Download Complex Analysis notes and mind map and more Study notes Mathematics in PDF only on Docsity!
h
A x B
C
3 COMPLEX
NUMBERS
Objectives
After studying this chapter you should
- understand how quadratic equations lead to complex numbers and how to plot complex numbers on an Argand diagram;
- be able to relate graphs of polynomials to complex numbers;
- be able to do basic arithmetic operations on complex numbers of the form a + ib ;
- understand the polar form (^) [ r , θ] of a complex number and its algebra;
- understand Euler's relation and the exponential form of a
complex number (^) re i^ θ^ ;
- be able to use de Moivre's theorem;
- be able to interpret relationships of complex numbers as loci in the complex plane.
3.0 Introduction
The history of complex numbers goes back to the ancient Greeks who decided (but were perplexed) that no number existed that satisfies
x^2 = − 1
For example, Diophantus (about 275 AD) attempted to solve what seems a reasonable problem, namely
'Find the sides of a right-angled triangle of perimeter 12 units and area 7 squared units.'
Letting AB = x , AC = h as shown,
then a rea = 12 x h
and perimeter = x + h + x^2 + h^2
Activity 1
Show that the two equations above reduce to
6 x^2 − 43 x + 84 = 0
when perimeter = 12 and area (^) = 7. Does this have real solutions?
A similar problem was posed by Cardan in 1545. He tried to solve the problem of finding two numbers, a and b , whose sum is 10 and whose product is 40;
i.e. a + b = 10 (1)
ab = 40 (2)
Eliminating b gives
a (10 − a ) = 40
or a^2 − 10 a + 40 = 0.
Solving this quadratic gives
a =
1 2
(10 ± −60 ) = 5 ± − 15
This shows that there are no real solutions, but if it is agreed to continue using the numbers
a = 5 + −15, b = 5 − − 15
then equations (1) and (2) are satisfied.
Show that equations (1) and (2) are satisfied by these values of x and y.
So these are solutions of the original problem but they are not real numbers. Surprisingly, it was not until the nineteenth century that such solutions were fully understood.
The square root of − 1 is denoted by i , so that
i = − 1
and a = 5 + 15 i , b = 5 − 15 i
are examples of complex numbers.
Activity 3
Solve the following equations, leaving your answers in terms of i :
(a) x^2 + x + 1 = 0 (b) 3 x^2 − 4 x + 2 = 0
(c) x^2 + 1 = 0 (d) 2 x − 7 = 4 x^2
The set of solutions to a quadratic equation such as
ax^2 + bx + c = 0
can be related to the intercepts on the x- axis when the graph of the function
f ( x ) = ax^2 + bx + c
is drawn.
Activity 4 Quadratic graphs
Using a graphics calculator, a graph drawing program on a computer, a spreadsheet or otherwise, draw the graphs of the following functions and find a connection between the existence or not of real solutions to the related quadratic equations.
(a) f ( x ) = x^2 − 1 (b) f ( x ) = x^2 − x − 6
(c) f ( x ) = x^2 − 2 x − 2 (d) f ( x ) = x^2 + x + 1
(e) f ( x ) = 3 x^2 − 4 x + 2 (f) f ( x ) = x^2 + 1
You should have noted that if the graph of the function either intercepts the x -axis in two places or touches it in one place then the solutions of the related quadratic equation are real, but if the graph does not intercept the x -axis then the solutions are complex.
If the quadratic equation is expressed as ax^2 + bx + c = 0 , then the
expression that determines the type of solution is b^2 − 4 ac , called the discriminant.
In a quadratic equation ax^2 + bx + c = 0, if:
b^2 − 4 ac > 0 then solutions are real and different
b^2 − 4 ac = 0 then solutions are real and equal
b^2 − 4 ac < 0 then solutions are complex
3.1 Complex number algebra
A number such as 3 + 4 i is called a complex number. It is the sum of two terms (each of which may be zero).
The real term (not containing i ) is called the real part and the coefficient of i is the imaginary part. Therefore the real part of 3 + 4 i is 3 and the imaginary part is 4.
A number is real when the coefficient of i is zero and is imaginary when the real part is zero.
e.g. 3 + 0 i = 3 is real and 0 + 4 i = 4 i is imaginary.
