Mathematical Database
COMPLEX NUMBERS
When we are solving quadratic equations with real roots, the roots of the equations exhibit
three cases: two distinct real roots, a double root or no real roots. To accommodate the case of no
real roots, i.e., to provide solutions to those quadratic equations, the concept of complex numbers
was invented. One may expect that as complex numbers are never encountered in our daily routines,
the studies of complex numbers are impractical. However, complex numbers are actually extremely
useful tools in many applications like factorization and proofs of trigonometric identities. In this set
of notes, we will illustrate the basic concepts and some of the applications of complex numbers with
examples.
1. Basic Terms and Concepts
As noted, the interests in complex numbers originated from studies of the roots of quadratic
equations. As an illustration, consider the equation x 2 + 1 = 0 . This equation does not admit real
roots because for all real number x, x 2 + 1 ≥ 0 + 1 = 1 > 0 . Yet, if we set i = −1 , then
i 2 + 1 = −1 + 1 = 0 is a root of the equation.
Definition 1.1. (Imaginary unit)
The imaginary unit is defined by i = −1 .
Based on the imaginary unit, we develop a set of numbers called the complex numbers. The
general form of a complex number z is z = a + bi, where a and b are real numbers.
Definition 1.2 (Real and imaginary parts of a complex number)
For a complex number z = a + bi, where a, b ∈
, a is called the real part of z and b is called the
imaginary part of z. These are denoted as Re(z) = a and Im(z) = b respectively.
Illustration. If z = 3 + 5i, then the real part of z is 3 and the imaginary part is 5. We write Re(z) = 3
and Im(z) = 5.
Illustration. All real numbers x can be represented in the form of a complex number by setting a =
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⊂
x and b = 0. Hence,
numbers.
, where
represents the set of real numbers and
the set of complex
In particular, if Im(z) = 0, then z is a real number. On the contrary, if Re(z) = 0 and Im ( z ) ≠ 0 ,
z is called a purely imaginary number.
In general, we cannot compare complex numbers with each other as in real numbers. For
example, we say 6 > 4, but we cannot say which of 6i and 4i is larger or smaller. However, we can
say that they are unequal. In fact, two complex numbers are said to be equal iff both their real parts
and imaginary parts are equal.
2. Arithmetic Operations
Basic arithmetic operations on complex numbers include addition, subtraction, multiplication
and division. For two complex numbers z1 = a1 + b1i and z2 = a2 + b2i , we have
z1 + z2 = ( a1 + b1i ) + ( a2 + b2i ) = ( a1 + a2 ) + ( b1 + b2 ) i ,
z1 − z2 = ( a1 + b1i ) − ( a2 + b2i ) = ( a1 − a2 ) + ( b1 − b2 ) i ,
z1 z2 = (a1 + b1i )(a2 + b2i ) = a1a2 + a1b2i + b1a2i + b1ib2i
= a1a2 + a1b2i + a2b1i + b1b2i 2 = (a1a2 − b1b2 ) + (a1b2 + a2b1 )i,
and if z2 ≠ 0 ,
z1 a1 + b1i ( a1 + b1i )( a2 − b2i ) ( a1a2 + b1b2 ) + ( a2b1 − a1b2 ) i
=
=
=
2
z2 a2 + b2i ( a2 + b2i )( a2 − b2i )
a2 2 − ( b2i )
=
( a1a2 + b1b2 ) + ( a2b1 − a1b2 ) i = a1a2 + b1b2 + a2b1 − a1b2 i.
a2 2 + b2 2
a2 2 + b2 2
a2 2 + b2 2
Example 2.1.
Suppose zk = 3− k + 2− k i for k = 0, 1, 2, … . Evaluate
∑
∞
k =0
zk .
Solution.
Evaluating the real and imaginary parts separately and applying the formula for geometric series,
we have
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∞
∞
∞
∞
k =0
k =0
k =0
k =0
∑ zk = ∑ ( 3− k + 2− k i ) = ∑ 3− k + i∑ 2− k =
1
1−
+
1
1
1
1−
3
2
i=
3
+ 2i .
2
Example 2.2.
Suppose zk = ak + bk i for k = 1, 2, where ak , bk ∈
. Let z1 = 3 z2 and z1 = (1 + 2i ) . Find a2 and b2 .
4
Solution.
We shall evaluate z1 first:
4
2
2
2
z1 = (1 + 2i ) = ⎡(1 + 2i ) ⎤ = (1 + 4i + 4i 2 ) = (1 − 4 + 4i ) = ( −3 + 4i )
⎣
⎦
= 9 − 24i + 16i 2 = 9 − 24i − 16 = −7 − 24i
2
2
Then, −7 − 24i = z1 = 3 z2 = 3 ( a2 + b2i ) = 3a2 + 3b2i . Equating the real and imaginary parts, we have
a2 = −
7
and b2 = −8 .
3
Example 2.3.
Express the complex number
1 + i tan θ
in the general form.
1 − i tan θ
Solution.
1 + i tan θ (1 + i tan θ )(1 + i tan θ ) 1 + 2i tan θ + ( i tan θ ) 1 − tan 2 θ + 2i tan θ
=
=
=
2
1 − i tan θ (1 − i tan θ )(1 + i tan θ )
1 + tan 2 θ
12 − ( i tan θ )
2
1 − tan 2 θ + 2i tan θ 1 − tan 2 θ 2i tan θ
=
=
+
= ( cos 2 θ − sin 2 θ ) + 2i sin θ cos θ
2
2
2
sec θ
sec θ
sec θ
= cos 2θ + i sin 2θ
Exercise
1
in the general form.
cos θ − i sin θ
1.
