3.complex numbers Further Mathematics Zimbabwe Zimsec Cambridge

55
Chapter 3 Complex Numbers
h
x BA
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 reiθ
;
• 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 area = 1
2
x h
and perimeter = x + h + x
2
+ h
2

56
Activity 1
Show that the two equations above reduce to
6x
2
− 43x + 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
−10a + 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.

57
Activity 2 The need for complex numbers
Solve if possible, the following quadratic equations by
factorising or by using the quadratic formula. If a solution is not
possible explain why.
(a) x2
−1 = 0 (b) x2
− x − 6 = 0
(c) x2
− 2x − 2 = 0 (d) x2
− 2x + 2 = 0
You should have found (a), (b) and (c) straightforward to solve
but in (d) a term appears in the solution which includes the
square root of a negative number and to obtain solutions you
need to use the symbol i = −1, or
i2
= −1
It is then possible to obtain a solution to (d) in Activity 2.
Example
Solve x2
− 2x + 2 = 0.
Solution
Using the quadratic formula
x =
−b ± b
2
− 4ac
2a
⇒ x =
− −2( ) ± −2( )2
− 4 1( ) 2( )
2 1( )
⇒ x =
2 ± − 4
2
But − 4 = 4 −1( ) = 4 −1 = 2 −1 = 2i
Therefore x =
2 ± 2i
2
⇒ x = 1 ± i
Therefore the two solutions are
x = 1+ i and x = 1− i
(using the definition of i).

58
Activity 3
Solve the following equations, leaving your answers in terms of i:
(a) x
2
+ x +1 = 0 (b) 3x
2
− 4x + 2 = 0
(c) x
2
+1 = 0 (d) 2x − 7 = 4x
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
− 2x − 2 (d) f x( ) = x
2
+ x +1
(e) f x( ) = 3x
2
− 4x + 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
− 4ac, called
the discriminant.
In a quadratic equation ax2
+ bx + c = 0, if:
b2
− 4ac > 0 then solutions are real and different
b
2
− 4ac = 0 then solutions are real and equal
b
2
− 4ac < 0 then solutions are complex

59
3.1 Complex number algebra
A number such as 3 + 4i 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 + 4i 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 + 0i = 3 is real and 0 + 4i = 4i 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 + 5i( )+ 2 − 3i( )
(b) 3 + 5i( )+ 6
(c) 7i − 4 + 5i( )

60
Solution
(a) 3 + 5i( )+ 2 − 3i( ) = 3 + 2 + 5 − 3( )i = 5 + 2i
(b) 3 + 5i( )+ 6 = 9 + 5i
(c) 7i − 4 + 5i( ) = 7i − 4 − 5i = −4 + 2i
Multiplication
Multiplication is straightforward provided you remember that
i2
= −1.
Example
Simplify in the form x + yi:
(a) 3 2 + 4i( )
(b) 5 + 3i( )i
(c) 2 − 7i( ) 3 + 4i( )
Solution
(a) 3 2 + 4i( ) = 3 2( )+ 3 4i( ) = 6 +12i
(b) 5 + 3i( )i = 5( ) i + 3i( )i = 5i + 3 i
2
( ) = 5i + −1( )3 = − 3 + 5i
(c) 2 − 7i( ) 3 + 4i( ) = 2( ) 3( ) − 7i( ) 3( ) + 2( ) 4i( ) − 7i( ) 4i( )
= 6 − 21i + 8i − −28( )
= 6 − 21i + 8i + 28
= 34 −13i
In general, if
z = a + bi, w = c + di,
then z w = (a + bi)(c + di)
= ac − bd + (ad + bc)i

61
Activity 5
Simplify the following expressions:
(a) 2 + 6i( )+ 9 − 2i( ) (b) 8 − 3i( )− 1+ 5i( )
(c) 3 7 − 3i( )+ i 2 + 2i( ) (d) 3 + 5i( ) 1− 4i( )
(e) 5 +12i( ) 6 + 7i( ) (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 + 7i
2 + 5i
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
=
1
i
×
−i
−i
=
1( ) −i( )
i( ) −i( )
=
−i
− −1( )
= −i
(b) The complex conjugate of 1+ i is 1− i, therefore
3
1+ i
=
3
1+ i
×
1− i
1− i
=
3 1− i( )
1+ i( ) 1− i( )
=
3 − 3i
2
=
3
2
−
3
2
i
Note: an alternative notation
often used for the complex
conjugate is z*.

62
(c) The complex conjugate of 2 + 5i is 2 − 5i therefore
4 + 7i
2 + 5i
=
4 + 7i
2 + 5i
×
2 − 5i
2 − 5i
=
43 − 6i
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 + 5i
6 − 5i
(d)
4i
1+ 2i( )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 + 5i = z − 1− i( )
(b) 1+ 2i( )z = 2 + 5i
Solution
(a) Writing z = x + iy,
4 + 5i = x + yi( )− 1− i( )
4 + 5i = x −1+ y +1( )i
Comparing real parts ⇒ 4 = x −1, x = 5
Comparing imaginary parts ⇒ 5 = y +1, y = 4
So z = 5 + 4i . In fact there is no need to introduce the real
and imaginary parts of z, since
4 + 5i = z − 1− i( )
⇒ z = 4 + 5i + 1− i( )
⇒ z = 5 + 4i
(b) 1+ 2i( )z = 2 + 5i
z =
2 + 5i
1+ 2i

