Download Chapter 10 Complex Numbers

Complex Numbers
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
10
Algebra
Chapter 10
Complex Numbers
10.1
Introduction
2
10.2
Geometrical Representation of a Complex Number
4
10.3
Polar Form of a Complex Number
4
10.4
Complex conjugate
8
10.5
Geometrical Applications
11
10.6
Transformation
19
10.7
DeMoivre's Theorem and nth Roots of a Complex Number
22
i  1
2
Prepared by K. F. Ngai
Page 1
Complex Numbers
10.1
A
Advanced Level Pure Mathematics
Introduction
Fundamental Concepts
(1) A complex number z is a number of the form a  bi where a, b are real numbers and i 2  1 .
(2) The set C of all complex numbers is defined by


C a  bi : a, b  R and i 2  1
where
a is called the real part of z and a  Re( z ) and
b is called the imaginary part of z and b  Im( z ) .
(3)
z is said to be purely imaginary if and only if Re( z )  0 and Im( z )  0 .
(4) When Im( z )  0 , the complex number z is real.
i 3  i 2  i  i ,
N.B.
Example
Solution
Solve
i4  i2  i2  1,
i 6  i 4  i 2  1 .
x 2  x  1  0 in terms of i .
B
Operations On Complex Numbers
Let
(1)
(2)
(3)
z1  a  bi and z 2  c  di . Then
z1  z 2  (a  c)  (b  d )i
z1  z 2  (a  c)  (b  d )i
z1 z 2 = (a  bi )(c  di )
= (ac  bd )  (ad  bc)i
(4)
i5  i 4  i  i ,
z1
a  bi c  di ac  bd (bc  ad )

 2
i , where z 2  0 .
=
=
c  di c  di c 2  d 2
c d2
z2
N.B. (i)
1 i
i
 2 
 i ;
i i
1
Prepared by K. F. Ngai
Page 2
Complex Numbers
(ii)
Example
Advanced Level Pure Mathematics
1
1
c  di
1

 2
 2
z2 .
2
z 2 c  di c  d
c d2
If z1  2  3i and z 2  1  4i , find
(a)
z1  2z 2
(b)
z 2  iz1
(c)
z1 z 2
(d)
z1
z2
Solution
Example
Express the following in the form of x  yi , where x, y are real numbers.
(a)
2 z
, where z  1  i
2 z
(b)
1  cos 2  i sin 2
1  cos 2  i sin 2
(c)
1
1  cos 2  i sin 2
Solution
Example
Solution
Find the square roots of the complex number 3  4i and 5  12i .
Prepared by K. F. Ngai
Page 3
Complex Numbers
10.2
Advanced Level Pure Mathematics
Geometrical Representation of a Complex Number
From the definition of complex numbers, a complex number z  a  bi is defined by the two real numbers
a and b . Hence, if we consider the real part a as the x  coordinate in the rectangular coordinates
system and the imaginary part b as the y  coordinate, then the complex number z can be represented
by the point (a, b) on the plane. This plane is called the complex plane or the Argand diagram. On this
plane, real numbers are represented by points on x  axis which is called the real axis; imaginary numbers
are represented by points on the y  axis which is called the imaginary axis. The number 0 is represented
by the origin O.
Any point (a, b) on this plane can be used to represent a complex number z  a  bi .
For example, as shown in Figure, the z1 , z 2 , z 3 represents respectively the complex numbers.
z1  3, z 2  2i, z3  4  3i
10.3
A.
Polar Form of a Complex Number
Polar Form
A complex number z  a  bi can be represented by a vector OP as shown in Figure.
Prepared by K. F. Ngai
Page 4
Complex Numbers
Advanced Level Pure Mathematics
The length of the vector OP , r  OP , is called the modulus of the complex number z , and it is denoted
by z . The angle between the vector OP and the positive real axis is defined to be the argument or
amplitude of z and is denoted by arg z or amp z .
arg z is infinitely many-valued, that is,
arg z    2k , where k  Z .
If arg z lies in the interval       , we call this value the principal value.
Theorem
Example
a  r cos
b  r sin 
r  a2  b2
tan  
b
a
Express the complex number z1  5  12i and z 2  3  3 3i in polar form.
Solution
B.
Use of Polar Form in Multiplication and Division
Theorem
Let z1  r1 (cos1  i sin 1 ) , z 2  r2 (cos 2  i sin  2 )
(1)
z1 z 2  z1 z 2 ,
Or
(2)
z1 z 2  r1r2 cos(1   2 )  i sin( 1   2 )
z1
z1
,

z2
z2
Or
arg( z1 z 2 )  arg z1  arg z 2 .
arg(
z1
)  arg z1  arg z 2
z2
z1 r1
 cos( 1   2 )  i sin(  1   2 )
z 2 r2
proof
Prepared by K. F. Ngai
Page 5
Complex Numbers
Example
Advanced Level Pure Mathematics
Let z1  1  3i and z 2  3  3i .
(a) By expressing z1 and z 2 in polar form, find z1 z 2 and
z1
.
z2
 iz
(b) Find the modulus and the principal value of the argument of  2
 z1



