Download 5.1 Fundamental Trig Identities

6.6 DeMoivre’s Theorem
I. Trigonometric Form of Complex
Numbers
A.) The standard form of the complex number z  a  bi
is very similar to the component form of a vector
v  ai  bj. If we look at the trigonometric form of v,
we can see
v  v  cos i  sin  j 
B.) If we graph the complex z = a + bi on the
complex plane, we can see the similarities with the
polar plane.
P (a, b)
z = a + bi
r
b
θ
a
C.) If we let a  r cos and b  r sin  then,
z  a  bi  r cos   r sin   i
where
r  z  a b ,
2
2
b
tan   and i  1
a
D.) Def. – The trigonometric form of a complex number z
is given by z  r cos   i sin 


or
z  rcis
Where r is the MODULUS of z and θ is the ARGUMENT
of z.
E.) Ex.1 - Find the trig form of the following:
1
3
2.) 
i
2 2
1.) 2i
r  0 2 2
2
2
2 
  tan   
0 2
z  0  2i
1
 r (cos   i sin  )



 2  cos  i sin 
2
2

 2cis

2
2
3
1 
r      
  1
2  2 
5
  tan 1  3 
3
1
3
z 
i
2 2
 r (cos   i sin  )
2


5
5 

 1 cos
 i sin

3
3


5
 cis
3
II. Products and Quotients
A.) Let z1  r1  cos. 1  i sin 1  and z2  r2  cos2  i sin 2 
Mult.Div. -
z1  z2  r1  r2  cos 1  2   i sin 1  2  
z1 r1
  cos 1   2   i sin 1   2  
z2 r2
DERIVE THESE!!!!
B.) Ex. 2 – Given z1  5  cos15
.   i sin15  &
z2  2  cos105  i sin105 
z1
find z1  z2 and
z2
z1  z2   5  2   cos 120   i sin 120  
 1
3 
 10   
i 
 2 2 
 5  5 3i
z1 5
  cos 90  i sin  90 
z2 2
5
 0  i
2
5
 i
2
III. Powers of Complex Numbers
A.) DeMoivre’s (di-’mȯi-vərz) Theorem –
If z = r(cosθ + i sinθ) and n is a positive integer, then,
z   r (cos   i sin  )  r n  cos n  i sin n 
n
n
Why??? – Let’s look at z2-
z  zz
2
z 2  r  cos  i sin    r  cos  i sin  
z 2  r 2  cos  i sin   cos  i sin  
z 2  r 2  cos 2   2i cos  sin   i 2 sin 2  
z 2  r 2  cos 2    1 sin 2    i  2 cos  sin  
z  r  cos 2  i sin 2 
2
2


3
B.) Ex. 3 – Find 1  i 3 by “Foiling”



 1 i 3 1 i 3 1 i 3


 1  2i 3  3 1  i 3


 2  2i 3 1  i 3
 2  6
 8




C.) Ex. 4– Now find 1  i 3
DeMoivre’s Theorem

r2

3
using

3
 

 
z   2  cos  i sin  
3
3 
 
3
3

 
  
z  2  cos 3    i sin 3   
3
 3 

3
3
z  8  cos   i sin  
3
z  8(1  0)  8
3
D.) Ex. 5 –Use DeMoivre’s Theorem to simplify
10
 3
3
i


2 
 2
2
 3
6

r  2 

 =
2
4
 2 
10
 6 
10
10 
z  
 i sin
  cos

4
4 
 2  
10
6 
 5
10
z  10  cos 
2 
 2
5

 5  
  i sin   

 2 
5
6
z10  10  i 
2
IV. nth Roots of Complex Numbers
A.) Roots of Complex Numbers –
v = a + bi is an nth root of z iff vn = z .
If z = 1, then v is an nth ROOT OF UNITY.
B.) If z  r  cos  i sin   , then the n distinct complex
numbers
n
 + 2 k
 + 2 k 

r  cos
 isin

n
n


Where k = 0, 1, 2, …, n-1 are the nth roots of
the complex number z.



th
C.) Ex. 6- Find the 4 roots of z  16cis   .
2





 
2
4
z1  16cis  
4
 
 4

z3  4 16cis  2
 4

 9 
z3  2cis 

8


 
z1  2cis  
8

 2

z2  4 16cis  2
 4

 5 
z2  2cis  
 8 










 6

z4  4 16cis  2
 4

 13 
z4  2cis 

 8 





V. Finding Cube Roots
A.) Ex. 7 - Find the cube roots of -1.
z 3  1
z 1  0
3
( z  1)  z 2  z  1  0
1  1  4(1)(1)
z  1 or z 
2(1)
1
3 1
3
z  1, 
i, 
i
2 2
2 2
Now....
Plot these points on the complex
plane. What do you notice
about them?
z  1  0i
z  1 cos   i sin  


1
3
z1  cos  i sin  
i
3
3 2 2
3
3
z2  cos
 i sin
 1  0i
3
3
5
5 1
3
z3  cos
 i sin
 
i
3
3 2 2
Equidistant from the origin and
equally spaced about the
origin.
1
z1
z2
2
-1
z3
VI. Roots of Unity
A.) Any complex root of the number 1 is also known
as a ROOT OF UNITY.
B.) Ex. 8 - Find the 6 roots of unity.
z  1 0i  cis0
0
z1  cis
6
1
0  2
z2  cis
6
 cis
0  4
z3  cis
6
2
 cis
3

3
1
3
 
i
2 2
1
3
 
i
2 2
0  6
z4  cis
6
 cis
0  8
z5  cis
6
4
 cis
3
0  10
z6  cis
6
5
 cis
3
 1
1
3
 
i
2 2
1
3
 
i
2 2