Express
2.
Suppose z = a + 2i is a root of the equation x 2 + 6 x + k = 0 , where both a and k are real. Find a
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and k.
3. The Argand Diagram, Modulus and Argument
Maybe it is too abstract to just talk about complex numbers. A better understanding of this
number system can be obtained by drawing diagrams. In this section, we will discuss the Argand
diagrams of complex numbers. The Argand diagrams are also of core importance in the application
of complex numbers on some geometry problems.
An Argand diagram consists of two axes: a horizontal one, like the x-axis in the rectangular
coordinate system, called the real axis, and a vertical one, like the y-axis, called the imaginary axis.
The point (x, y) in an Argand diagram represents the complex number x + yi.
In figure 1, the complex numbers represented by the points A, B, C and D are 1 + i, 2 – 3i, 3 –
2i and -3 + i respectively. By convention the number zero is represented by the origin O. Another
way to interpret Argand diagrams is by way of vectors. For example, the complex numbers
mentioned above are said to be associated with the vectors OA , OB , OC and OD respectively.
Readers might have noticed that 1 + i + 2 – 3i = 3 – 2i, which in the notation of vectors is
represented by OA + OB = OA + AC = OC .
Im
3
2
A
D
-4
1
-3
-2
-1
0
1
2
3
4 Re
-1
-2
C
-3
B
-4
Figure 1: Argand Diagram
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From the perspectives of the Argand diagram and vectors, we can define some more properties
of complex numbers. The length of the vector in the Argand diagram, r, is called the modulus of the
associated complex number, while the angle the vector made with positive real axis, θ , is said to be
an argument of the complex number. Hence we have, if z ≠ 0 ,
x + yi = r = x 2 + y 2 ,
x = r cos θ and y = r sin θ .
An immediate result of the above is that tan θ =
y
.
x
Since cos θ = cos (θ + 2kπ ) and sin θ = sin (θ + 2kπ ) for any integer k, there are infinitely
many possible values for the argument of a complex number. And this leads to the following
definition.
Definition 3.1.
The principal value of the argument of a complex number z = x + yi is the angle θ such that
r cos θ = x , r sin θ = y and −π < θ ≤ π .
¾
We usually denote the principal argument by Arg z, and the set of arguments by arg z = Arg z +
2kπ .
Example 3.1.
Find the modulus and the principal value of the argument of the complex number 1 + i.
Solution.
We now have x = y = 1. Hence, modulus = r = 12 + 12 = 2 .
π
3π
⎛ y⎞
in the interval ( −π , π ] . We
and −
Besides, tan −1 ⎜ ⎟ = tan −1 1 has only two possible values,
4
4
⎝x⎠
π
π
⎛ 3π ⎞
⎛ 3π ⎞
have r cos = 1 = x and r sin = 1 = y , but r cos ⎜ − ⎟ = −1 ≠ x and r sin ⎜ − ⎟ = −1 ≠ y .
4
4
⎝ 4 ⎠
⎝ 4 ⎠
Hence, the principal value of the argument of the complex number 1 + i is
π
4
.
Exercise
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1.
Find the moduli and principal arguments of the following complex numbers:
3 +i
(a)
(b) 1 − 3i
−3 − 3i
(c)
(d) sin θ − i cos θ , where 0 ≤ θ ≤ π2 .
2.
π
π ⎞⎛ π
π⎞
⎛
(a) Evaluate ⎜ cos + i sin ⎟⎜ cos + i sin ⎟ .
6
6 ⎠⎝
3
3⎠
⎝
(b) Suppose z1 = cos
π
+ i sin
π
and z2 = cos
π
+ i sin
π
. Draw the points Z1 and Z 2 ,
6
6
2
2
representing z1 and z2 respectively, on the Argand diagram. What is the relationship
between OZ1 and OZ 2 ?
4. Other Forms of Representation
The concepts of modulus and argument are important in dealing with representations of
complex numbers. In this section, we will introduce the polar form and the exponential form, both
of which are associated with representing complex numbers in terms of their moduli and arguments.
The Polar Form
Definition 4.1. (The polar form)
Suppose r and θ are respectively the modulus and argument of the complex number z. Then the
polar form of z is r ( cos θ + i sin θ ) , or abbreviated as rcisθ .
Illustration. The polar form of the complex number 1 + i is
2 cis
π
4
π
π⎞
⎛
2 ⎜ cos + sin ⎟ , or in short,
4
4⎠
⎝
.
Example 4.1.
Express, in polar form, the following complex numbers:
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(a)
3 +i
(b) 1 + sin θ − i cos θ for −
π
2
≤θ ≤
3π
.
2
Solution.
(a) Let z = 3 + i . Then r =
( 3)
2
tan θ =
+ 12 = 3 + 1 = 2 .
1
5π
π
(rejected) or
⇒θ = −
6
6
3
π
π⎞
⎛
Hence, the polar form of z is rcisθ = 2 ⎜ cos + i sin ⎟ .
6
6⎠
⎝
(b) Let z = 1 + sin θ − i cos θ . Then,
⎡
⎛π
⎞
⎛π
⎞
⎛π θ ⎞ ⎤
⎛π θ ⎞
⎛π θ ⎞
z = 1 + cos ⎜ − θ ⎟ − i sin ⎜ − θ ⎟ = 1 + ⎢ 2 cos 2 ⎜ − ⎟ − 1⎥ − 2i sin ⎜ − ⎟ cos ⎜ − ⎟
⎝2
⎠
⎝2
⎠
⎝ 4 2⎠ ⎦
⎝ 4 2⎠
⎝ 4 2⎠
⎣
⎛π θ ⎞⎡ ⎛π θ ⎞
⎛ π θ ⎞⎤
⎛π θ ⎞⎡ ⎛θ π ⎞
⎛ θ π ⎞⎤
= 2 cos ⎜ − ⎟ ⎢cos ⎜ − ⎟ − i sin ⎜ − ⎟ ⎥ = 2 cos ⎜ − ⎟ ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ .
⎝ 4 2 ⎠⎣ ⎝ 4 2 ⎠
⎝ 4 2 ⎠⎦
⎝ 4 2 ⎠⎣ ⎝ 2 4 ⎠
⎝ 2 4 ⎠⎦
3π
π π θ π
⎛π θ ⎞
, we have − ≤ − ≤ , i.e., 2 cos ⎜ − ⎟ ≥ 0 . Therefore, the polar
2
2
2 4 2 2
⎝ 4 2⎠
⎛π θ ⎞⎡ ⎛θ π ⎞
⎛ θ π ⎞⎤
form of z is 2 cos ⎜ − ⎟ ⎢ cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ .
⎝ 4 2 ⎠⎣ ⎝ 2 4 ⎠
⎝ 2 4 ⎠⎦
Since −
π
≤θ ≤
The polar form of representation is especially useful when we perform multiplication and
division of complex numbers.
Theorem 4.1.
Let z1 and z2 be two complex numbers. Then for arg ( zi ) ≠ 0 ,
1.
2.
z1 z2 = z1 z2 and arg ( z1 z2 ) = arg ( z1 ) + arg ( z2 ) .
If z2 ≠ 0 ,
z
⎛z ⎞
z1
= 1 and arg ⎜ 1 ⎟ = arg ( z1 ) − arg ( z2 ) .
z2
z2
⎝ z2 ⎠
Proof. Let z1 = r1cisθ1 and z2 = r2 cisθ 2 be the polar form representations.
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For property 1, using the above notations,
z1 z2 = r1 ( cos θ1 + i sin θ1 ) ⋅ r2 ( cos θ 2 + i sin θ 2 )
= r1r2 ( cos θ1 cos θ 2 + i sin θ1 cos θ 2 + i cos θ1 sin θ 2 + i 2 sin θ1 sin θ 2 )
= r1r2 ⎡⎣( cos θ1 cos θ 2 − sin θ1 sin θ 2 ) + i ( sin θ1 cos θ 2 + cos θ1 sin θ 2 ) ⎤⎦
= r1r2 ⎡⎣cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) ⎤⎦ = r1r2 cis (θ1 + θ 2 ) .
Hence, z1 z2 = r1r2 = z1 z2 and arg ( z1 z2 ) = θ1 + θ 2 = arg ( z1 ) + arg ( z2 ) .
For property 2, let z1 ' =
z1
. Then using the results of the above proof, we have
z2
z1 ' z2 = z1 ' z2 =
On the other hand, z1 ' z2 =
⎛z ⎞
z1
z2 and arg ( z1 ' z2 ) = arg ( z1 ') + arg ( z2 ) = arg ⎜ 1 ⎟ + arg ( z2 ) .
z2
⎝ z2 ⎠
z1
⋅ z2 = z1 . Hence,
z2
z 'z
z
⎛z ⎞
z1
= 1 2 = 1 and arg ⎜ 1 ⎟ = arg ( z1 ' z2 ) − arg ( z2 ) = arg ( z1 ) − arg ( z2 ) .
z2
z2
z2
⎝ z2 ⎠
Q.E.D.
Definition 4.2. (The exponential form)
Suppose r and θ are respectively the modulus and argument of the complex number z. Then the
exponential form of z is reiθ .
Theorem 4.1. can be applied to the exponential form easily with the law of indices. For
property 1, by definition 4.2, z1 z2 = r1eθ1 ir2 eθ2 = r1r2 ei(θ1 +θ2 ) . Hence, z1 z2 = r1r2 = z1 z2 and
arg ( z1 z2 ) = θ1 + θ 2 = arg ( z1 ) + arg ( z2 ) Similarly, for property 2,
z1 r1eiθ1 r1 i(θ1 −θ2 )
=
= e
. So,
z2 r2 eiθ2 r2
z
⎛z ⎞
z1 r1
= = 1 and arg ⎜ 1 ⎟ = θ1 − θ 2 = arg ( z1 ) − arg ( z2 ) .
z2 r2 z2
⎝ z2 ⎠
Theorem 4.2.
1.
eiθ + e −iθ = 2 cos θ .
2.
eiθ − e− iθ = 2i sin θ .
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Proof. For property 1, eiθ + e −iθ = ( cos θ + i sin θ ) + ⎡⎣cos ( −θ ) + i sin ( −θ ) ⎤⎦ = 2 cos θ .
For property 2, eiθ − e− iθ = ( cos θ + i sin θ ) − ⎡⎣ cos ( −θ ) + i sin ( −θ ) ⎤⎦ = 2i sin θ .
Q.E.D.
Example 4.2.
Simplify the fraction
1 + sin θ + i cos θ
.
1 − sin θ + i cos θ
Solution.
Notice that
⎛π
⎞
⎛π
⎞
⎛π
⎞ i⎜ −θ ⎟
sin θ + i cos θ = cos ⎜ − θ ⎟ + i sin ⎜ − θ ⎟ = e ⎝ 2 ⎠ ,
⎝2
⎠
⎝2
⎠
⎛
π⎞
π⎞
π ⎞ i⎜θ − ⎟
⎛π
⎞
⎛π
⎞
⎛
⎛
sin θ − i cos θ = cos ⎜ − θ ⎟ − i sin ⎜ − θ ⎟ = cos ⎜ θ − ⎟ + i sin ⎜ θ − ⎟ = e ⎝ 2 ⎠ .
