cos 13 - i sin 13  .
5
5 
cos 5 + i sin 5  cos 2 - i sin 2 
cos 3
Solution :
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2.
 1 + sin  + i cos  
Show that 

 1 + sin  - i cos  
n




= cos n -   + i sin n -   .
2

2

Solution :
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Remarks :
zn =
1.
∴
2.
z
n
cos n
zn = z
+ i sin n 
= z
n
cos n + i sin n
n
if 0  z < 1 , then
lim
n
zn =
lim
n
z
n
=0
P.M./Complex/p.13

lim
n
zn = 0
imaginary
axis
x z
x z2
where z n is the distance of the point representing z n
x
from the origin .
n
x z
0
x z3
real
axis
Application of De Moivre's Theorem to Trigonometry
I.
1 - zn
1 + z + z2 + ... + z n -1 =
1- z


if z < 1 , then
zr =
r=0
1
1 - z

(i.e.
 zr
converges if z < 1 )
r=0
Example

 1
To find   
r=0  2 
r
cos r .
Consider
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II. Express cosine or sine of multiple angles in terms of cos  , sin  , i.e.
cos m = f (cos, sin )
sin m  = f (cos, sin )
.
Example
(a)
Prove that cos 5 = 16 cos5  - 20 cos3  + 5 cos  .
2
4
6
8
1
(b)
Hence, deduce that cos
 cos
 cos
 cos
=
5
5
5
5
16
Solution :
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P.M/Complex/p.14
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(b) ________________________________________________________________________
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4
5
2
5
unit
circle
0
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6
5
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8
5
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Classwork
(a) Prove that cos 6 = 32 cos6  - 48 cos4  + 18 cos2 - 1.
(b) Hence, show that the roots of the equation 64x3 - 96x2 + 36x - 3 = 0 are

5
7

5
7
and deduce that sec2
cos2
, cos2
, cos2
+ sec2
+ sec2
= 12 .
18
18
18
18
18
18
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III. Express a power of sin  or cos  in terms of sines or cosines of
multiple angles.
P.M/Complex/p.15
1.
De Moivre's Theorem
2.
Binomial Theorem
3.
Let z = cos  + i sin 
z n = cos n + i sin n
 cos n =
z n + z-n
2

 z =
z = 1
 n
; z -n = z
and
1
z
= cos n - i sin n
sin n  =
z n - z-n
2i
Example
Show that 32 cos6 = cos 6 + 6 cos 4 + 15 cos 2 + 10 .
Solution :
64 cos6 = _______________________________________________________________
= _______________________________________________________________
= _______________________________________________________________
= _______________________________________________________________
= _______________________________________________________________
Advantage : easier to evaluate
e.g. 32

cos 6 d

cosn x dx
= _________________________________________________________
= _________________________________________________________
Classwork
(a) Show that 16 sin 5 = sin 5 - 5 sin 3 + 10 sin  .
(b) Hence, solve the equations
(i)
16 sin 5 = sin 5 ;
1
(ii)
.
cos 5 + 5 cos 3 + 10 cos  =
2
Solution :
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P.M/Complex/p.16
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W.35 Ex.A Q.4
H.W. Ex.A Q.2 , 3 , 5
The n th Roots of Unity
zn - 1 = 0
- a polynomial of degree n , so there should be n roots
1 = cos 2k + i sin 2k
where k is any integer
 2k 
 2k 
Consider z = cos 
+ i sin 

 .
k
 n 
 n 
Raising both sides to power n, we get

 2k 
 2k  
z
= cos 
+ i sin 


k
 n 
 n  

n
n
= cos 2k + i sin 2k = 1
 z k is a solution of z n - 1 = 0 .
Since k can be any integer, there seems to be an infinite number of roots (more than n roots) for the
give equation.
However, if n is used to divide k, then
k = nq + r

for some integer q and r = 0, 1, 2, ... , n-1 .
 2k 
 2k 
 2(nq + r) 
 2(nq + r) 
cos 
 + i sin 

 + i sin 
 = cos 




 n 
 n 
n
n
2r 
2r 


= cos  2q +
 + i sin  2q +



n 
n 
 2r 
 2r 
= cos 
 + i sin 

 n 
 n 
As r = 0, 1, 2, ... , n-1 , it shows that there are only n roots (n
values of zk ) for the given equation.
 2k 
 2k 
+ i sin 
 z = cos 


k
 n 
 n 
or
i
e
2 k
n
z
3
z
2
n
2
2
n
2
n
for k = 0, 1, 2, ... , n-1 .
n th roots of a complex number :
z
2
n
1
z
0
z n-1
P.M./Complex/p.17
The n th roots of a complex number w are the n values of z which satisfy the equation
zn - w = 0
 (*)
Let w = r (cos  + i sin ) .
1



+ i sin  .
Consider z' = r n  cos

n
n

(z' ) n = r cos  + i sin   = w
∴
z' is a solution of the equation (*) .
Let z0 , z1 , ... , zn-1 be the n th roots of 1.
The roots of z n - w = 0 are z' z0 , z' z1 , ... , z' zn-1 .