63
z =
2 + 5i
1+ 2i
×
1− 2i
1− 2i
z =
12 + i
5
=
12
5
+
1
5
i
Activity 7
(a) Solve the following equations for real x and y
(i) 3 + 5i + x − yi = 6 − 2i
(ii) x + yi = (1− i)(2 + 8i).
(b) Determine the complex number z which satisfies
z(3 + 3i) = 2 − i .
Exercise 3A
1. Solve the equations:
(a) x
2
+ 9 = 0 (b) 9x
2
+ 25 = 0
(c) x
2
+ 2x + 2 = 0 (d) x
2
+ x +1 = 0
(e) 2x
2
+ 3x + 2 = 0
2. Find the quadratic equation which has roots
2 ± 3i .
3. Write the following complex numbers in the
form x + yi .
(a) (3+ 2i) + (2 + 4i) (b) (4 + 3i) − (2 + 5i)
(c) (4 + 3i) + (4 − 3i) (d) (2 + 7i) − (2 − 7i)
(e) (3+ 2i)(4 − 3i) (f) (3+ 2i)
2
(g) (1+ i)(1− i)(2 + i)
4. Find the value of the real number y such that
(3+ 2i)(1+ iy)
is (a) real (b) imaginary.
5. Simplify:
(a) i3
(b) i4
(c)
1
i
(d)
1
i
2
(e)
1
i
3
6. If z = 1+ 2i , find
(a) z
2
(b)
1
z
(c)
1
z
2
7. Write in the form x + yi :
(a)
2 + 3i
1+i
(b)
−4 + 3i
−2 − i
(c)
4i
2 − i
(d)
1
2 + 3i
(e)
3− 2i
i
(f)
p + qi
r + si
8. Simplify:
(a)
(2 + i)(3− 2i)
1+ i
(b)
(1− i)
3
(2 + i)
2
(c)
1
3+ i
−
1
3− i
9. Solve for z when
(a) z(2 + i) = 3− 2i (b) (z + i)(1− i) = 2 + 3i
(c)
1
z
+
1
2 − i
=
3
1+ i
10. Find the values of the real numbers x and y in
each of the following:
(a)
x
1+ i
+
y
1− 2i
= 1
(b)
x
2 − i
+
yi
i +3
=
2
1+ i

64
11. Given that p and q are real and that 1+ 2i is a
root of the equation
z2
+ ( p + 5i)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 + 2i (b) z = 4 − 5i (c) z = −2 −i
Solution
The Argand diagram is shown opposite.
Activity 8
Let z1= 5 + 2i , z 2= 1+ 3i, z 3= 2 − 3i, z 4= −4 − 7i .
(a) Plot the complex numbers z1, z 2 , z 3, z 4 on an Argand
diagram and label them.
(b) Plot the complex numbers z1+ z 2 and z1− z 2 on the same
Argand diagram. Geometrically, how do the positions of
the numbers z1+ z 2 and z1− z 2 relate to z1 and z 2 ?
2
1
–1
–2
–3
–4
–5
–1–3 1 2 3 4 5
z = 3 + 2i
z = 4 – 5i
z = –2 – i
–2
imaginary
real
Imaginary
Real
12. The complex numbers u, v and w are related by
1
u
=
1
v
+
1
w
.
Given that v = 3+ 4i, w = 4 − 3i , find u in the
form x + iy.

65
3.4 Polar coordinates
Consider the complex number z = 3 + 4i 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 = 32
+ 42
= 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 cosθ, b = rsinθ .
Therefore
z = a + bi = r cosθ + r isinθ = r cosθ + isinθ( )
where z = r,andarg z( ) = θ .
It is more usual to express the angle θ in radians. Note also that
it is convention to write the i before sinθ , i.e. isinθ is
preferable to sinθ i .
In the diagram opposite, the point A could be labelled 2 3, 2( )
or as 2 3 + 2i .
The angle that OA makes with the positive x-axis is given by
θ = tan
−1 2
2 3





 = tan
−1 1
3





 .
Therefore θ =
π
6
or 2π +
π
6
or 4π +
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
0.5
1.0
1.5
2.0
0.5 1.0 1.5 2.0 2.5 3.0 3.5O
Imaginary
O
1
2
3
4
1 2 3 4
A

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Using polar coordinates the point A could be labelled with its
polar coordinates r,θ[ ] as 4,
π
6





. Note the use of squared
brackets when using polar coordinates. This is to avoid
confusion with Cartesian coordinates.
Thus 2 3 + 2i = 4 cos
π
6





 + isin
π
6











 .
Important note: if you are expressing a + ib in its polar form,
where a and b are both positive, then the formula θ = tan−1 b
a
is
quite sufficient. But in other cases you need to think about the
position of a + ib in the Argand diagram.
Example
Write z = −1− i in polar form.
Solution
Now z = a + ib where a = −1 and b = −1 and in polar form the
modulus of z = z = r = 1
2
+1
2
= 2 and the argument is
5π
4
or 225°( ): its principal value is −
3π
4
.
Hence z = 2,
−3π
4




in polar coordinates. (The formula
tan−1 b
a
would have given you
π
4
.)
Activity 9
(a) Write the following numbers in r,θ[ ] form:
(i) 7 + 2i (ii) 3 − i (iii) −4 + 6i (iv) − 3 − i
(b) Write the following in a + bi form:
(remember that the angles are in radians)
(i) 3,
π
4