2⎠
2⎠
⎝2
⎠
⎝2
⎠
⎝
⎝
Hence,
⎡ i⎛⎜ θ2 − π4 ⎞⎟ i⎛⎜ π4 −θ2 ⎞⎟ ⎤
⎛θ π ⎞
⎞
e
⎢e ⎝ ⎠ + e ⎝ ⎠ ⎥
i ⎜ −θ ⎟
− ⎟
⎛π
⎞ 2 cos ⎜
⎝2 ⎠
−
i
θ
⎜
⎟
1 + sin θ + i cos θ 1 + e
⎢⎣
⎥⎦
2 4⎠
⎝
⎝2 ⎠
=
= ⎛θ π ⎞ ⎛ π θ ⎞
=e
⎛ π⎞
⎛θ π ⎞
i⎜θ − ⎟
1 − sin θ + i cos θ
i⎜ − ⎟ ⎡ i⎜ − ⎟
i⎜ − ⎟ ⎤
⎛π θ ⎞
2
2i sin ⎜ − ⎟
1 − e ⎝ ⎠ e ⎝ 2 4 ⎠ ⎢e ⎝ 4 2 ⎠ − e ⎝ 2 4 ⎠ ⎥
⎝ 4 2⎠
⎢⎣
⎥⎦
⎛θ π ⎞
− ⎟
⎛π
⎞ cos ⎜
⎛π
⎞
⎛π
⎞
π
i ⎜ −θ ⎟
i ⎜ −θ ⎟
2 4⎠
⎛ θ π ⎞ i⎜⎝ 2 −θ ⎟⎠ i i 2
⎛θ π ⎞
⎝
⎝2 ⎠
⎝2 ⎠
=e
⋅ i cot ⎜ − ⎟ = e
⋅ e cot ⎜ − ⎟
=e
⎛θ π ⎞
⎝2 4⎠
⎝2 4⎠
−i sin ⎜ − ⎟
⎝2 4⎠
⎛θ π ⎞
⎛θ π ⎞
= ei(π −θ ) cot ⎜ − ⎟ = cot ⎜ − ⎟ cis (π − θ ) .
⎝2 4⎠
⎝2 4⎠
⎛π
⎛π θ ⎞
i⎜ − ⎟
⎝ 4 2⎠
Exercise
1.
Simplify the fraction
3 + 3 3i
.
1− i
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2.
Prove that ( cos θ − i sin θ ) = cos 3θ − i sin 3θ .
3
5. Complex Conjugates
Definition 5.1. (Complex conjugate)
Suppose z = a +bi is a complex number, where a, b ∈
. Then the complex conjugate of z is defined
as z = a − bi .
Thus the complex conjugate of a complex number is the mirror image of the original number
across the real axis. There are some nice properties associated with complex conjugate pairs.
Theorem 5.1.
For any complex number z, we have
1.
z = z.
2.
arg ( z ) = − arg ( z ) .
3.
z + z = 2 Re ( z ) .
4.
z − z = 2i Im ( z ) .
5.
zz = z .
6.
z = z.
2
Proof. The theorem follows easily from the definition and its proof is left as an exercise.
Q.E.D.
An implication of this theorem is that if z is real, then z − z = 2i Im ( z ) = 0 . Likewise, if z is
purely imaginary, then z + z = 2 Re ( z ) = 0 . The converses of these are also true, with the only
exception that the number zero is real but 2Re(0) = 0. These facts are often useful in proving some
of the properties of some numbers in questions.
Example 5.1.
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Suppose w = 1 and z = w +
1
. Prove that z is a real number.
w
Solution.
By property 5 of theorem 5.1, we have ww = w = 12 = 1 . Hence, w =
2
z = w+
1
as w ≠ 0 . Therefore,
w
1
= w + w = 2 Re ( w ) must be a real number.
w
Theorem 5.2.
For complex numbers z1 and z2 , we have
1.
z1 + z2 = z1 + z2 .
2.
z1 − z2 = z1 − z2 .
3.
z1 ⋅ z2 = z1 ⋅ z2 .
4.
⎛ z1 ⎞ z1
⎜ ⎟ = , if z2 ≠ 0 .
⎝ z2 ⎠ z 2
Proof. The proofs of the first two properties are left as an exercise.
For property 3, we let z1 = a1 + b1i and z2 = a2 + b2i . Then z1 = a1 − b1i and z2 = a2 − b2i . Hence,
z1 ⋅ z2 = ( a1 − b1i )( a2 − b2i ) = ( a1a2 − b1b2 ) − ( a1b2 + a2b1 ) i
= ( a1a2 − b1b2 ) + ( a1b2 + a2b1 )i = ( a1 + b1i )( a2 + b2i ) = z1 z2 .
For property 4, using previous results,
⎛z ⎞
z
z2 ⋅ ⎜ 1 ⎟ = z2 ⋅ 1 = z1 .
z2
⎝ z2 ⎠
And the theorem follows.
Q.E.D.
Example 5.2.
Let z = x + yi. If z − 6 = 5 and z = 5 , find the possible values of x and y.
Solution.
From property 5 of theorem 5.1 and property 3 of theorem 5.2,
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(
)
52 = z − 6 = ( z − 6 ) z − 6 = ( z − 6 )( z − 6 ) ,
2
25 = zz − 6 z − 6 z + 36 = z − 6 ( z + z ) + 36
2
= 52 − 6 ⋅ 2 Re ( z ) + 36.
