Download Complex Number 3

EXPANSION FOR COS AND SINE
LOCI IN COMPLEX NUMBER
Expansion of Sin and Cosine
Expand this!
2  3 j
4
Expansion of Sin and Cosine
Theorem 6: (Binomial Theorem)
If n  N then
 a  b   a n  n C1a n 1b  n C2 a n  2b 2  ...  n Cr a n r b r  ...  n Cnb n
n
Expansion of Sin and Cosine
Example 1.20
Expand using binomial theorem, then write in standard form of
complex number:
a)
b)
 2  3 j 3
 cos   j sin  4
Answer:
a )  46  9 j
b)
 cos
4
 
  6 cos 2  sin 2   sin 4   j 4 cos3  sin   4 cos  sin 3 

Expansion of Sin and Cosine
Example 1.19:
State cos 5  in terms of cosines.
De
Moivre’s
Theorem
?
Expansion of Sin and Cosine
Theorem 5:
If z  cos   j sin 
a)
zn 
b)
zn 
1
n
z
1
z
n
, then:
 2 cos n
 2 j sin n
Theorem 6: (Binomial Theorem)
If n  N, then:
a  b n  a n  n C1a n1b  n C2 a n2b 2  ...  n Cr a nr b r  ...  n Cnb n
Expansion of Sin and Cosine
Example 1.19:
State cos 5  in terms of cosines.
Solution:
By applying Theorem 5 and Binomial Theorem
5
1
5

z


2
cos



 
z

1
Theorem 5
Expansion of Sin and Cosine
Expand LHS using Binomial Theorem:
5
1
1

5 5
4 1 
5
3 1 
 z    z  C1 z    C2 z  
z

z
z
3
4
2
5
1
5
2 1 
5
1 1 
5
 C3 z    C 4 z    C5  
z
z
z
1
1 1
5
3
 z  5 z  10 z  10  5 3  5
z
z
z
1
 5 1  3 1

  z  5   5 z  3   10 z  
z  
z 
z


 2 cos 5  52 cos 3   102 cos  
 2 cos 5  10 cos 3  20 cos   2 
Expansion of Sin and Cosine
Expand RHS:
2 cos  
5
 32 cos 5    3
Equate (2) and (3):
32 cos 5   2 cos 5  10 cos 3  20 cos 
1
5
5
5
cos   cos 5  cos 3  cos 
16
16
8
Expansion of Sin and Cosine
Example 1.21:
By using De Moivre’s theorem and Binomial theorem, prove
that:
sin 4  4 cos 3  sin   4 cos  sin 3 
Expansion of Sin and Cosine
Solution:
By applying DMT:
 cos  j sin  4  cos 4  j sin 4 1
Expand LHS using Binomial theorem:
 cos  j sin  4   cos 4   6 cos 2  sin 4   sin 4  

 j 4 cos3  sin   4 cos  sin 3 

 2
Expansion of Sin and Cosine
Equate (1) and (2), then compare:

cos 4  j sin 4  cos 4   6 cos 2  sin 4   sin 4 

 j 4 cos3  sin   4 cos  sin 3 


Real part: cos 4  cos 4   6 cos 2  sin 4   sin 4 
3
3
Imaginary part: sin 4  4 cos  sin   4 cos  sin 
Expansion of Sin and Cosine
Example 1.22:
Using appropriate theorems, state the following in terms of sine
and cosine of multiple angles :
a 
b 
cos 3
sin 5 
Answer:
a 
cos 3  cos 3   3 cos  sin 2 
b 
1
5
5
sin   sin 5  sin 3  sin 
16
16
8
5
Loci in the Complex Number
Since any complex number, z = x+iy correspond to point (x,y) in
complex plane, there are many kinds of regions and geometric
figures in this plane can be represented by complex equations or
inequations.
Definition 1.9
 A locus in a complex plane is the set of points that have
specified property.
 A locus in a complex plane could be a straight line, circle,
ellipse and etc.
Loci in the Complex Number
i) | z − a |= b

Loci in the Complex Number
Example 1.22:
Equation of circle with center at the origin and radius, r
z  z0  r
x
x2  y2  r 2
P on circumference:
z  z0  r
r
P outside circle:
y
O0,0 
P  x, y 
z  z0  r
P inside circle:
z  z0  r
Loci in the Complex Number
Example 1.23:
What is the equation of circle in complex plane with radius 2
and center at 1+j
Solution:
z  1  j   2
Distance from
center to any
point P must be
the same
x
 x  12   y  12  4
r
z 0 1,1
y
P  x, y 
Loci in the Complex Number

Loci in the Complex Number
Example 1.23:
Find the equation of locus if:
z j  z2
Loci in the Complex Number
Solution:
 x  yj   j
x  j  y  1
  x  yj   2
  x  2   jy
x 2   y  1 
2
 x  2 2  y 2
x 2   y  1   x  2   y 2
2
2
x2  y 2  2 y  1  x2  y 2  4x  4
y  2 x 
3
2
* Locus eq : A straight line eq with m = -2
Loci in the Complex Number
x
y  2 x 
3
4
 3
 0, 
 2
0,1
3 
 ,0 
4 
2,0
P  x, y 
Distance from
point (0,-1) and
(2,0) to any point
P must be the
same
y
Loci in the Complex Number
Example 1.24:
Find the equation of locus if:
i)
z2j
1
z 1
ii )
z 3 2 j  5
Loci in the Complex Number
i)
z2j
z2j
1
1
z 1
z 1
z  2 j  z  1  x  yj  2 j  x  yj  1
x   y  2  j   x  1  yj
 x 2   y  2 2   x  12  y 2

 x    y  2
2
2
 
2
 x  1  y 2
2

2
 x 2   y  2 2   x  12  y 2
x2  y 2  4 y  4  x2  2 x  1  y 2
x2  y 2  4 y  4  x2  2 x  1  y 2  0
4 y  2 x  3  0
y
1
3
x .
2
4
Loci in the Complex Number
ii )
z 3 2 j  5
x  yj  3  2 j  5
x  3   y  2 j  5
 x  32   y  2 2  5
 x  3  2   y  2  2  52
* Locus eq : A circle of radius 5 and centre (3,-2) and.
Loci in the Complex Number

Loci in the Complex Number
Interpret the following loci complex plane.
(i) Arg z 

4
(ii) Arg ( z  1) 

4
(i) Arg z 

represents a half ray as shown in the following diagram
4
Note that the point z = 0 is not included as arg z is not
defined for z = 0.
(i) Arg (z  1) 

4