 (ii) 5, π[ ] (iii) 6, 4.2[ ]
(iv) 2,
−2π
3




x1 2–1–2
–2
–1
1
2
y
–1 – i

67
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
z1 = r1,θ 1[ ] z 2 = r2 ,θ 2[ ]
Note that, in a + bi form,
z1 = r1 cosθ1 + i r1sinθ1
and z 2 = r2 cosθ2 + i r2 sinθ2
So z1 z 2 = r1 cosθ1 +i r1 sinθ1( ) r2 cosθ2 +i r2 sinθ2( )
= r1r 2 cosθ1 + isinθ1( ) cosθ2 +isinθ2( )
= r1r 2 cosθ1 cosθ2 − sinθ1 sinθ2( )[
+ sinθ1 cosθ2 + cosθ1 sinθ2( )i].
Simplify the expressions in the brackets.
Using the formulae for angles,
z1 z 2 = r1r 2 cos θ1 + θ2( )+ isin θ1 + θ2( )[ ]
or, in polar notation
z1 z 2 = r1r 2, θ1 + θ2[ ].
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 z1 = 3, 0.7[ ], z 2 = 2,1.2[ ] and z 3 = 4, − 0.5[ ],
(a) find z1 × z2 and z1 × z3
(b) show that 1, 0[ ]× z1= z1

68
(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 E5
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 D6
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 E5
2 0.25π
= B5*cos(C5*A5)
= B5*sin(C5*A5)
1
2
–1
–2
–3
–4
–5
–1–2–3 1 2 3 4–4

69
Exercise 3B
1. Mark on an Argand diagram the points
representing the following numbers:
(a) 2 (b) 3i (c) −i (d) 1+ 2i (e) 3− i
(f) −2 + 3i
2. The points A, B, C and D represent the numbers
z1 , z 2, z 3, and z 4 and O is the origin.
(a) If OABC is a parallelogram, and z1 = 1+ i ,
z 2 = 4 + 5i , find z 3 .
(b) Find z 2 and z 4 when ABCD is a square and
(i) z1 = 2 + i, z 3= 6 + 7i
(ii) z1 = 6 − 2i, z 3= 6i
3. Find the modulus and argument of
(a) 1− i (b) 1+ 3i (c) 3− 3i (d) 3+ 2i
4. Show that
(a) z = z (b) argz = −argz
and illustrate these results on an Argand
diagram.
5. Find the modulus and argument of z1 , z 2, z1 z 2
and
z1
z2
when z1= 1+ i and z 2= 3 + i. What do
you notice?
6. Write in the form a + bi
(a) 4,
π
3





 (b) 5,
π
2






(c) 3 2, −
3π
4





 (d) 4, 13π[ ]
7. Write in polar form
(a) 1+ i (b) −2 + i (c) −5 (d) 4i (e) 3+ 4i
(f) −3− 4i (g) 3− 4i (h) −3+ 4i
8. In this question, angles are in radians.
(a) (i) Plot the following complex numbers on
an Argand diagram and label them:
z1 = 4,0[ ], z2 = 3,
π
2



, z3 = 2,
−π
2




z4 = 3,
π
3



, z5 = 2,
5π
3




(ii) Let the complex number z = 1,
π
2




Calculate z × z1, z × z2 , etc. and plot the
points on the same diagram as in (i).
What do you notice?
(b) Repeat (a) (ii) using z = 1,
π
3




(c) In general, what happens when a complex
number is multiplied by 1,θ[ ]? Make up some
examples to illustrate your answer.
(d) Repeat (a) (ii) using z = 0.5,
π
2




(e) In general, what happens when a complex
number is multiplied by 0.5,
π
2



? Make up
some examples to illustrate your answer.
(f) Repeat (e) for 3,
π
3




(g) Describe what happens when a complex
number is multiplied by 3,
π
3



 . Make up
some examples to illustrate your answer.
3.6 De Moivre's theorem
An important theorem in complex numbers is named after the
French mathematician, Abraham de Moivre (1667-1754).
Although born in France, he came to England where he made
the acquaintance of Newton and Halley and became a private
teacher of Mathematics. He never obtained the university
position he sought but he did produce a considerable amount of
research, including his work on complex numbers.

70
The derivation of de Moivre's theorem now follows.
Consider the complex number z = cos
π
3
+ isin
π
3





 .
Then z
2
= cos
π
3
+ isin
π
3





 × cos
π
3
+ isin
π
3






= cos2 π
3
− sin2 π
3
+ 2i cos
π
3
sin
π
3
= cos
2π
3
+ isin
2π
3
or with the modulus/argument notation
z = 1,
π
3






and z
2
= 1,
π
3





 × 1,
π
3





 = 1,
2π
3





.
Remember that any complex number z = x + yi 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 = cosθ + i sinθ
and z2
= cosθ + i sinθ( )2
= cos2
θ − sin2
θ + 2i cosθ sinθ
= cos2θ + isin2θ
i.e. z2
= 1,θ[ ]2
= 1,2θ[ ].
Activity 12
(a) Use the principle that, with the usual notation,
r1,θ1[ ]× r 2,θ2[ ]= r1 r 2,θ1 + θ2[ ]
to investigate cos
π
6
+ isin
π
6






n
when n = 0, 1, 2, 3, ..., 12.

