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Complex Numbers
If we wish to work with  1 , we need to extend the set of real
numbers
Definitions
i is a number such that i2 = -1
i 
C is the set of numbers Z, of the form z  a  ib where a and b
are real numbers.
a is called the real part of Z and we write a = R(z) of a = Re(z)
b is called the imaginary part of Z and we write b = i(z) or b = Im(z)
Given z1  a  bi
and z2  c  di
Addition is defined by:
z1  z2  (a  bi)  (c  di)
 (a  c)  (b  d )i
Multiplication is defined by: z1 z2  (a  bi)(c  di)
 ac  adi  bci  bdi 2
 (ac  bd )  (ad  bc)i
We may write a + bi or a + ib, whichever we find more convenient
a) Given z1  3  2i and z2  4  3i find
(i) z1  z2
(ii)
z1  z2
(iii) z1 z2
(i) z1  z2  (3  2i)  (4  3i)
(ii) z1  z2  (3  2i)  (4  3i)
 7  5i
 1  i
(iii) z1 z2  (3  2i)(4  3i)
 12  9i  8i  6i 2
 6  17i
b) Solve the equation z 2  2 z  5  0
Using the quadratic formula
 b  b 2  4ac
z
2a
2  4  20
z
2
2
 16
 
2
2
16 1
 1
2
 1 2i
Page 90 Exercise 1
Questions 1, 2, 3, 6, 7, 8
Complex Conjugate
When z  a  bi , then its complex conjugate is denoted by
z  a  bi
Note:
z z  a b
2
2
This is useful when we wish to carry out a division.
(4  2i )
a) Calculate
(2  3i )
(4  2i) (4  2i) (2  3i)


(2  3i) (2  3i) (2  3i)
8  12i  4i  6i 2

49

14  8i
13
14 8
  i
13 13
b) Calculate
5  12i
Let a  bi  5  12i where a, b  
2
2
2

a

b
 2abi  5  12i
(
a

bi
)

5

12
i
Then
Equating parts we get: 2ab  12
 ab  6
6
a
b
2
2
also a  b  5
2
6
    b 2  5
b
36 2
 2 b  5
b
 36  b 4  5b 2
 b 4  5b 2  36  0
b 4  5b 2  36  0
 (b 2  9)(b 2  4)  0
 b 2  4 or  9
Since b  , b  2
 a  3 or  3
 5  12i  3  2i or  3  2i
Page 91 Ex 2
Questions
1(a), (b), (c), 2(c), (e)
3(a), (b), (f), 5(a), (b)
TJ Exercise 2.
TJ Exercise 1 - if needed.
Argand Diagrams
The complex number z  x  yi is
represented on the plane by the
point P(x,y). The plane is referred
to as “The Complex Plane”, and
diagrams of this sort are called
Argand Diagrams.
y
p
r

y
x
x
Any point on the x-axis represents a purely Real Number
Any point on the y-axis represents a purely imaginary number
z  x  iy is represented by OP
OP is considered a vector which has been rotated of the x-axis.
The length of OP, r, is called the modulus of z denoted z
The size of the rotation is called the amplitude or argument of z.
It is often denoted Arg z.
This angle could be   2n where n is any integer.
We refer to the value of Arg z which lies in the range -< as
the principal argument. It is denoted arg z, lower case ‘a’.
r  x2  y 2
y
  tan 1 ( y x)      
p
r

y
x
x
By simple trigonometry: x  r cos
y  r sin 
Thus x  iy  z can be re written as
z  r (cos  i sin  )
This is referred to as the Polar form of z.
a) Find the modulus and argument of the complex number z  3  4i
z  3 4 5
2
2
Arg z  tan 1 (4 3)  n radians
 0.927  n radians
Since (3,4) lies in the first quadrant, n = 0
arg z  0.927 radiansto 3 significant figures
b) Find the modulus and argument of the complex number z  3  4i
z  3 4 5
2
2
Arg z  tan 1 (4 3)  n radians
 0.927  n radians
Since (-3,-4) lies in the third quadrant, n = -1
arg z  0.927  
 2.21radiansto 3 significant figures
c) Express z  2  2i in the form r (cos  i sin )
0
r  z  2 2
2
2
  arg z  tan 1 ( 2 2)  450
2 2
 z  2 2(cos 45  i sin 45 )
0
0
0
(2,2) is in Q1
Page 94 Exercise 3
Questions
3a, b, d, e, i
6a, b, f
7a, b, c
Loci-Set of points on the complex plane
a) Given z  x  iy , draw the locus of the point
which moves on the complex plane so that
z 4
(i)
(ii)
z 4
z  4  x2  y 2  4
 x 2  y 2  16
y
(i)
(ii)
4
-4
4
-4
This is a circle, centre the origin radius 4
y
4
x
-4
4
-4
x
b) If z  x  iy
(i ) Find the equation of the locus z  2  3
(ii ) Draw the locus on an argand diagram.
(a) z  2  3  x  2  iy  3
 ( x  2) 2  y 2  3
 ( x  2)2  y 2  9
y
This is a circle centre (2, 0) radius 3 units.
-1
5
x
c) if z  x  iy, find the equation of the locus
arg z 
arg z 

