Transcript Applications of De Moivre`s theorem

Chapter 40
4/13/2015
By Chtan FYHS-Kulai
1
In mathematics, de
Moivre‘s formula, named
after Abraham de
Moivre.
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The formula is important
because it connects complex
numbers and trigonometry.
The expression "cos x + i sin x"
is sometimes abbreviated to
"cis x".
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3
By expanding the left hand
side and then comparing the
real and imaginary parts
under the assumption that x is
real, it is possible to derive
useful expressions for cos(nx)
and sin(nx) in terms of cos(x)
and sin(x).
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4
Furthermore, one can use a
generalization of this
formula to find explicit
expressions for the n-th
roots of unity, that is,
complex numbers z such
that zn = 1.
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5
De Moivre’s theorem
For all values of n, the value, or
one of the values in the case
where n is fractional, of
n
cos  i sin  
is
cos n  i sin n
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Proofing of
De Moivre’s
Theorem
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Now, let us prove this important
theorem in 3 parts.
1. When n is a positive
integer
2. When n is a negative
integer
3. When n is a fraction
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Case 1 : if n is a positive integer
n2
cos  i sin  
 cos  i sin  cos  i sin  
2
 cos   2i sin  cos  sin 
 cos 2  i sin 2
2
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2
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9
n3
cos  i sin  
2
 cos  i sin  cos  i sin  
 cos  i sin  cos 2  i sin 2 
3
 cos cos 2  i sin  cos 2  i cos sin 2  sin  sin 2
 cos  2   i sin   2 
 cos3  i sin 3
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n4
cos  i sin  
3
 cos  i sin  cos  i sin  
 cos  i sin  cos3  i sin 3 
4
 cos cos3  i sin  cos3  i cos sin 3  sin  sin 3
 cos  3   i sin   3 
 cos 4  i sin 4
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Continuing this process, when n is
a positive integer,
cos  i sin  
n
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 cosn  i sin n
By Chtan FYHS-Kulai
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Case 2 : if n is a negative integer
Let n=-m where m is positive integer
cos  i sin  
m
 cos  i sin  
n
1

m
cos  i sin  
1

cos m  i sin m
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
1
cos m  i sin m 

cosm  i sin m  cosm  i sin m 
cos m  i sin m

2
2
cos m  sin m
 cos m  i sin m
 cos m   i sin  m 
 cos n  i sin n
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Case 3 : if n is a fraction equal
to p/q, p and q are integers
cos  i sin  
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p
q
p
p
 cos   i sin 
q
q
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15
Raising the RHS to power q we have,
q

p
p 
 cos   i sin    cos p  i sin p
q
q 

but,
q is an integer
cos  i sin  
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p
 cos p  i sin p
 p is an integer
By Chtan FYHS-Kulai
16
q

p
p 
p
  cos   i sin    cos   i sin  
q
q 

p

p
p 
 cos   i sin    cos  i sin   q
q
q 

Hence, De Moivre’s Theorem applies when
n is a rational fraction.
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Proofing by
mathematical
induction
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

Let S  n | rcos  i sin    r n cosn  i sin n 
n
1. z  z  r cos  i sin    r cos1  i sin 1 .
1
1
T hus 1 S .
2. Assume z  r cosk  i sin k .
k
k
Now,
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z
k 1
 z z
k
 r cos k  i sin k   r cos  i sin  
k
 r  r cosk     i sin k   
k
r
k 1
 cosk  1  i sin k  1 
Thus k  S  k  1 S .
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The hypothesis of Mathematical
Induction has been satisfied ,
and we can conclude that
S  N.
Hence,
z  r cos  i sin    r cos n  i sin n 
n
n
n
n  N
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e.g. 1
Let z = 1 − i. Find
Soln:
z
10
.
First write z in polar form.
z  1   1
2
2
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arg  z   
Polar form :

4
  
  
z  2  cos    i sin   
 4 
  4
Applying de Moivre’s Theorem gives :
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 
 

 
 


z  2  cos10      i sin 10     
 4 
 4 

 
 10 
 10  
5
 2  cos 
  i sin 
 
 4 
  4 
  5 
 5  
 32 cos 
  i sin 
 
 2 
  2 
10
10
  
  
 32 cos    i sin   
 2 
  2
 32i
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It can be verified directly that 1  i 10  32i
By Chtan FYHS-Kulai
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Properties of
1
z and
z
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If
z  cos   i sin 
1
then  z 1
z
1
 cos  i sin  
 cos    i sin   
 cos  i sin 
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z  cos   i sin 
Hence,
1
 cos   i sin 
z
1
z   2 cos 
z
1
z   2i sin 
z
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Similarly, if
z  cosn  i sin n
n
1
 cos n  i sin n
n
z
Hence,
1
z  n  2 cos n
z
n
1
z  n  2i sin n
z
n
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We have,
1
z   2 cos 
z
Maximum value of cosθ is 1,
minimum value is -1.
Hence, normally
1
2 z  2
z
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What happen, if
the value of
1
z 
z
is more than 2 or less
than -2 ?
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e.g. 2
Given that z  cos   i sin 
1
Prove that z  n  2 cos n
z
n
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e.g. 3
If
1
z   1 , find
z
(i)
(ii)
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1
z  3
z
1
4
z  4
z
3
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Do take note of the
following :
4
1

4
4
4


z


2
cos


2
cos



z

1
z  4  2 cos 4
z
4
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By Chtan FYHS-Kulai
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e.g. 4
4
4
1
1