71
(b) In the same way as in (a), investigate
3cos
π
6
+3isin
π
6






n
for n = 0, 1, 2, ..., 6.
You should find from the last activity that
cosθ + isinθ( )n
= cos nθ( )+ isin nθ( ).
In r,θ[ ] form this is r,θ[ ]n
= r
n
,nθ[ ] 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 cosθ + isinθ( )n
= cosnθ + isin nθ for n = 3 and n = 4 .
Activity 14
Show that
coskθ + isin kθ( ) cosθ + isinθ( ) = cos k +1( )θ + isin k +1( )θ .
Hence show that if
cosθ + isinθ( )k
= coskθ + isin kθ
then cosθ + isinθ( )k+1
= cos k +1( )θ( )+ isin k +1( )θ( ).
The principle of mathematical induction will be used to prove
that cosθ + isinθ( )n
= cos nθ( )+ isin nθ( ) for all positive integers.
Let S(k) be the statement
' cosθ + isinθ( )k
= coskθ + isin kθ '.
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
14) 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,

72
cosθ + isinθ( )n
= cosnθ + isin nθ
If n is a negative integer, then let m = −n
cosθ + isinθ( )n
= cosθ + isinθ( )−m
=
1
cosθ + isinθ( )m
where m is positive and, from the work above,
cosθ + isinθ( )m
= cosmθ + isin mθ( ).
Therefore cosθ + isinθ( )n
=
1
cosmθ + isin mθ( )
Activity 15
Show that
1
cosmθ + isin mθ( )
= cosmθ − isin mθ
and hence that cosθ + isinθ( )n
= cosnθ + isin nθ when n is a
negative integer.
Hint : multiply top and bottom by cosmθ − isin mθ( ) and use
the fact that sin −A( ) = −sin A( ).
When n is a rational number, i.e. n =
p
q
where p and q are
integers, then as q is an integer
cos
p
q





θ + isin
p
q





θ








q
= cos pθ + isin pθ( )
Since p is an integer
cos pθ + isin pθ = cosθ + isinθ( )p
,
and hence
cos
p
q





θ + isin
p
q





θ








q
= cosθ + isinθ( )p

73
Thus cos
p
q





θ + isin
p
q





θ








= cosθ + isinθ( )
p
q
Therefore cosnθ + isin nθ = cosθ + isinθ( )n
for any rational
number n and clearly this leads to
r cosθ + isinθ( )( )n
= rn
cosnθ + isin nθ( )
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 cos3θ = cos
3
θ − 3cosθ sin
2
θ .
Solution
By de Moivre's theorem:
cos3θ + isin3θ = cosθ + isinθ( )3
Comparing real parts of the equation above you obtain
cos3θ = cos
3
θ − 3cosθ sin
2
θ
Example
Simplify the following expression:
cos2θ + isin2θ
cos3θ + isin3θ
Solution
cos2θ + isin2θ
cos3θ + isin3θ
=
cosθ + isinθ( )2
cosθ + isinθ( )3
=
1
cosθ + isinθ( )1
= cos
3
θ + 3cos
2
θ isinθ( )+ 3cosθ isinθ( )2
+ isinθ( )3
= cos3
θ + 3i cos2
θ sinθ − 3cosθ sin2
θ − isin3
θ
= cos
3
θ −3cosθ sin
2
θ + i(3cos
2
θ sinθ − sin
3
θ)

74
= cosθ + isinθ( )−1
= cos −θ( )+ isin −θ( )( )
= cosθ − isinθ
Exercise 3C
1. Use de Moivre's theorem to prove the trig
identities:
(a) sin2θ = 2sinθ cosθ
(b) cos5θ = cos
5
θ −10cos
3
θ sin
2
θ + 5cosθ sin
4
θ
2. If z = cosθ + isinθ then use de Moivre's theorem to
show that:
(a) z +
1
z
= 2cosθ (b) z
2
+
1
z
2
= 2cos2θ
(c) z
n
+
1
z
n
= 2cosnθ
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.
3. Simplify the following expressions:
(a)
cos5θ + isin5θ
cos2θ − isin2θ
(b)
cosθ − isinθ
cos4θ − isin4θ
1
0.8
0.6
0.4
0.2
0.5 1–0.5–1
1
0.8
0.6
0.4
0.2
0.5 1–0.5–1
B7 C7
D7 E7
= B5^3
= C5*3- 2* int C5*3
2( )( )
= B7*cos(A5*C7)
= B7*sin(A5*C7)

75
3.8 Solutions of z3
= 1
Write down one solution of z
3
= 1.
De Moivre's theorem can be used to find all the solutions
of z
3
= 1.
Let z = r,θ[ ]
then z
3
= r,θ[ ]3
= r
3
,3θ[ ]
and you can express 1 as 1 = 1,2nπ[ ] where n is an integer.
Then r
3
,3θ[ ]= 1,2nπ[ ]
Therefore r
3
= 1 and 3θ = 2nπ
i.e. r = 1 and θ =
2nπ
3
The solutions are then given by letting n = 0, 1, 2, ...
If n = 0, z1 = 1,0[ ]= 1
If n = 1, z2 = 1,
2π
3