2
Solution
Prepared by K. F. Ngai
Page 6
Complex Numbers
Example
Advanced Level Pure Mathematics
Prove that if z  1 and z  1 , then
1 z
is purely imaginary.
1 z
Solution
Example
Show that z1  z 2 and z1 z 2 are both real , then either z1 and z 2 are both real or z1  z 2 .
Solution
Prepared by K. F. Ngai
Page 7
Complex Numbers
10.4
A.
Advanced Level Pure Mathematics
Complex conjugate
Complex Conjugate
Definition
Let z  a  bi , where a, b  R . The complex conjugate of z , denoted by z is defined as
z  a  bi
Theorem
Properties of Complex Conjugate
Let z be a complex number. Then
(1)
z is real if and only if z  z .
(2)
zz
(3)
(4)
z  z
(5) arg z   arg z ( z  0)
(6)
z  z  2 Re( z )
(7)
zz  z
2
z  z  2i Im( z )
proof
Prepared by K. F. Ngai
Page 8
Complex Numbers
Theorem
Advanced Level Pure Mathematics
Properties of Complex Conjugate
(Continued)
Let z1 and z 2 be two complex numbers. Then
(1)
z1  z 2  z1  z 2
(2)
z1 z 2  z1  z 2
z
(3)  1
 z2
Example
 z1
 
 z2
( z 2  0)
Prove that, for any complex numbers z1 and z 2 .
2
z1  z 2  z1  z 2
2
2
2
 2( z1  z 2 )
Solution
Example
Let z 
1 i
. Find Re( z ) .
2i
Solution
Example
Prove that if z  1 and z  1 , then
1 z
is purely imaginary.
1 z
Solution
Prepared by K. F. Ngai
Page 9
Complex Numbers
Example
Advanced Level Pure Mathematics
Let u, v and w be complex numbers with modulus equal to 1.
Show that if u  v  w  0 , then uv  vw  wu  0 .
Solution
B.
Roots of Polynomials with Real Coefficients Occurs in Conjugate Pairs
Let f ( x)  a n x n  a n 1 x n 1    a1 x  a0  0 ( an  0 ) be a polynomial with real coefficients and
degree n (  2 ). If z  a  bi ( b  0 ) is a root of this polynomial, then z  a  bi is also a root.
Prepared by K. F. Ngai
Page 10
Complex Numbers
10.5
Advanced Level Pure Mathematics
Geometrical Applications
Addition
Subtraction
Scalar multiplication
Vectors
 
 
(ai  bj )  (ci  dj )


= (a  c)i  (b  d ) j
 
 
(ai  bj )  (ci  dj )


= (a  c)i  (b  d ) j
 


 (ai  bj )  ai  bj ,   R
Complex Numbers
(a  bi)  (c  di)
= (a  c)  (b  d )i
(a  bi)  (c  di)
= (a  c)  (b  d )i
 (a  bi )  a  bi,   R
In the set C of all complex numbers, if z  a  bi is regarded as a vector v  ai  bj ; then as far as the
above three operations are concerned complex numbers behave similar to those of vectors.
Geometrical Meaning of the Difference of Two complex Numbers
Suppose the complex numbers representing the points Z and P on the Argand diagram be z  x  yi
and p  a  bi representing the points Z and P on the Argand diagram respectively.
(1) The complex number z  p represents the vector PZ ;
(2) The modulus z  p represents the length of PZ ;
(3) arg( z  p) represents the angle between the vector PZ and the positive x-axis.
Usually, if the point Z on the Argand diagram is represented by the complex number z , we use Z (z ) to
denote it. Therefore, for any four points Z1 ( z1 ) , Z 2 ( z 2 ) , P1 ( p1 ) and P2 ( p2 ) on the Argand diagram, the
angle  between the vectors P1 Z1 and P2 Z 2 , as shown in Figure, is given by