Having introduced a complex number, the ways in which they can be combined, i.e. addition, multiplication, division etc., need to be defined. This is termed the algebra of complex numbers. You will see that, in general, you proceed as in real numbers, but using
i^2 = − 1
where appropriate.
But first equality of complex numbers must be defined.
If two complex numbers, say
a + bi , c + di
are equal , then both their real and imaginary parts are equal;
a + bi = c + di ⇒ a = c and b = d
Addition and subtraction
Addition of complex numbers is defined by separately adding real and imaginary parts; so if
z = a + bi , w = c + di
then z + w = ( a + c ) + ( b + d ) i.
Similarly for subtraction.
Example
Express each of the following in the form x + yi.
(a) (^) ( 3 + 5 i ) + (^) ( 2 − 3 i )
(b) (^3 +^5 i ) +^6
(c) 7 i − (^) ( 4 + 5 i )
Activity 5
Simplify the following expressions:
(a) ( 2 + 6 i ) + ( 9 − 2 i ) (b) ( 8 − 3 i ) − ( 1 + 5 i )
(c) 3 7( − 3 i ) + i ( 2 + 2 i ) (d) ( 3 + 5 i ) ( 1 − 4 i )
(e) ( 5 + 12 i ) ( 6 + 7 i ) (f) ( 2 + i )^2
(g) i^3 (h) i^4
(i) (1 − i ) 3 (j) (1 + i ) 2 + (1 − i ) 2
(k) (2 + i ) 4 + (2 − i ) 4 (l) (^) ( a + ib ) ( a − ib )
Division
The complex conjugate of a complex number is obtained by changing the sign of the imaginary part. So if z = a + bi , its complex conjugate, z , is defined by
z = a − bi
Any complex number a + bi has a complex conjugate a − bi
and from Activity 5 it can be seen that (^) ( a + bi ) ( a − bi ) is a real
number. This fact is used in simplifying expressions where the denominator of a quotient is complex.
Example
Simplify the expressions:
(a) 1 i
(b) 3 1 + i
(c)
4 + 7 i 2 + 5 i
Solution
To simplify these expressions you multiply the numerator and denominator of the quotient by the complex conjugate of the denominator.
(a) The complex conjugate of i is − i , therefore
1 i
i
×
− i − i
( ) − 1 ( i )
( ) − i ( i )
− i
= − i
(b) The complex conjugate of 1 + i is 1 − i , therefore
3 1 + i
1 + i
×
1 − i 1 − i
3 1( − i )
(^1 +^ i ) (^1 −^ i )
3 − 3 i 2
i
Note: an alternative notation often used for the complex conjugate is z *.
(c) The complex conjugate of 2 + 5 i is 2 − 5 i therefore
4 + 7 i 2 + 5 i
=
4 + 7 i 2 + 5 i
×
2 − 5 i 2 − 5 i
=
43 − 6 i 29
=
43 29
−
6 29
i
Activity 6 Division
Simplify to the form a + ib
(a)
4 i
(b)
1 − i 1 + i
(c)
4 + 5 i 6 − 5 i
(d)
4 i
( 1 + 2 i )^2
3.2 Solving equations
Just as you can have equations with real numbers, you can have equations with complex numbers, as illustrated in the example below.
Example
Solve each of the following equations for the complex number z.
(a) 4 + 5 i = z − ( 1 − i )
(b) (^) ( 1 + 2 i ) z = 2 + 5 i
Solution
(a) Writing z = x + iy ,
4 + 5 i = (^) ( x + y i ) − (^) ( 1 − i )
4 + 5 i = x − 1 + (^) ( y + (^1) ) i
Comparing real parts ⇒ 4 = x − 1 , x = 5
Comparing imaginary parts ⇒ 5 = y + 1 , y = 4
So z = 5 + 4 i. In fact there is no need to introduce the real and imaginary parts of z , since
4 + 5 i = z − ( 1 − i )
⇒ z = 4 + 5 i + ( 1 − i )
⇒ z = 5 + 4 i
(b) ( 1 + 2 i ) z = 2 + 5 i
z =
2 + 5 i 1 + 2 i
- Given that p and q are real and that 1 + 2 i is a root of the equation z^2 +^ (^ p^ +^5 i ) z^ +^ q (2^ −^ i )^ =^0 determine: (a) the values of p and q ; (b) the other root of the equation.