 = cos
2π
3
+ isin
2π
3
= −
1
2
+
3
2
i
If n = 2, z3 = 1,
4π
3





 = cos
4π
3
+ isin
4π
3
= −
1
2
−
3
2
i
What happens if n = 3, 4, ... ?
Activity 17 Cube roots of unity
Plot the three distinct cube roots of unity on an Argand diagram.
What do you notice?
1
1–1
–1n = 2
n = 1

76
Activity 18
Use de Moivre's theorem to find all solutions to the following
equations and plot the results on an Argand diagram.
(a) z4
= 1 (b) z3
= 8 (c) z3
= i
3.9 Euler's theorem
You have probably already met the series expansion of ex
,
namely
e
x
= 1+ x +
x2
2!
+
x3
3!
+
x4
4!
+K
Also the series expansions for cosθ and sinθ are given by
cosθ = 1−
θ2
2!
+
θ4
4!
−
θ6
6!
+K
sinθ = θ −
θ3
3!
+
θ5
5!
−
θ7
7!
+K
Activity 19
(a) For each of the following values of θ , use the series for ex
with x replaced by iθ to calculate (to 4 d.p.) the value of e
iθ
.
(Write your answer in the form a + bi .)
(i) θ = 0 (ii) θ = 1 (iii) θ = 2 (iv) θ = −0.4
(b) Calculate cosθ and sinθ for each of the values in (a).
(c) Find a connection between the values of eiθ
, cosθ and sinθ
for each of the values of θ given in (a) and make up one
other example to test your conjecture.
(d) To prove this for all values of θ , write down the series
expansions of e
iθ
, cosθ and sinθ and show that
e
iθ
= cosθ + isinθ .

77
The previous activity has shown that
eiθ
= cosθ + isinθ
which is sometimes known as Euler's theorem.
It is an important result, and can be used to derive de Moivre's
theorem in a simple way. If z is any complex number then in
polar form
z = x + yi = r cosθ + r isinθ
= r cosθ + isinθ( )
= re
iθ
, using Euler's theorem.
Thus zn
= reiθ
( )
n
= rn
eniθ
= rn
e
i nθ( )
or r cosθ + irsinθ( )n
= rn
cos nθ( )+ isin nθ( )( )
⇒ cosθ + isinθ( )n
= cos nθ( )+ isin nθ( )
which is de Moivre's theoem.
What assumptions about complex number algebra have been
made in the 'proof' above?
One interesting result can be obtained from Euler's theorem by
putting θ = π . This gives
e
iπ
= cosπ + isin π
= −1+ i × 0 .
So eiπ
+1 = 0
This is often referred to as Euler's equation, since it connects
the five most 'famous' numbers
0, 1, π , e, i
with a '+ ' and '= ' sign!
Try substituting other values of θ in Euler's theorem and see
what equation is derived.
r y
x
θ

78
3.10 Exponential form of a
complex number
When a complex number z has modulus r, which must be non-
negative, and argument θ, which is usually taken such that it
satisfies −π < θ ≤ π, you have already shown that it can be
represented in the forms
(i) r cosθ + isinθ( )
(ii) r, θ[ ]
(iii) reiθ
Expression (iii) is referred to as the exponential form of a
complex number.
Activity 20
Write each of the following complex numbers in the exponential
form.
(a) 2 cos
π
3
+ i cos
π
3




(b) 5,
2π
3



 (c) 1− i 3
3.11 Solving equations
You have already investigated the solutions of the equation
z3
= 1 and similar equations using a spreadsheet and by using de
Moivre's theorem. A similar approach will now be used to solve
more complicated equations.
Example
Write down the modulus and argument of the complex number
4 − 4i.
Solve the equation z5
= 4 − 4i, expressing your answers in the
exponential form.
Solution
4 − 4i = 42
+ −4( )2
{ } = 4 2
As before it is often helpful to make a small sketch of an
Argand diagram to locate the correct quadrant for the argument.

79
So arg 4 − 4i( ) =
−π
4
Therefore the complex number 4 − 4i can be expressed as
4 2,
−π
4




It is quite convenient to work using the polar form of a complex
number when solving z5
= 4 − 4i.
Let z = r, θ[ ], then z5
= r5
, 5θ[ ].
So as to obtain all five roots of the equation, the argument is
considered to be 2nπ −
π
4
where n is an integer.
Equating the results
r5
, 5θ[ ]= 4 2, 2nπ −
π
4




r5
= 4 2 ⇒ r = 2
5θ = 2nπ −
π
4
⇒ θ = 8n −1( )
π
20
Now choose the five appropriate values of n so that θ lies
between −π and π.
n = −2 ⇒ θ =
−17π
20
n = −1 ⇒ θ =
−9π
20
n = 0 ⇒ θ =
−π
20
n = 1 ⇒ θ =
7π
20
n = 2 ⇒ θ =
15π
20
or
3π
4
The solutions in exponential form are therefore
2 e
− 17π
20
i
, 2e
− 9π
20
i
, 2e
− π
20 , 2e
7π
20
i
and 2e
3π
4
i
.
real
imaginary
4 – 4i
4
– 4
Real
Imaginary