( z' z k
∴

n
= (z' ) n ( z k ) n = w  1 = w )
The n th roots of w are
2 k
1 
i
i
n
n
r e e n
for k = 0, 1, 2, ... , n-1
  2 k
1
i
n
= r e n
1
= rn

 2k +  
 2k +   
 + i sin 

cos 
 n  
n 

Equation reducible to the form z n - w = 0 :
(z + 1)11 - (z - 1)11 = 0
e.g.
Solution :
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
(z + 1)11 - (z - 1)11 = _______________________________________________________
= _______________________________________________________
= _______________________________________________________
= _______________________________________________________
Example :
Factorize z2n + z2n-1 + ... + z + 1 into real quadratic factors.
P.M/Complex/p.18
Solution :
z2n+1 - 1
z2n + z2n-1 + ... + z + 1 =
z - 1
Suppose z2n + z2n-1 + ... + z + 1 = 0 ,

z2n+1 - 1 = 0 and z  1

z = ___________________________________________________________
z2n + z2n-1 + ... + z + 1
∴
= _______________________________________________________________
= _______________________________________________________________
= _______________________________________________________________
Example :
Let n be a positive integer. By solving the equation x2n + 1 = 0 , show that
x n + x -n =
n 
(2k - 1) 
  x+ x -1 - 2 cos 2n  .
k=1
n 
(2k - 1) 
Hence, deduce that 22n-1   sin 2
 = 1 and
4n 
k=1 
cos n =
n 

(2k - 1) 
 1 - sin 2 2 csc 2 4n  .
k=1
Solution :
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∴ x2n + 1 = ______________________________________________________________
= ______________________________________________________________
= ______________________________________________________________
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P.M/Complex/p.19
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Note :
n th roots of unity :
 2k 
 2k 
 z = cos 
+ i sin 


k
 n 
 n 
zn = 1
i
Let w = e
2
n
or
i
e
2 k
n
for k = 0, 1, 2, ... , n-1
 zk = w k .
 The n th roots of unity are 1 , w, w 2 , ... , w n-1.
For z n - 1 = 0 ,
sum of roots = -
coefficient of z n -1
coefficient of z n
 1 + w + w 2 + ... + w n-1 = 0 .
Example :
x3 - 1 = 0
Let w = e
i
2
3 .
 1 + w + w2 = 0
Example :
n C0
Solution :
i
+ n C3x3 + n C6 x6 + ... = ?
2
3
 1 + w + w2 = 0 and w 3 = 1
Let w = e
(1 + x) n = n C0 + n C1x + n C2 x2 + n C3x3 + n C4 x4 + ...
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Therefore,
nC0 + n C3 + n C6 + ...
=
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=
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=
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=
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=
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Classwork
P.M/Complex/p.20
nC2 + n C5 + n C8 + ... = ?
(1 + x) n = n C0 + n C1x + n C2 x2 + n C3x3 + n C4 x4 + ...
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Therefore,
n C 2 + n C 5 + n C 8 + ...
=
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=
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=
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=
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=
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W.35 Ex.B Q.2, 4, 5
H.W. : Ex.B Q.1, 7, 9
Ex.C Q.2
Primitive Root
An n th root of unity z is said to be primitive if there doesn't exist a positive integer m less than n for
which z m = 1 .
z9 = 1
e.g.
Consider z3 =
2
i
For z3 = e 3
∴

2
i
3
zk =
e
,
i
e
2k
9
and z 5 =
z33 =
ei2 
for k = 0, 1, 2, ... , 8
10 
i
9 .
e
= 1.
There exists a positive integer n less than 9 such that z
Let n be a positive integer such that z5n = 1 

 10 
 10 
cos 
 n + i sin 
 n=1
 9 
 9 

 10 
cos 
 n =1
 9 

 10 

 n = 2k
 9 

5n
= k
9
∴
9 divides n.
∴
9<n
and
e
n
3
 10 
i
n
 9 
=1 .
=1
 10 
sin 
n = 0
 9 
for some integer k
 z5 is a primitive root of unity.
Theorem :
e
i
2k
n is a primitive n th root of 1 if k, n are relatively prime to each other .
P.M./Complex/p.21
 2k
 i
Assume 1 =  e n




cos





2km
= 1
n
m
=
and
2km
is an even integer
n
e
i
2km
2km
n
= cos
n
sin

2km
.
n
+ i sin
2km
= 0
n
km
is also an integer
n
If k, n are relatively prime to each other, then n divides m.
 n  m
Therefore, if k and n are relatively prime to each other, then
W.35 Ex.C Q.1,3 W.36 Q.1 W.36.5 Q.4 , 9 - 11 , 15
H.W. : W.36 Q.2, 3
W.36.5 Q. 5 ,6 , 7 , 12 , 13 , 14 ,17
END
e
i
2k
n is a primitive n th root of 1 .