=
=
=
1   2
arg( z1  p1 )  arg( z 2  p2 )
z  p1
arg( 1
)
z 2  p2
Prepared by K. F. Ngai
Page 11
Complex Numbers
Theorem
Advanced Level Pure Mathematics
Angle between Two Line Segments
Let Z1 ( z1 ), Z 2 ( z 2 ), P1 ( p1 ) and P2 ( p2 ) be four points on the Argand diagram. If  is the angle between
the line segments P1 Z1 and P2 Z 2 , then
  arg(
z1  p1
)
z 2  p2
 is considered to be positive if it is obtained by rotating anti-clockwise the vector P2 Z 2 representing the
denominator to the vector P1 Z1 representing the numerator.
Collinear
Theorem
Let P1 , P2 , P3 be three distinct points in the Argand diagram representing respectively the
complex numbers z1 , z 2 , z 3 . Then P1 , P2 , P3 are collinear if and only if
z 3  z1
,
z 2  z1
Where  is a non-zero real number.
Proof
Equation of a Circle
Let A be a point in the complex plane and the complex number corresponding to it be a . The equation of
the circle with A as centre and radius r is given by
z  a  r,
where z is the complex number corresponding to any point P on the circle.
This equation is then rewritten as
za
2
=
r2
( z  a)( z  a) =
r2
z z  az  a z  aa =
r2
Prepared by K. F. Ngai
Page 12
Complex Numbers
Advanced Level Pure Mathematics
z z  az  a z  ( r 2  a a )
This equation is in the form z z  az  a z  k , where k is a real constant.
Example
Find the centre and radius of the circle with equation
z z  (1  2i ) z  (1  2i ) z  4 in the
complex plane.
Solution
Theorem
Given that Z1 ( z1 ), Z 2 ( z 2 ) and P( p) are three points on the Argand diagram. Then
z  p

(1) Z 2 PZ1  arg 1
z

p
 2

(2) The three points Z1 , Z 2 , P are collinear if and only if
(3)
Example
z1  p
is real.
z2  p
Z1 P and Z 2 P are mutually perpendicular if and only if
z1  p
is purely imaginary.
z2  p
1
3
i . If the points A, B and C are
Let z1 be a non-zero complex number and w   
2 2
respectively represented by the complex numbers z1 , z 2  wz1 and z 3  w 2 z1 , show that
ABC is an equilateral triangle.
Solution
Prepared by K. F. Ngai
Page 13
Complex Numbers
Example
Advanced Level Pure Mathematics
Let z1 and z 2 be two non-zero complex numbers. Prove that if z1  z 2  z1  z 2 , then
arg z1  arg z 2  n 