80
Activity 21
Show that 1+ i is a root of the equation z4
= −4 and find each of
the other roots in the form a + bi where a and b are real.
Plot the roots on an Argand diagram. By considering the
diagonals, or otherwise, show that the points are at the vertices
of a square. Calculate the area of the square.
Activity 22
Given that k ≠ 1 and the roots of the equation z3
= k are α, β
and γ , use the substitution z =
x − 2( )
x +1( )
to obtain the roots of the
equation
x − 2( )3
= k x +1( )3
Exercise 3D
1. By using de Moivre's theorem, find all solutions
to the following equations, giving your answers
in polar form. Plot each set of roots on an
Argand diagram and comment on the symmetry.
(a) z4
= 16 (b) z3
= −27i (c) z5
= −1
2. Find the cube roots of
(a) 1+ i (b) 2i − 2
giving your answers in exponential form.
3. Using the answers from Question 1(a), determine
the solutions of the equation
x +1( )4
= 16 x −1( )4
giving your answers in the form a + bi.
3.12 Loci in the complex plane
Suppose z is allowed to vary in such a way that z −1 = 2. You
could write z = x + iy and obtain
x −1( )2
+ y2
{ } = 2
or x −1( )2
+ y2
= 4
4. Using the results from Question 1(b), solve the
equation
1+ 27i x +1( )3
= 0
giving your answers in the form a + bi.
5. Solve the equation z3
= i z −1( )3
giving your
answers in the form a + bi.
Plot the solutions on an Argand diagram and
comment on your results.
6. Determine the four roots of the equation
z − 2( )4
+ z +1( )4
= 0
and plot them on an Argand diagram.

81
You can immediately identify this as the cartesian equation of a
circle centre 1, 0( ) and radius 2. In terms of the complex plane,
the centre is 1+ 0i.
This approach could be adopted for most problems and the
exercise is simply one in algebra, lacking any geometrical feel for
the locus.
Instead, if ω is a complex number, you can identify z − ω as the
distance of z from the point represented by ω on the complex
plane. The locus z −1 = 2 can be interpreted as the set of points
that are 2 units from the point 1+ 0i; in other words, a circle
centre 1+ 0i and radius 2.
Activity 23
Illustrate the locus of z in the complex plane if z satisfies
(a) z − 3 + 2i( ) = 5 (b) z − 2 + i = 1+ 3i
(c) z + 2i = 2 (d) z − 4 = 0
Activity 24
Describe the path of a point which moves in a fixed plane so that
it is always the same distance from two fixed points A and B.
Illustrate the locus of z in the case when z satisfies
z + 3 = z − 4i .
You would probably have had some difficulty in writing down a
cartesian equation of the locus in Activity 24, even though you
could describe the locus geometrically.
Activity 25
Describe the locus of z in the case where z moves in such a way
that
z = z + 2 − 2i .
Now try to write down the cartesian equation of this locus which
should be a straight line.
By writing z = x + iy, try to obtain the same result algebraically.
Real
Imaginary
1
Real
Imaginary

82
Activity 26
Investigate the locus of P when P moves in the complex plane
and represents the complex number z which satisfies
z +1 = k z −1
for different values of the real number k.
Why does k = 1 have to be treated as a special case?
Example
The point P represents the complex number z on an Argand
diagram. Describe the locus geometrically and obtain a
cartesian equation for the locus in the cases
(a) z = z − 4
(b) z + z − 4 = 6
(c) z = 2 z − 4
Solution
(a) From your work in Activity 25, you should recognise this
as a straight line. In fact, it is the mediator, or
perpendicular bisector, of the line segment joining the
origin to the point 4 + 0i .
It should be immediately obvious that its cartesian equation
is x = 2 ; however, writing
z = x + iy
z = x + iy = x − 4 + iy
Squaring both sides gives
x2
+ y2
= x − 4( )2
+ y2
leading to
0 = −8x +16
or x = 2 .
(b) You may be aware of a curve that is traced out when the
sum of the distances from two fixed points is constant. You
could try using a piece of string with its ends fastened to
two fixed points. The curve is called an ellipse.
A sketch of the locus is shown opposite.
real
imaginary
4– 1 5
Imaginary
Real
real
imaginary
42
Imaginary
Real

83
You can obtain a cartesian equation by putting z = x + iy
x + iy + x − 4 + iy = 6
So x2
+ y2
+ x − 4( )2
+ y2
( ) = 6
⇒ x − 4( )2
+ y2
= 6 − x2
+ y2
( )[ ]
2
x2
− 8x +16 + y2
= 36 −12 x2
+ y2
( ) + x2
+ y2
12 x2
+ y2
( ) = 20 + 8x
3 x2
+ y2
( ) = 5 + 2x
9 x2
+ y2
( )= 25 + 20x + 4x2
5x2
− 20x + 9y2
= 25
5 x − 2( )2
+ 9y2
= 45
x − 2( )2
9
+
y2
5
= 1
(c) You should have discovered in Activity 26 that the locus
will be a circle when the relationship is of this form. It is
called the circle of Apollonius.
You could possibly sketch the locus without finding the
cartesian equation.
Let z = x + iy
x + iy = 2 x − 4 + iy
x2
+ y2
( ) = 2 x − 4( )2
+ y2
( )
x2
+ y2
= 4 x2
− 8x +16 + y2
( )
0 = 3x2
+ 3y2
− 32x + 64
In order to find the centre and radius you can complete the
square
x2
+ y2
−
32
3
x +
64
3
= 0
x −
16
3