Expand z   and  z  
z
z


By puttingz  cos  i sin  , deduce that
1
4
4
cos   sin   cos 4  3.
4
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Applications of
De Moivre’s
theorem
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We will consider three
applications of De Moivre’s
Theorem in this chapter.
1. Expansion of cosn , sin n , tann .
2. Values of cos  i sin  
1
.
q
3. Expressions for cos  , sin  in
terms of multiple angles.
n
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n
36
Certain trig identities can be
derived using De Moivre’s theorem.
In particular, expression such as
cos n , sin n , tann
can be expressed in terms of :
cos , sin  , tan
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e.g. 5
Use De Moivre’s Thorem to
find an identity for
cos5 , sin 5
in terms of cos , sin  .
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e.g. 6
Find all complex cube
roots of 27i.
Soln:
We are looking for complex number z with
3
the property z  27i
Strategy : First we write 27i in polar form :-
27i  0  27i  27  27; arg 27i  
2
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
2
39



27i  27 cos  i sin 
2
2

Now suppose z  cos   i sin 
Satisfies z  27i . Then, by De
Moivre’s Theorem,
3



r cos3  i sin 3   27i  27 cos  i sin 
2
2

3
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Thus r  27  r  3
3
cos 3  cos

2
This means :
; sin 3  sin
3 

2

2
 2k
where k is an integer.
Possibilities are : k=0, k=1, k=2
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k  0, 3 


, 
2
6

   3 1  3 3 3

 z1  3 cos  i sin   3
i  
 i
6
6  2
2
2
2


5
k  1, 3   2 ,  
2
6
5
5

 z 2  3 cos  i sin
6
6

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3
1
3 3 3
 
 i   
 i
  3 
2
2
2
  2
By Chtan FYHS-Kulai
42

3
k  2, 3   4 ,  
2
2
3
3 

 z3  3 cos  i sin
  30  i   3i
2
2 

z1
z2
z3
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In general : to find the complex
nth roots of a non-zero complex
number z.
1. Write z in polar form :
z  r (cos  i sin  )
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2. z will have n different nth roots
(i.e. 3 cube roots, 4 fourth roots, etc.)
3. All these roots will have the same
1 the positive real nth
modulus
rn
roots of r) .
4. They will have different arguments
:    2   2  2    n  1 2 
n
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,
n
,
n
,,
By Chtan FYHS-Kulai
n
45
5. The complex nth roots of z are
given (in polar form) by

 
 
z1  r  cos   i sin  
n
 n 

1
n

   2 
   2
z2  r  cos
  i sin
 n 
 n


 


   4 
   4
z2  r  cos
  i sin
 n 
 n


 

1
n
1
n
…etc
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e.g. 7 Find all the complex
fourth roots of -16.
Soln:
Modulus = 16
Argument = ∏
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16  16cos  i sin  
Fourth roots of 16 all have modulus :
1
4
16  2
and possibilities for the arguments
are :
   2   2  2    3  2 
,
,
,
4
4
4
4
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Hence, fourth roots of -16 are :



z1  2 cos  i sin   2  2i
4
4

3
3 

z2  2 cos  i sin    2  2i
4
4 

5
5 

z3  2 cos  i sin    2  2i
4
4 

7
7 

z4  2 cos  i sin
   2  2i
4
4 

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e.g. 8
Given that
1 i
and
z
2
z  z m0
50
25
find the value of m.
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e.g. 9
Solve z  1  0 , hence prove
that

3 1
5
cos
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5
 cos
5
By Chtan FYHS-Kulai

2
51
e.g. 10
Find the cube roots of -1.
show that they can be
2
denoted by  1,  ,  and
prove that
2
   1  0
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e.g. 11
Solve the following equations,
giving any complex roots in the
form r cos  i sin  
(i)
x 1  0
6


 


1
,

cos

i
sin



3
3



(ii) x  4 x  8  0
6
3
 

 
3
3

2
cos

i
sin
,
2
cos

i
sin




12
12 
4
4

 
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 7
 7


 i sin
, 2  cos
12
12





53
e.g. 12
Prove that
1
sin   35 sin   21 sin 3  7 sin 5  sin 7 
64
7
Hence find  35sin   64sin  d
7
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54
e.g. 13
4t  4t
Show that tan 4 
2
4
1  6t  t
3
where t  tan 
Use your result to solve the
equation
t  4t  6t  4t  1  0
4
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3
2
By Chtan FYHS-Kulai
55
e.g. 14
Use De Moivre’s Theorem to
find 32cos6   cos6 d

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56
e.g. 15
1 and  are t wo of t hefift h root sof unit y,
and  has a posit iveacut e argument .
If u     and v    
4
2
3
provet hat u  v  uv  1 and u  v  5.
1
Deduce t hat cos 72 
4
4/13/2015
By Chtan FYHS-Kulai


5 1 .
57
e.g. 16
Solve t heequat ion
x  2 x  4  0.
6
4/13/2015
3
By Chtan FYHS-Kulai
58
e.g. 17
P roveby induct ion t hatif n is a
posit iveint eger:
cos  i sin    cos n  i sin n .
n
n
Evaluat e1  i   1  i  when n  20.
n
4/13/2015
By Chtan FYHS-Kulai
59
e.g. 18
Factorize x  x  1 .
14
4/13/2015
By Chtan FYHS-Kulai
7
60
e.g. 19
6
2
Express sin  cos  in
terms of multiple angles
and hence evaluate

2

sin  cos d
6
2
0
4/13/2015
By Chtan FYHS-Kulai
61
e.g. 20
Express cos 5
in
terms of cos  and
hence evaluate tan 6 in
terms of tan  .
4/13/2015
By Chtan FYHS-Kulai
62
The end
4/13/2015
By Chtan FYHS-Kulai
63