2
, where n is a non-negative integer.
Solution
Prepared by K. F. Ngai
Page 14
Complex Numbers
Example
Advanced Level Pure Mathematics
Suppose the vertices P, Q and R of an equilateral triangle represent the complex numbers
z1 , z 2 and z 3 respectively.
(a) Show that z1 z 2  z 2 z3  z3 z1  z1  z 2  z3 .
2
2
2
(b) If z1 , z 2 and z 3 are the roots of the equation z 3  3 pz 2  3qz  r  0, show that
p2  q .
Solution
Prepared by K. F. Ngai
Page 15
Complex Numbers
Theorem
Advanced Level Pure Mathematics
More Properties on Moduli
Let z1 and z 2 be complex numbers. Then
(1) Re( z)  z , Im( z)  z
(2)
Corollary
z1  z 2  z1  z 2
(Triangle Inequality)
z1  z 2    z n  z1  z 2   z n
Proof
This property can be proved by using mathematical induction on n .
Example
Let z1 and z 2 be complex numbers.
Prove that z1  z 2  z1  z 2
Solution
Loci
When a variable complex number z has to satisfy some specific conditions, there is a set of points in the
Argand diagram representing all the possible values of z . The graph of this set of points is called the locus
of the complex number z .
Example
Interpret the following loci in the Argand diagram.
(a)
z  (a  ib)  r
(b)
z i
1
z 1
(c)
z
2
z i
(a, b, r  R)
Solution
Prepared by K. F. Ngai
Page 16
Complex Numbers
Example
If
Advanced Level Pure Mathematics
z i
is pure imaginary, interpret the locus of z in the Argand diagram.
zi
Solution
Prepared by K. F. Ngai
Page 17
Complex Numbers
Example
Advanced Level Pure Mathematics
Let  be a complex constant and k a real constant.
Show that the equation  z   z  k represents a straight line.
Solution
Prepared by K. F. Ngai
Page 18
Complex Numbers
10.6
Advanced Level Pure Mathematics
Transformation
Translation or Displacement
Definition
Let b be a fixed complex number. The function f ( z )  z  b, z  C is called a
translation.
Example
Given a translation f defined by f ( z )  z  1  2i, z  C.
(a) Plot the point f (2  3i ) .
(b) Sketch the image of the set S  z  C : z  2  1 under f .
Solution
Enlargement
Definition
Example
Let p be a fixed real number, the function f ( z )  pz, z  C is called an enlargement.
1
Given two enlargements defined by f ( z )  2 z and g ( z )   z , z  C .
2
(a) Plot the point f (1  2i ) .
(b) Sketch the image of the (i) ellipse (E ) in Figure A under f .
(ii) triangle (T) in Figure B under g .
Solution
Prepared by K. F. Ngai
Page 19
Complex Numbers
Advanced Level Pure Mathematics
Rotation
Definition
Let  be a fixed real number. The function f ( z )  z (cos   i sin  ), z  C is called a
rotation and  is the angle of rotation.
Example
Given a rotation f ( z )  z (cos 120  i sin 120), z  C.
(a) Plot the point f ( 2  i ) .
(b) Sketch the image of region R in figure under f .
Solution
Example
Let w  3z 
1
. If the locus of z on the z  plane is a unit circle centred at the origin, i.e.
z
z  1, show that the locus of the points represented by w on the w  plane is an ellipse.
Solution
Prepared by K. F. Ngai
Page 20
Complex Numbers
Example
Advanced Level Pure Mathematics
The complex numbers z and w are represented by points P and Q in an Argand
diagram. If z (1  w)  w and P describes the line 2x  1  0 , prove that Q describes a
circle whose centre is at the origin.
Solution
Prepared by K. F. Ngai
Page 21
Complex Numbers
10.7
Advanced Level Pure Mathematics
DeMoivre's Theorem and nth Roots of a Complex Number
For any real number 1 , 2 , (cos1  i sin 1 )(cos  2  i sin  2 )  cos(1   2 )  i sin( 1   2 )
In particular, if   1   2 , we have (cos   i sin  ) 2  cos 2  i sin 2 .
For any positive integer n , by induction on n , the result may be generalized as
(cos   i sin  ) n  cos n  i sin n
 ()
and this is known as the DeMoivre's Theorem for integral index.
For any negative integer n , we may let n  m . Then
(cos  i sin  ) n
=
(cos   i sin  )  m
=
1
(cos   i sin  ) m
=
1
cos m  i sin m
=
cos m  i sin m
cos 2 m  sin 2 m
=
cos( m )  i sin( m )
=
cos n  i sin n
Hence, () also holds for negative integers n .
For any rational number n . Put n 
p
, where p, q  Z and no loss of generality if q is taken as to be
q
positive. Then
(cos n  i sin n ) q =

cos n  i sin n
cos nq  i sin nq
=
cos p  i sin p
=
(cos   i sin  ) p
p
q
=
(cos   i sin  )
=
(cos  i sin  ) n
In general, for any real number  , positive number r and rational number n , we have
Prepared by K. F. Ngai
Page 22
Complex Numbers
Advanced Level Pure Mathematics
[r (cos   i sin  )] n  r n (cos n  i sin n )
The nth roots of a complex number w are the n values of z which satisfy the equation z n  w . If we
write w  cos  i sin  and assuming that the equation is satisfied by z  cos   i sin  , then
1
(cos   i sin  ) n  cos   i sin 
 ()
cos  i sin   (cos   i sin  ) n  cos n  i sin n
By equating the real parts and imaginary parts on both sides, we have
cos  cos n
sin   sin n
n  2k  

2k  
n
 ()
where k  Z .
For k  0,1,2,, n  1 , since

n

2k   2n  2



   2 
n
n
n
n
We obtain the n distinct complex roots for () with the values of  obtained in () .
For k  0 or k  n  1, the root obtained is equal to one of the roots mentioned above. Hence, the equation
() has only n distinct complex roots.
Theorem
DeMoivre's Theorem for Rational Index
Let n be a positive integer and  be a real number. Then
1
n
(cos   i sin  )  cos
2k  
2k  
 i sin
, where k  0,1,2,, n  1
n
n
Using the DeMoivre's Theorem, we will have the following properties.
(i)
z m z n  z m n
(ii)
zm
 z mn
zn
(iii)
z0  1
(iv)
( z m ) n  z mn
Prepared by K. F. Ngai
Page 23
Complex Numbers
Advanced Level Pure Mathematics
Application of DeMoivre's Theorem to Trigonometry
A
Direct application of DeMoivre's Theorem and the binomial theorem, we are able to express
(i)
multiple angles such as sin n and cos n in terms of sin  and cos , and
(ii)
powers of sin  and cos back again into multiple angles.
Example
Verify that cos 3  4 cos 3   3 cos 
< Express cos n , sin n in terms of powers of sin  and cos >
Solution
Example
(a) Show that sin 4  4 sin  (2 cos 3   cos )
(b) Prove
sin 6
 32 cos 5   32 cos 3   6 cos  .
sin 
For what values of  that the result is not true?
Solution
Prepared by K. F. Ngai
Page 24
Complex Numbers
Example
Advanced Level Pure Mathematics
Prove that cos 6  32 cos 6   48 cos 4   18 cos 2   1.
Hence show that the roots of the equation 64 x 3  96 x 2  36 x  3  0 are cos 2 (
and cos 2 (
7
) , and deduce that
18
sec 2 (