2
+ y2
=
256
9
−
64
3
=
64
9
Centre of circle is at
16
3
+ 0i and radius is
8
3
.
real
imaginary
42 6 Real
Imaginary

84
Activity 27
By recognising the locus
z − 2 = 3 z −10
as the circle of Apollonius, use the idea of simple ratios to
determine the coordinates of the centre and the radius of the
circle.
Check your answer by finding the cartesian equation of the
circle.
Activity 28
By folding a piece of paper, create an angle of 45° and cut it
out. Now mark two fixed points on a piece of paper and
explore the locus of the vertex as you keep the two sides of the
cut-out in contact with the fixed points as shown.
You should find that P moves on the arc of a circle.
Alternatively, when you have a circle and two fixed points A
and B, if you choose a sequence of points P1, P2 ,P3, ... on the
circumference, what do you notice about the angles
AP1B, AP2B, AP3B, etc.?
This is an example of the constant angle locus.
Example
The point P represents z in the complex plane. Find the locus of
P in each of the cases below when z satisfies
(a) argz =
5π
6
(b) arg z − 2 + 3i( ) =
−π
4
(c) arg
z −1
z +1




=
π
4
Solution
(a) The locus is a half-line starting at the origin making an
angle
5
6
π with the real axis.
45o
P
fixed
point
fixed
point
Discover the
locus of the
vertex P
B
P3
P2
P1
A
Fixed
point Fixed
point
real
imaginary
locus
Real
Locus
Imaginary

85
(b) The half-line to be considered here is one which starts at the
point 2 − 3i.
It makes an angle of
π
4
below the real axis as shown opposite.
(c) You need to make use of the fact that
arg
z −1
z +1




= arg z −1( )− arg z +1( )
One possible solution for z is shown in the second diagram
opposite.
By the results of Activity 28 you can see that the locus of z is
the major arc of a circle passing through 1+ 0i and −1+ 0i.
Since the angle at the centre of the circle is twice that on the
circumference, it can be seen that the centre of the circle is at
0 + i and hence the radius of the circle is 2.
The problem can be tackled algebraically but there are
difficulties that can creep in by assuming
arg x + iy( ) = tan−1 y
x




Nevertheless, you can obtain the cartesian equation of the full
circle of which the locus is only part.
Let z = x + iy
arg
z −1
z +1




= arg
x −1+ iy
x +1+ iy






= arg
x −1( )+ iy{ } x +1( )− iy{ }
x +1( )2
+ y2






= arg
x2
−1+ y2
( )+ 2iy
x +1( )2
+ y2








=
π
4
Taking tangents of both sides
2y
x2
−1+ y2 = 1
⇒ x2
+ y2
− 2y = 1
x2
+ y −1( )2
= 2
from which we see the centre is 0 + i and the radius is 2.
Note: this approach does not indicate whether the locus is the
major or minor arc of the circle and so the first approach is
recommended.
real
imaginary
1– 1
i
real
imaginary
1– 1
arg (z + 1)
arg (z – 1)
z
Imaginary
Imaginary
Real
Real
Real1– 1
– 2
– 1
1
Imaginary
– 3
– 4
3 42
2 – 3i
Imaginary
Real
π
4

86
Exercise 3E
1. Sketch the locus of z described by
(a) z + 3− 4i = 5 (b) z + 2 = z − 5 + i
(c) z + 3i = 3 z − i (d) z − 2 + z − 3+ i = 0
2. Describe geometrically and obtain a cartesian
equation for the locus of z in each of the
following cases.
(a) z − 3
2
= 100 (b) z −1
2
+ z − 4
2
= 9
(c) z −1 + z − 4 = 5 (d) z −1 − z − 4 = 1
3.13 Miscellaneous Exercises
1. Find the modulus and argument of the complex
numbers
z1 = 1+ i and z2 = 1− 3i.
Hence find in the form z = r, θ[ ] where
−π < θ ≤ π and r > 0 , the complex numbers
(a) z1 z2 (b)
z1
z2
(c)
z2
z1
(d) z1
2
(e) z2
3
(f)
z1
2
z2
4
2. Express the numbers 1, 3i, − 4, z = 2 + 5i
in the r, θ[ ] form. Hence express in the r, θ[ ]
form
(a)
1
z
(b) 3zi (c)
z
3i
(d) −4z (e)
−4
z
3. Find 3 + i in the r, θ[ ] form. Hence find
(a) 3 + i( )
3
(b) 3 + i( )
8
in the form a + bi.
(c) Find the least value of the positive integer n
for which 3 + i( )
n
is
(i) purely real
(ii) purely imaginary.
4. Find in the form a + bi
(a) 1+ 3i( )
5
(b) 3 − i( )
10
(c) 1− i( )7
by making use of de Moivre's theorem.
3. Describe geometrically and sketch the region on
the complex plane for which
(a) 2 < z − 3+ i ≤ 5 (b)
−π
4
≤ arg z − 2i( )≤
π
3
4. Sketch the loci for which
(a) arg
z +1
z − i