Hence, 12 Re ( z ) = 36 ⇒ x = Re ( z ) = 3 .
On the other hand, 5 = z = x 2 + y 2 = 32 + y 2 . Thus, y 2 = 52 − 32 = 42 . The possible values of y
are therefore ±4 .
We started our discussion of complex numbers on solving algebraic equations. In fact, there is
a nice relationship between the complex conjugates in solving equations.
Theorem 5.3.
If z is a zero of a real polynomial (polynomial with real coefficients), then so is z .
Proof. Suppose z is a zero of the real polynomial an x n + an −1 x n −1 +
5.2, ( z ) = z ⋅ z ⋅
k
⋅z = z⋅z⋅
a0 . From property 3 of theorem
⋅ z = z k for positive integer k. Besides, since the ak ’s are real, we
have ak = ak . Then,
an ( z ) + an −1 ( z )
n
n −1
+
a0 = an z n + an −1 z n −1 +
= an z n + an −1 z n −1 +
a0 = an z n + an −1 z n −1 +
a0
a0 = 0 = 0
Thus, z is also a zero of the polynomial.
Q.E.D.
This theorem implies that for a real polynomial, either both members of a complex conjugate
pair are the zeros, or both are not. In fact, a real polynomial can only have an even number of
complex zeros.
Example 5.3.
Given that 3 + i is a root of the equation x 4 − 5 x3 + 10 x 2 + ax + b = 0 , where a and b are real
numbers. Find a and b.
Solution.
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Since 3 + i is a root, its conjugate, 3 – i, is also a root. By factor theorem, x 4 − 5 x3 + 10 x 2 + ax + b is
divisible by both x – (3 + i) and x – (3 – i), i.e., by ( x − 3 − i )( x − 3 + i ) = x 2 − 6 x + 10 .
Upon long division by x 2 − 6 x + 10 , the remainder of x 4 − 5 x3 + 10 x 2 + ax + b
( a + 26 ) x + ( b − 60 ) . Thus, we have a + 26 = 0 and b – 60 = 0, i.e., a = –26 and b = 60.
is
Exercise
1.
Prove theorem 5.1.
2.
Prove the first two properties of theorem 5.2.
3.
Let k be a real constant and z be a complex number with z = 1 . Prove that z + k = kz + 1 .
4.
Given that 1 + 3i is a root of the equation x 4 + 3x 2 + ax + b = 0 . Find a and b and factorize
x 4 + 3x 2 + ax + b .
6. De Moivre’s Theorem
Performing multiplications and divisions on complex numbers can be very tedious and errorprone, as we have to deal with both the real and imaginary parts. If we employ the polar form and
exponential form, the operations will be much simpler. Furthermore, if the operations involve
rational powers of complex numbers, the operations can be eased with the De Moivre’s theorem.
Theorem 6.1. (De Moivre’s Theorem for Integral Index)
For any integer n, ( cos θ + i sin θ ) = cos nθ + i sin nθ .
n
Proof. We shall prove the theorem for non-negative n by induction first. For n = 0 or 1, the result is
obvious.
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Suppose for n = k, ( cisθ ) = ciskθ . Then when n = k+1, by theorem 4.1,
k
( cisθ )
k +1
= ( cisθ ) cisθ = ciskθ cisθ = cis ( k + 1)θ .
k
So, by mathematical induction, the theorem is true for all non-negative integer n.
For negative n, let n = –m . So m is positive. Then,
( cisθ )
n
= ( cisθ )
−m
=
1
( cisθ )
m
=
1
cis0
=
= cis ( 0 − mθ ) = cis ( − mθ ) = cisnθ .
cismθ cismθ
Thus the theorem is true for all integer n.
Q.E.D.
Example 6.1.
Redo Example 2.3 using De Moivre’s theorem.
Solution.
If cos θ = 0 , then θ = kπ +
π
sin 2θ = sin ( 2kπ + π ) = 0 . So,
2
. Hence, we have tan θ = ∞ , cos 2θ = cos ( 2kπ + π ) = −1 and
1 + i tan θ
= −1 = cos 2θ + i sin 2θ .
1 − i tan θ
For cos θ ≠ 0 ,
1 + i tan θ (1 + i tan θ )( cos θ ) cos θ + i sin θ
cos θ + i sin θ
=
=
=
1 − i tan θ (1 − i tan θ )( cos θ ) cos θ − i sin θ cos ( −θ ) + i sin ( −θ )
=
−1
cisθ
2
= cisθ ⎡⎣cis ( −θ ) ⎤⎦ = cisθ ( cisθ ) = ( cisθ ) = cis2θ
cis ( −θ )
= cos 2θ + i sin 2θ .
The De Moivre’s theorem for integral index is often used in proving trigonometric identities.
In fact, many identities can be proved with the help of some direct results of the theorem: for
z = cisθ ,
zn +
1
= cisnθ + cis ( − nθ ) = 2 cos nθ ,
zn
zn −
1
= cisnθ − cis ( − nθ ) = 2i sin nθ .
zn
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Example 6.2.
Show that cos 5θ = cos5 θ − 10 cos3 θ sin 2 θ + 5cos θ sin 4 θ .
Hence prove that cos 5θ = 16 cos5 θ − 20 cos3 θ + 5cos θ .
Solution.