3


 tan 1 ( y ) 
x
3
3
y
y
x
x
 tan 
3
 3
 y  3x
This is a straight line through the origin gradient
3
Page 96 Exercise 4
TJ Exercise 7
Questions 1a, b, d, f, j
3a, b,
4a, c
Polar Form and Multiplication
The polar form of z is z  r (cos  i sin  )
Consider z1 z2 where z1  a(cos A  i sin A) and
z2  b(cos B  i sin B)
z1 z2  ab(cos A  i sin A)(cos B  i sin B)
 ab(cos A cos B  i 2 sin A sin B  i cos A sin B  iSinA cos B)
 ab(cos A cos B  sin A sin B  i(sin A cos B  cos A sin B))
 ab(cos( A  B)  i sin( A  B))
Hence, z1 z2  z1  z2
and
Arg ( z1 z2 )  Arg z1  Arg z2
Note arg(z1z2) lies in the range (-, ) and adjustments have to be made
by adding or subtracting 2  as appropriate if Arg(z1z2) goes outside
that range during the calculation.
Note: cos  i sin   cos   i sin  
Also, if z  r (cos  i sin  ) then
1 1
 (cos   i sin   )
z r
1 1
1
So,

and arg   arg z
z
z
z
z1
1
 z1 
z2
z2
z1
Hence
 z1  z2
z2
Arg
z1
 Arg z1  Arg z2
z2
a) Simplify 3(cos
 12(cos

3



2



 i sin )  4(cos  i sin )
3
3
2
2
 i sin


 )
3 2
5
5
 12(cos
 i sin )
6
6
Now turn to page 96 Exercise 5 Questions 1 and 2.
Let us now look at question 3 on page 99.
z  r (cos


 i sin )
3
3
2
2
(a) z  zz  r (cos
 i sin )
3
3
(b) z 3  z 2 z  r 3 (cos   i sin  )
2
2
2
2
(c) z  z z  r (cos
 i sin
)
3
3
4
3
4
This leads to the pattern:
4 2
Because

3
3
If z  r (cos  i sin  )
then z n  r n (cos n  i sin n )
De Moivre’s Theorem
If z  r (cos  i sin  )
then z n  r n (cos n  i sin n )
a) Given z  1  i 3 find
z  1 3
2
(i) z 2
3
tan  
1
 
(ii) z 5

3
(iii) z 7


z  2  cos  i sin 
3
3

 1
3
 4   i   2  i 2 3
2 
 2
2
2
(i ) z  4  cos
 i sin 
3
3 

5
5 

 

5
5
(ii) z  2  cos  i sin   32  cos
 i sin
  16  i16 3
3
3 
3
3 


2
7
7 




(iii ) z  2  cos
 i sin   128  cos  i sin   64  i63 3
3
3 
3
3


7
7
b) Given z  2  i
z  4 1
 5
find z 4 . Round your answer to the nearest integer
1
   0.464 to 3 decimal places
tan  
2
z  5  cos 0.464  i sin 0.464 
z 
4
 5   cos1.856  i sin1.856  25 0.281  0.960i   7  24i
4
Page 101 Exercise 6 questions 1 to 3, 4g, h, i, j.
Roots of a complex number
5
5
Consider z1  cos  i sin
6
6
and z2  cos

6
10
10




2
z1  cos
 i sin
 cos
 i sin
 cos  i sin
6
6
3
3
3
3
z22  cos

3
6
 i sin

 i sin

3
2
It would appear that if z  cos

3
 i sin

3
then z  z 2


5
5
z  cos  i sin and cos  i sin
the solutions being  radians apart.
6
6
6
6
z 2  cos