3.3 Argand diagram
Any complex number z = a + bi can be represented by an ordered pair ( a , b ) and hence plotted on xy -axes with the real part measured along the x -axis and the imaginary part along the y -axis. This graphical representation of the complex number field is called an Argand diagram , named after the Swiss mathematician Jean Argand (1768-1822).
Example
Represent the following complex numbers on an Argand diagram:
(a) z = 3 + 2 i (b) z = 4 − 5 i (c) z = − 2 − i
Solution
The Argand diagram is shown opposite.
Activity 8
Let z (^) 1 = 5 + 2 i , z (^) 2 = 1 + 3 i , z (^) 3 = 2 − 3 i , z (^) 4 = − 4 − 7 i.
(a) Plot the complex numbers z (^) 1 , z (^) 2 , z (^) 3 , z (^) 4 on an Argand diagram and label them.
(b) Plot the complex numbers z (^) 1 + z (^) 2 and z (^) 1 − z (^) 2 on the same Argand diagram. Geometrically, how do the positions of the numbers z (^) 1 + z (^) 2 and z (^) 1 − z (^) 2 relate to z (^) 1 and z (^) 2?
2 1
–3 –1 1 2 3 4 5
z = 3 + 2 i
z = 4 – 5 i
z = –2 – i
imaginary
real
Imaginary
Real
- The complex numbers u , v and w are related by 1 u
= 1 v
.
Given that v = 3 + 4 i , w = 4 − 3 i , find u in the form x + iy.
3.4 Polar coordinates
Consider the complex number z = 3 + 4 i as represented on an Argand diagram. The position of A can be expressed as coordinates (3, 4), the cartesian form, or in terms of the length and direction of OA.
Using Pythagoras' theorem, the length of OA = 3 2 + 4 2 = 5.
This is written as z = r = 5. z is read as the modulus or
absolute value of z.
The angle that OA makes with the positive real axis is
θ = tan−^1
4 3
^
=
53.13° (or 0.927 radians ).
This is written as arg ( ) = z 53.13°. You say arg ( ) z is the
argument or phase of z.
The parameters z and arg ( ) z are in fact the equivalent of polar
coordinates r , θ as shown opposite. There is a simple connection between the polar coordinate form and the cartesian or rectangular form ( a , b ):
a = r cos θ, b = r sin θ.
Therefore
z = a + bi = r cos θ + r i sin θ = r (cos θ + i sin θ)
where z = r , and arg ( ) = z θ.
It is more usual to express the angle θ in radians. Note also that it is convention to write the i before sin θ , i.e. i sin θ is
preferable to sin θ i.
In the diagram opposite, the point A could be labelled (^) ( 2 3, 2)
or as 2 3 + 2 i.
The angle that OA makes with the positive x -axis is given by
θ = tan−^1
=^ tan
Therefore θ =
π 6
or 2 π +
π 6
or 4 π +
π 6
or ... etc. There is an
infinite number of possible angles. The one you should normally use is in the interval −π <θ ≤π , and this is called the principal argument.
Real
( a , b )
r (^) b
a x
y
θ
A
0
O 0.5 1.0 1.5 2.0 2.5 3.0 3.
Imaginary
O
1
2
3
4
1 2 3 4
A
3.5 Complex number algebra
You will now investigate the set of complex numbers in the
modulus/argument form, [ r , θ].
Suppose you wish to combine two complex numbers of the form
z (^) 1 = (^) [ r 1 , θ (^1) ] z (^) 2 = (^) [ r 2 , θ (^2) ]
Note that, in a + bi form,
z (^) 1 = r 1 cos θ 1 + i r 1 sin θ 1
and z (^) 2 = r 2 cos θ 2 + i r 2 sin θ 2
So z (^) 1 z (^) 2 = (^) ( r 1 cos θ 1 + i r 1 sin θ 1 ) ( r 2 cos θ 2 + i r 2 sin θ 2 )
= r (^) 1 r (^) 2 (cos θ 1 + i sin θ 1 ) (cos θ 2 + i sin θ 2 )
= r (^) 1 r (^) 2 [( cos θ 1 cos θ 2 − sin θ 1 sin θ 2 )
- (^) (sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) i ].