According to the De Moivre’s theorem, ( cos θ + i sin θ ) = cos 5θ + i sin 5θ . However, by the
5
binomial theorem,
( cos θ + i sin θ )
5
= cos5 θ + 5cos 4 θ ( i sin θ ) + 10 cos3 θ ( i sin θ ) + 10 cos 2 θ ( i sin θ ) + 5cos θ ( i sin θ ) + ( i sin θ )
2
3
4
5
= cos5 θ + 5i cos 4 θ sin θ − 10 cos3 θ sin 2 θ − 10i cos 2 θ sin 3 θ + 5cos θ sin 4 θ + i sin 5 θ .
Equating the real parts of both equations, we get cos 5θ = cos5 θ − 10 cos3 θ sin 2 θ + 5cos θ sin 4 θ .
Hence, using the identity sin 2 θ + cos 2 θ = 1 , we have
cos 5θ = cos5 θ − 10 cos3 θ sin 2 θ + 5cos θ sin 4 θ
= cos5 θ − 10 cos3 θ (1 − cos 2 θ ) + 5cos θ (1 − cos 2 θ )
2
= 16 cos5 θ − 20 cos3 θ + 5cos θ .
Example 6.3.
Let z = cisθ . By using the relations z +
sin 2 θ cos 4 θ = −
1
1
= 2 cos θ and z − = 2i sin θ , deduce that
z
z
1
1
1
1
cos 6θ − cos 4θ + cos 2θ + .
32
16
32
16
Solution.
Substituting with the given relations,
2
4
1
1
( 2i sin θ ) ( 2 cos θ ) = ⎛⎜ z − ⎞⎟ ⎛⎜ z + ⎞⎟
z⎠ ⎝
z⎠
⎝
−64sin 2 θ cos 4 θ = z 6 + 2 z 4 − z 2 − 4 − z −2 + 2 z −4 + z −6
2
4
= ( z 6 + z −6 ) + 2 ( z 4 + z −4 ) − ( z 2 + z −2 ) − 4
= 2 cos 6θ + 4 cos 4θ − 2 cos 2θ − 4.
Hence,
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sin 2 θ cos 4 θ = −
1
1
1
1
cos 6θ − cos 4θ + cos 2θ + .
32
16
32
16
The De Moivre’s theorem can be generalized to rational index.
Theorem 6.2. (De Moivre’s Theorem for Rational Index)
For any positive integer n,
1
( cisθ ) n = cis
2 kπ + θ
n
where k = 0, 1, 2, … , n – 1.
1
Proof. We let rcisα = ( cisθ ) n , where r is the modulus of the complex number. Then
r n cisnα = ( rcisα ) = cisθ . Taking the modulus on both sides, we get r n = 1 , i.e., r = 1 as r must be
n
positive. So cisθ = cisnα , implying cos θ = cos nα and sin θ = sin nα .
Therefore, nα = θ + 2kπ , where k = 0, 1, 2, … , n – 1. Thus, α =
1
( cisθ ) n = rcisα = cis
2kπ + θ
, i.e.,
n
2 kπ + θ
.
n
Q.E.D.
Furthermore, if m and n are relatively prime integers and n is positive, we have
( cisθ )
m
n
1
2kπ + mθ
= ⎡( cisθ ) ⎤ = ( cismθ ) n = cis
⎣
⎦
n
m
1
n
where k = 0, 1, 2, … , n – 1.
With the De Moivre’s theorem for rational index, we can easily evaluate the nth roots of
complex numbers without dealing with tedious multiplications. We will illustrate this with an
example.
Example 6.4.
Find the cube roots of the complex number 1 + 3i .
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Solution.
First of all, we transform the complex number into its polar form:
⎛1
3 ⎞
π
i ⎟⎟ = 2cis .
1 + 3i = 2 ⎜⎜ +
3
⎝2 2 ⎠
Thus,
(1 + 3i )
1
3
1
1
π ⎞3
⎛
⎛ π ⎞3
= ⎜ 2cis ⎟ = 2 3 ⎜ cis ⎟ = 2 3 cis
3⎠
3⎠
⎝
⎝
1
1
2 kπ +
π
3
3
for k = 0, 1, 2, i.e.,
(
1 + 3i
)
1
3
1
= 2 3 cis
π
9
1
or 2 3 cis
1
7π
13π
.
or 2 3 cis
9
9
The nth roots of the number 1 are given the special name the nth roots of unity.
Definition 6.1. (nth roots of unity)
The n distinct roots of the equation z n = 1 are called the nth roots of unity.
Suppose w is a nth root of unity. Thus wn = 1 = cis0 . Hence, w = cis
2 kπ + 0
2 kπ
, where k
= cis
n
n
= 0, 1, 2, … , n – 1. Furthermore, if w ≠ 1 ,
1 + w + w2 +
wn −1 =
wn − 1
0
=
=0.
w −1 w −1
Example 6.5.
Factor the polynomial z 7 − 1 into a product of linear and quadratic polynomials with real
coefficients.
Solution.
The roots of the equation z 7 − 1 = 0 are the 7th roots of unity:
1, cis
2π
4π
6π
8π
10π
12π
, cis
, cis
, cis
, cis
and cis
.
7
7
7
7
7
7
Then,
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2π ⎞ ⎛
4π ⎞⎛
6π ⎞⎛
8π ⎞ ⎛
10π ⎞ ⎛
12π ⎞
⎛
z 7 − 1 = ( z − 1) ⎜ z − cis
⎟ ⎜ z − cis
⎟⎜ z − cis
⎟⎜ z − cis ⎟ ⎜ z − cis
⎟ ⎜ z − cis
⎟
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠
⎝
2π ⎞ ⎛
4π ⎞⎛
6π ⎞⎛
−6π ⎞ ⎛
−4π ⎞ ⎛
−2π ⎞
⎛
= ( z − 1) ⎜ z − cis
⎟ ⎜ z − cis
⎟⎜ z − cis
⎟⎜ z − cis
⎟ ⎜ z − cis
⎟ ⎜ z − cis
⎟
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠⎝
7 ⎠
⎝
3
3
−2 k π ⎞
2 kπ
2 kπ ⎞ ⎛
⎛ 2
⎞
⎛
z
−
cis
=
z
−
1
+ 1⎟ .