3
 i sin

3


5
5
z  cos  i sin and cos  i sin
6
6
6
6
y
z2

6
z2
5
6
z1
x
2
radians apart.
The solutions are  radians apart, or think of it as
2
By De Moivre’s theorem, when finding the nth root of a complex
number we are effectively dividing the argument by n. We should
therefore study arguments in the range (-n, n) so that we have all
the solutions in the range (-, ) after division.
If z  r  cos  i sin   then the n solutions of the equation z1n  z are given by
  2 k 
  2 k  



z1  r  cos 
  i sin 
  where k  0,1, 2,...., n  1
 n 
 n 

1
n
The position vectors of the solution will divide the circle of radius r,
centre the origin, into n equal sectors.
a) Solve the equation z 3  4  i 4 3.
4 3  3  

z

8
cos

i
sin
z  16  (16  3)  64  8 arg( z )  tan 



3
3
4
3




3
3
1
 

   2  0  

2

0

 3

3
  

 i sin 
 2  cos  i sin 
For k = 0 z  8  cos 



3
3
9
9
 


 



 
1
3

 




2

1


2

1

1
 3

3
   7
7 
3
 i sin 
 2  cos  i sin 
For k = 1 z  8  cos 



3
3
9
9 
 


 



 
4 3  3  

z  8  cos  i sin 
z  16  (16  3)  64  8 arg( z )  tan 

3
 3
 4  3
3
3
1

 




2

2


2

2

1
 3

3
   13
13 
3
 i sin 
 2  cos
 i sin
For k = 2 z  8  cos 




3
3
9
9

 


 



 
5 
5  



 2  cos 
  i sin 

 9 
 9 

5
5 

 2  cos  i sin 
9
9 


 
7
7  
5
5 

z   cos  i sin  ,  cos
 i sin  ,  cos  i sin  ,
9
9 
9
9  
9
9 

b) Solve the equation z 5  1
z5  1
arg z  0
  2 k 
  2 k  



z  1 cos 
  i sin 
  , for k  0,1, 2,3, 4.
 5 
 5 

k  0, gives  cos0  i sin 0   1
2
2
k  1, gives  cos
 i sin 
5
5 

4
4 

k  2, gives  cos
 i sin 
5
5 

4
4 

k  3, gives  cos
 i sin 
5
5 

2
2 

k  4, gives  cos
 i sin 
5
5 

Page 106 Exercise 7: Question 2 plus a selection from 1
Polynomials
In 1799 Gauss proved that every polynomial equation with complex
coefficients, f(z) = 0, where z  C, has at least one root in the set of
complex numbers. He later called this theorem the fundamental theorem
of algebra. In this course we restrict ourselves to real coefficients but the
fundamental theorem still applies since real numbers are also complex.
If the root of a ploynomial equation is non-real, ie of the form r  cos  i sin  , 
r sin   0, then its conjugate r  cos  i sin   is also a root.
A polynomial of degree n will have n complex roots.
A polynomial of degree n with real coefficients, can be reduced to a
product of real linear factors and real irreducible quadratic factors which
will have complex solutions.
a) Show that z  1  2i is a root of the equation
f ( z )  z 4  6 z 3  18z 2  30 z  25
b) Hence find all the other roots.
z  1  2i
 z  1 4  5
z  5  cos1.107  i sin1.107 
arg z  1.107radians
We need to find z2, z3 and z4
And substitute them into the
Original equation.
z  5  cos 2.214  i sin 2.214   3  4i
2
2
z  5  cos3.321  i sin 3.321  11  2i
3
3
z  5  cos 4.428  i sin 4.428  7  24i
4
4
 f ( z )   7  24i   6  11  2i   18  3  4i   30 1  2i   25
 7  24i  66  12i  54  72i  30  60i  25
0
Thus z  1  2i is a root.
If z  1  2i is a root, then the conjugate 1  2i is also a root.
Thus  z  1  2i   and  z- 1-2i   are complex factors of z.
Multiplying the complex factors to find the real quadratic
 z  1  2i  z  1  2i 
 z 2  z  2 zi  z  1  2i  2 zi  2i  4i 2
 z2  2z  5
z 2 4z 5
z 2  2 z  5 z 4  6 z 3  18 z 2  30 z  25
z 4  2 z 3  5z 2
Using Division
4z 3 13z 2 30z
4 z 3  8 z 2  20 z
5 z 2  10 z  25
5 z 2  10 z  25
0
Hence the complimentary real factor is z 2  4 z  5
Using
4  16-20
2
 2i
Hence all four roots are:
1  2i, 1  2i, 2  i, 2  i.
Page 108 Exercise 8 Questions 2, 3, 4, 5 and 6
Review on Page 110