=
3π
2
(b) arg z − 2( )3
=
π
2
(c) arg z + 2( )− arg z − 3( ) =
π
3
(d) arg
z − 5 + 7i
z +1+ i




=
π
2
5. Simplify 1+ i( )10
− 1− i( )10
.
Given that n is a positive integer, show that
1+ i( )4n
− 1− i( )4n
= 0.
6. Given that z =
3
2
+
1
2
i, simplify z2
, z3
, z4
and
illustrate each of these numbers as points on an
Argand diagram.
7. Show that the three roots of z3
= 1 can be
expressed in the form 1, ω, ω 2
.
Hence show that 1+ ω + ω 2
= 0.
Using this relation and the fact that ω3
= 1,
simplify the following
(a) 1+ ω( )7
(b) 1− ω( ) 1− ω 2
( )
(c)
ω5
1+ ω
(d) 1− ω + ω 2
( )
4
(e) ω − ω 2
( )
5
(f)
1+ ω 2
( ) 1− ω( )
1+ ω( )
8. The roots of the equation z2
+ 4z + 29 = 0 are z1
and z2 . Show that z1 = z2 and calculate, in
degrees, the argument of z1 and the argument
of z2 .
In an Argand diagram, O is the origin and z1 and
z2 are represented by the points P and Q.
Calculate the radius of the circle passing
throught the points O, P and Q. (AEB)

87
9. Sketch on an Argand diagram the loci given by
z −1− 2i = 5
z − 5 + i = z + 3− 5i
Show that these loci intersect at the point z1
where z1 = −2 − 2i, and at a second point z2 . Find
z2 in the form a + bi , where a and b are real.
Express z1 in the form r cosα + isinα( ) where
r > 0 and − π < α ≤ π, giving the value of r and
the value of α . Show that z1 is a root of the
equation z4
+ 64 = 0.
Express z4
+ 64 in the form
z2
+ Az + B( ) z2
+ Cz + D( )
where A, B, C and D are real, and find these
numbers. (AEB)
10. (a) Find the modulus and argument of the
complex number
3 + i
1+ i 3
giving the
argument in radians between −π and π.
(b) Find the value of the real number λ in the
case when
3 + iλ
1+ i 3
is real. (AEB)
11. The complex number u = −10 + 9i
(a) Show the complex number u on an Argand
diagram.
(b) Giving your answer to the nearest degree,
calculate the argument of u.
(c) Find the complex number v which satisfies
the equation
uv = −11+ 28i.
(d) Verify that u + v = 8 2. (AEB)
12. (a) The complex number z satisfies the
equation z +1 = 2 z −1. The point P
represents z on an Argand diagram. Show
that the locus of P is a circle with its centre
on the real axis, and find its radius.
(b) Find the four roots of the equation
z +1( )4
+ 4 z −1( )4
= 0,
expressing the roots z1 , z2 , z3 and z4 in the
form a + bi.
Show that the points on an Argand diagram
representing z1 , z2 , z3 and z4 are the
vertices of a trapezium and calculate its area.
(AEB)
13. Let z be the complex number −1+ 3i.
(a) Express z2
in the form a + bi.
(b) Find the value of the real number p such that
z2
+ pz is real.
(c) Find the value of the real number q such that
Arg z2
+ qz( )=
5π
6
. (AEB)
14. Use the method of mathematical induction to
prove that
cosθ + isinθ( )n
= cosnθ + isinnθ,
where n is a positive integer.
Deduce that the result is also true when n is a
negative integer.
Show that
2cosnθ = zn
+ z−n
,
where z = cosθ + isinθ.
By considering z + z−1
( )
4
, show that
cos4
θ ≡
1
8
cos4θ + 4cos2θ + 3( ).
Hence evaluate cos4
0
π
6
∫ 2θ dθ. (AEB)
15. You are given the complex number
ω = cos
2π
5
+ isin
2π
5
.
(a) Write down the value of ω5
and prove that
1+ ω + ω 2
+ ω3
+ ω 4
= 0.
Simplify ω + ω4
( ) ω2
+ ω3
( ).
Form a quadratic equation with integer
coefficients having roots
ω + ω4
( ) and ω2
+ ω3
( )
and hence prove that
cos
2π
5
=
−1+ 5
4
.
(b) In an Argand diagram the point P is
represented by the complex number z.
Sketch and describe geometrically in each
case, the locus of the point P when
(i) z − ω = z −1
(ii) arg
z − ω
z −1




=
π
5
.
(AEB)

88
16. (a) Use de Moivre's theorem to show that
3 − i( )
n
= 2n
cos
nπ
6
− isin
nπ
6




,
where n is an integer.
(i) Find the least positive integer m for
which 3 − i( )
m
is real and positive.
(ii) Given that 3 − i( ) is a root of the
equation z9
+16 1+ i( )z3
+ a + ib = 0,
find the values of the real constants
a and b.
(b) The point P represents a complex number z
on an Argand diagram and
z − ω6
= 3 z − ω3
,
where ω = 3 − i.
Show that the locus of P is a circle and find
its radius and the complex number
represented by its centre.
(AEB)

3.complex numbers Further Mathematics Zimbabwe Zimsec Cambridge

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