18
)  sec 2 (

18
) , cos 2 (
5
)
18
5
7
)  sec 2 ( )  12 .
18
18
Solution
Prepared by K. F. Ngai
Page 25
Complex Numbers
B
Advanced Level Pure Mathematics
If z  cos  i sin  , we have
z  z 1  2 cos ,

z  z 1  2i sin  ,

1
( z  z 1 ) ;
2
1
sin   ( z  z 1 ) .
2i
cos  

As z n  cos n  i sin n and z  n  cos n  i sin n

z n  z  n  2 cos n ,

z n  z  n  2i sin n ,

1 n
( z  z n )
2
1
sin n  ( z n  z  n ) .
2i
cos n 
Example
Solution
Express cos 4  and sin 4  in terms of functions of multiple angles.
Example
Solution
Prove 16 sin 5   sin 5  5 sin 3  10 sin 
Prepared by K. F. Ngai
Page 26
Complex Numbers
Example
Advanced Level Pure Mathematics
(a) Prove that 32 cos 6   cos 6  6 cos 4  15 cos 2  10 .
(b) Prove that 32 cos 4  sin 2   2  cos 2  2 cos 4  cos 6 .
Solution
Example
By expanding (1  i ) 2n , show that
n
n
(1) k C 22kn  2 n cos
,

2
k 0
n 1
 (1) C
k
k 0
2n
2 k 1
 2 n sin
n
2
Solution
Prepared by K. F. Ngai
Page 27
Complex Numbers
Example
Advanced Level Pure Mathematics
(a) Show that cos 5  16 cos 5   20 cos 3   5 cos .
(b) Using (a), or otherwise, solve 16 cos 4   20 cos 2   5  0 for values of  between
0 and 2 . Hence find the value of cos 2

10
cos 2
3
.
10
Solution
The nth roots of a Complex Number
1
If w n  z  r (cos   i sin  ) , then w  r n (cos
Example
Solution
  2k
n
 i sin
  2k
n
) , k  0,1,2, n  1 .
Find the three cube roots of  8 and locate them in the complex plane.
Prepared by K. F. Ngai
Page 28
Complex Numbers
Example
Advanced Level Pure Mathematics
Find the fifth roots of  1 .
Also, interpret the result in the Argand diagram.
Solution
nth of Unity
Theorem
nth of Unity of Their Properties
Let n be a positive integer. Then the equation z n  1 has n distinct roots given by
2k
2k
z k  cos
 i sin
(k  0,1,2,, n  1)
n
n
These roots are called the nth roots of unity. If we denote one of them by  ( 1) , then we have
 n =1
1     2     n 1 = 0
Proof
Prepared by K. F. Ngai
Page 29
Complex Numbers
Example
Advanced Level Pure Mathematics
Let n be a positive integer and   cos
2
2
 i sin
, find the values of
n
n
(a) 1     2     n 1 ,
(b)     2  3     n 1 ,
(c)
(1   )(1   2 )(1   3 )(1   n1 )
(d) 1   m   2 m   3m     ( n1) m , where m  Z .
Solution
Prepared by K. F. Ngai
Page 30
Complex Numbers
Advanced Level Pure Mathematics
Solution of Equations
Example
Solve the equation z 8  2 z 4 cos 4  1  0. When 0   

4
, show that the roots occur in
Conjugate pairs.
Solution
Prepared by K. F. Ngai
Page 31
Complex Numbers
Example
Advanced Level Pure Mathematics
Let n be a positive integer. By solving the equation x 2 n  1  0 , show that
n
2k  1
x n  x  n   ( x  x 1  2 cos
)
2n
k 1
Hence deduce that
n
n
2k  1

2k  1
2 2 n 1  sin 2
  1 and cos n   ( 1  sin 2 csc 2
)
4n
2
4n
k 1
k 1
Solution
Prepared by K. F. Ngai
Page 32
Complex Numbers
Advanced Level Pure Mathematics
Prepared by K. F. Ngai
Page 33