= ( z − 1) ∏ ⎜ z − cis
(
)
∏
⎟⎜
⎟
⎜ z − 2 zcos
7 ⎠
7
7 ⎠⎝
⎠
k =1 ⎝
k =1 ⎝
Exercise
1.
Solve the equation z 4 + 4 z 2 + 5 = 0 .
2.
Find the fourth roots of the number
3.
By considering the real parts of the equation cos 2nθ + i sin 2nθ = ( cisθ )
n
3 −i .
2n
, prove that
cos 2nθ = ∑ ( −1) C22kn cos 2 n − 2 k θ sin 2 kθ .
k
k =0
7. Applications
As shown in the last section, complex numbers find their use in proving trigonometric
identities. Besides, it is also useful in solving geometry problems. This application arises from the
geometric representation of complex numbers on the Argand diagram.
For instance, the complex number z = x + yi can represent the vector OZ = xi + y j . On the
other hand, the resultant vector from the subtraction of vectors can also be represented by the
subtraction of the corresponding complex numbers. The modulus of the resulting complex number
is the length of the resultant vector and the argument represents the angle between the resultant
vector and the x-axis.
In this section, we will illustrate the applications of complex numbers with a couple of
examples.
Example 7.1.
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If the points A1 , A2 , B1 and B2 are associated with the complex numbers a1 , a2 , b1 and b2
respectively, express the angle between the line segments A1 B1 and A2 B2 in terms of a1 , a2 , b1 and
b2 .
Solution.
Refering to Figure 2,
angle between line segments A1 B1 and A2 B2
= θ = θ1 − θ 2
= angle between A1 B1 and x -axis − angle between A2 B2 and y -axis
= arg ( a1 − b1 ) − arg ( a2 − b2 )
⎛ a −b ⎞
= arg ⎜ 1 1 ⎟
⎝ a2 − b2 ⎠
y
A1
A2
θ
B2
B1
θ2
θ1
x
Figure 2
Example 7.2.
Let O = (0, 0) and A = (6, 8). Prove that for every point P (other than O and A) on the circle
represented by x 2 + y 2 − 6 x − 8 y = 0 , we have OP ⊥ PA .
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Solution.
Let p = x + yi. Then OP is represented by p and PA is represented by p – (6 + 8i). Then, following
from the result of Example 7.1,
OP ⊥ PA
⇔ angle between OP and PA is
π
2
or
3π
2
⎛
⎞ π
3π
p
⇔ arg ⎜⎜
⎟⎟ = or
2
⎝ p − ( 6 + 8i ) ⎠ 2
p
⇔
is purely imaginary
p − ( 6 + 8i )
⇔
⎡
⎤
⎛
⎞
p
p
p
+⎢
⎟⎟ = 0.
⎥ = 2 Re ⎜⎜
−
+
6
8
p − ( 6 + 8i ) ⎣ p − ( 6 + 8i ) ⎦
p
i
(
)
⎝
⎠
Notice that
⎡
⎤
p
p
p
p
x + yi
x − yi
+⎢
+
=
+
⎥=
p − ( 6 + 8i ) ⎣ p − ( 6 + 8i ) ⎦ p − ( 6 + 8i ) ⎡ p − ( 6 + 8i ) ⎤ ( x − 6 ) + ( y − 8 ) i ( x − 6 ) − ( y − 8 ) i
⎣
⎦
=
2 x ( x − 6 ) − 2 y ( y − 8) i 2
( x − 6 ) + ( y − 8)
2
2
=
2 ( x2 + y2 − 6x − 8 y )
( x − 6 ) + ( y − 8)
2
2
= 0.
Hence, OP ⊥ PA .
We conclude this set of notes with a discussion on the triangle inequality, a well-known
theorem which can be readily applied to complex numbers as well.
Theorem 7.1 (Triangle Inequality)
For any two complex numbers z1 and z2 , we have z1 + z2 ≤ z1 + z2 . Equality holds iff Z1 , Z 2 and
O are collinear, and Z1 and Z 2 are on the “same side” of O.
Proof. In a triangle Z1Z 2O , the sum of lengths of any two sides is greater than the length of the
third side, i.e., we have Z1O + Z 2 ' O ≥ Z1Z 2 ' , where equality holds iff Z1 , Z 2 ' and O are collinear,
and Z1 and Z 2 ' are on the opposite side of O. In terms of complex numbers,
z1 + z2 ' ≥ z1 − z2 ' .
Taking z2 ' = − z2 , we have z1 + z2 ≤ z1 + − z2 = z1 + z2 . Note that Z 2 , Z 2 ' and O are always
collinear and Z 2 and Z 2 ' are on the opposite side of O. Hence, equality holds iff Z1 , Z 2 and O are
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collinear and Z1 and Z 2 are on the “same side” of O.
Q.E.D.
¾
In fact, for n ≥ 2 , we have z1 + z2 +
+ zn ≤ z1 + z2 +
+ zn . This follows from induction
on n and theorem 7.1.
Example 7.3.
Suppose wn z + wn −1 z 2 +
+ wz n = 1 for some w such that w ≤ 1 . Prove that z >
1
.