Simplify the expressions in the brackets.
Using the formulae for angles,
z (^) 1 z (^) 2 = r (^) 1 r (^) 2 [cos ( θ 1 +θ 2 ) + i sin( θ 1 +θ 2 )]
or, in polar notation
z (^) 1 z (^) 2 = (^) [ r (^) 1 r (^) 2 , θ 1 +θ 2 ].
For example, (^) [3, 0.5 ] × (^) [4, 0.3 ] = (^) [12, 0.8 ].
That is, the first elements of the ordered pairs are multiplied and the second elements are added.
Activity 10
Given that z 1 = (^) [3, 0. 7 ], z (^) 2 = (^) [2, 1.2 ] and z (^) 3 = (^) [ 4, − 0.5],
(a) find z 1 × z 2 and z 1 × z 3
(b) show that (^) [1, 0 ] × z (^) 1 = z (^) 1
(c) (i) find a complex number z = [ r , θ] such that
z × z 2 = [1, 0 ].
(ii) find a complex number z = [ r , θ] such that
z × z 3 = [1, 0 ].
(d) for any complex number (^) [ r , θ] show that
1 r
, −θ
× [ r , θ] = (^) [1, 0 ] ( r > 0 ).
Activity 11
Use a spreadsheet package to plot numbers on an Argand diagram by entering numbers and formulae into cells A5 to E as shown opposite.
Cells D5 and E5 calculate the x and y coordinates respectively of the complex number whose modulus and argument are in cells B5 and C5 (the argument is entered as a multiple of π ).
A second number can be entered in cells B6 and C6 and its ( x , y ) coordinates calculated by using appropriate formulae in cells D and E6.
This can be repeated for further numbers (the spreadsheet facility 'FILL DOWN' is useful here).
Use the appropriate facility on your spreadsheet to plot the ( x , y ) values.
Label rows and columns if it makes it easier.
Experiment with different values of r and θ.
An example is shown in the graph opposite and the related spreadsheet below.
A5 B5 C5 (^) D5 E
π 2 0.
= B5* cos(C5* A5)
= B5* sin(C5* A5)
1
2
–4 –3 –2 –1 1 2 3 4
The derivation of de Moivre's theorem now follows.
Consider the complex number z = cos
π 3
π 3
^
^
Then z^2 = cos
π 3
π 3
^
×
cos
π 3
π 3
^
= cos 2
π 3
− sin 2
π 3
π 3
sin
π 3
= cos
2 π 3
2 π 3
or with the modulus/argument notation
z = 1,
π 3
and z^2 = 1,
π 3
×^ 1,
π 3
=^ 1,
2 π 3
^
^
Remember that any complex number z = x + y i can be written in
the form of an ordered pair [ r , θ] where r = x^2 + y^2 and
θ = tan−^1
y x
^
^
If the modulus of the number is 1, then z = cos θ + i sin θ
and (^) z^2 = (cos θ + i sin θ)^2
= cos 2 θ − sin 2 θ + 2 i cos θ sin θ
= cos 2 θ + i sin 2 θ
i.e. z^2 = [1, θ]^2 = [1, 2 θ].
Activity 12
(a) Use the principle that, with the usual notation,
[^ r^ 1 ,^ θ 1 ] ×^ [ r^ 2 ,^ θ 2 ] =^ [ r^ 1 r^ 2 ,^ θ 1 +θ 2 ]
to investigate cos
π 6
π 6
^
n when n = 0, 1, 2, 3, ..., 12.
(b) In the same way as in (a), investigate
3cos
π 6
π 6
^
n
for n = 0, 1, 2, ..., 6.
You should find from the last activity that
( cos θ + i sin θ) n^ = cos ( n θ) + i sin ( n θ).
In (^) [ r ,^ θ] form this is (^) [ r ,^ θ] n^ =^ [ r^ n^ ,^ n^ θ] and de Moivre's theorem
states that this is true for any rational number n.
A more rigorous way of deriving de Moivre's theorem follows.
Activity 13
Show that (cos θ + i sin θ) n^ = cos n θ + i sin n θ for n = 3 and n = 4.
Activity 14
Show that
( cos k θ + i sin k θ) (cos θ + i sin θ) = cos ( k + 1 )θ + i sin ( k + 1 )θ.
Hence show that if
( cos θ + i sin θ) k^ = cos k θ + i sin k θ
then ( cos θ + i sin θ) k^ +^1 = cos (^) (( k + 1 )θ) + i sin (^) (( k + 1 )θ).
The principle of mathematical induction will be used to prove
that (cos θ + i sin θ) n^ = cos ( n θ) + i sin ( n θ) for all positive integers.
Let S ( k ) be the statement
' (cos θ + i sin θ) k^ = cos k θ + i sin k θ '.
As S (1) is true and you have shown in Activity 14 that S ( k )
implies S k ( + 1 ) then S (2) is also true. But then (again by Activity
- S (3) is true. But then ... Hence S( n ) is true for n = 1, 2,3,K. This is the principle of mathematical induction (which you meet more fully later in the book). So for all positive integers n ,
Thus cos
p q
θ + i sin
p q
θ
= (cos θ + i sin θ)
pq
Therefore cos n θ + i sin n θ = ( cos θ + i sin θ) n^ for any rational
number n and clearly this leads to
(^ r^ (^ cos^ θ +^ i^ sin^ θ))
n
= r n^ ( cos n θ + i sin n θ)
3.7 Applications of
de Moivre's theorem
There are many applications of de Moivre's theorem, including the proof of trigonometric identities.
Example
Prove that cos 3 θ = cos 3 θ − 3cos θ sin 2 θ.
Solution
By de Moivre's theorem:
cos 3 θ + i sin 3 θ = (cos θ + i sin θ)^3
Comparing real parts of the equation above you obtain
cos 3 θ = cos 3 θ − 3cos θ sin 2 θ
Example
Simplify the following expression:
cos 2 θ + i sin 2 θ cos 3 θ + i sin 3 θ
Solution
cos 2 θ + i sin 2 θ cos 3 θ + i sin 3 θ
(cos θ +^ i^ sin^ θ)^2
( cos θ + i sin θ)^3
(cos θ + i sin θ)^1
= cos 3 θ + 3cos 2 θ ( i sin θ) + 3cos θ ( i sin θ)^2 + ( i sin θ)^3
= cos 3 θ + 3 i cos 2 θ sin θ − 3cos θ sin 2 θ − i sin 3 θ
= cos 3 θ − 3cos θ sin 2 θ + i (3cos 2 θ sin θ − sin 3 θ )
= (cos θ + i sin θ) −^1
= (^) (cos (− θ) + i sin ( −θ))
= cos θ − i sin θ
Exercise 3C
- Use de Moivre's theorem to prove the trig identities: (a) sin 2 θ = 2sin θ cos θ (b) cos5 θ = cos 5 θ −10 cos 3 θ sin 2 θ + 5cos θ sin 4 θ
- If z = cos θ + i sin θ then use de Moivre's theorem to show that:
(a) z + 1 z
= 2 cos θ (b) z^2 + 1 z^2
= 2 cos2 θ
(c) z n^ + 1 z n^
= 2 cos n θ
Activity 16
Make an educated guess at a complex solution to the equation
z^3 = 1 and then use the facilities of the spreadsheet to raise it to the power 3 and plot it on the Argand diagram. If it is a solution of the equation then the resultant point will be plotted at distance 1 unit along the real axis. The initial spreadsheet layout from Activity 11 can be adapted. In addition, the cells shown opposite are required.
What does the long formula in cell C7 do? Is it strictly necessary in this context?
Below are two examples of the output from a spreadsheet using these cells – the first one is not a cube root of 1 but the second is.
- Simplify the following expressions:
(a) cos5^ θ +^ i sin 5^ θ cos2 θ − i sin 2 θ
(b) cos^ θ −^ i sin^ θ cos 4 θ − i sin 4 θ
1
–1 –0.5 0.5 1
1
–1 –0.5 0.5 1
B7 C7 (^) D7 E
= B5^ 3
= C5* 3 - 2 * int( ( C5*3 2 ))
= B7 * cos(A5* C7)
= B7 * sin(A5* C7)