2
Solution.
Suppose z ≤
1
, applying the Triangle Inequality,
2
1 = 1 = wn z + wn −1 z 2 +
= wn z + wn −1 z 2 +
+ wz n ≤ wn z + wn −1 z 2 +
+ w zn = w z + w
n
1
1
1 1 1
≤ 1n ⋅ + 1n −1 ⋅ 2 + + 1⋅ n = + 2 +
2
2
2
2 2
1⎛
1 ⎞
⎜1 − n ⎟
1
2
2 ⎠
= ⎝
= 1 − n < 1.
1
2
1−
2
which is a contradiction. Therefore, we must have z >
+
n −1
+ wz n
z +
2
+w z
n
1
2n
1
.
2
Exercise
1.
For each of the following conditions, find the locus of z (take z = x + yi):
a.
z − ( 2 + 3i ) = 6 .
b.
2z
1
is purely imaginary, where z ≠ − .
3z + 1
3
c.
z −1 = z + 1 + 1.
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a −b x− y
a −b x− y
=
=
or
.
c −b z − y
c −b z − y
2.
Prove that ∆ABC ∼ ∆XYZ iff
3.
Suppose z0 , z1 , … , zn −1 are complex numbers such that zk = cis
n −1
Prove that for any complex number z,
∑ z−z
k =0
k
2kπ
for k = 0, 1, …, n – 1.
n
≥n.
8. Solutions to Exercise
Arithmetic Operations
1.
cos θ + i sin θ .
2.
Since a + 2i is a root of x 2 + 6 x + k = 0 ,
0 = ( a + 2i ) + 6 ( a + 2i ) + k = ( a 2 + 4ai + 4i 2 ) + ( 6a + 12i ) + k
2
= ( a 2 − 4 + 6a + k ) + ( 4a + 12 ) i.
Equating the real and imaginary parts of both sides, we have a 2 − 4 + 6a + k = 0 and 4a + 12 =
0. Therefore, a = –3 and k = 13.
The Argand diagram, modulus and argument
1.
(a) modulus = 2, principal value of argument =
π
6
(b) modulus = 2, principal value of argument = −
.
π
3
.
(c) modulus = 3 2 , principal value of argument = −
(d) modulus = 1, principal value of argument = θ −
2.
π
2
3π
.
4
.
(a) i
(b) OZ 2 is obtained by rotating OZ1 by
π
3
about O anticlockwise.
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Other forms of representation
1.
Converting to the polar form and applying theorem 4.1,
⎛1
3 ⎞
π
6⎜ +
i⎟
6cis
2
2
3 + 3 3i
⎡π
⎝
⎠ =
3
=
= 3 2cis ⎢
1 ⎞
1− i
⎛ 1
⎛ π⎞
⎣3
2cis ⎜ − ⎟
2⎜
i⎟
−
⎝ 4⎠
2 ⎠
⎝ 2
2.
7π
⎛ π ⎞⎤
− ⎜ − ⎟ ⎥ = 3 2cis
.
12
⎝ 4 ⎠⎦
Converting into the exponential form,
( cos θ − i sin θ )
3
= ⎡⎣ cos ( −θ ) + i sin ( −θ ) ⎤⎦ = ⎡⎣ ei( −θ ) ⎤⎦ = ei( −3θ )
= cos ( −3θ ) + i sin ( −3θ ) = cos 3θ − i sin 3θ .
3
3
Complex conjugates
1.
Applying property 5 of theorem 5.1,
(
)
z + k = ( z + k ) z + k = ( z + k )( z + k ) = zz + k ( z + z ) + k 2
2
= z + k ( z + z ) + k 2 ⋅1 = 1 + k ( z + z ) + k 2 z
2
2
= 1 + k ( z + z ) + k 2 zz = ( kz + 1)( kz + 1) = ( kz + 1)( kz + 1)
= kz + 1 .
2
2.
Since 1 + 3i is a root, 1 – 3i is also a root. Then x 4 + 3x 2 + ax + b is divisible by
⎡⎣ x − (1 + 3i ) ⎤⎦ ⎡⎣ x − (1 − 3i ) ⎤⎦ = x 2 − 2 x + 10 .
However, x 4 + 3x 2 + ax + b = ( x 2 − 2 x + 10 )( x 2 + 2 x − 3) + ( a − 26 ) x + ( b + 30 ) . Hence, a = 26
and b = –30. Thus, x 4 + 3x 2 + ax + b = ( x 2 − 2 x + 10 )( x 2 + 2 x − 3) = ( x − 1)( x + 3) ( x 2 − 2 x + 10 ) .
De Moivre’s theorem
1.
First, consider the equation as a quadratic equation in z 2 , i.e., ( z 2 ) + 4 z 2 + 5 = 0 . Then by the
2
quadratic formula, we get z 2 = −2 + i or − 2 − i . So the roots of the equation are the square
roots of the complex numbers – 2 + i and – 2 – i. Applying theorem 6.2 to find these square
1
1
1
1
roots, we get z = 5 4 cis1.34 or 5 4 cis1.80 or 5 4 cis4.48 or 5 4 cis4.94 .
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2.
Note
⎛ 3 1 ⎞
⎛ −π
− i ⎟⎟ = 2cis ⎜
3 − i = 2 ⎜⎜
⎝ 6
⎝ 2 2 ⎠
(
3 −i
)
⎞
⎟ . Hence,
⎠
1
4
⎛ −π
+ 2 kπ
⎜
= 2 cis ⎜ 6
4
⎜
⎝
⎞
⎟
⎟
⎟
⎠
1
4
for k = 0, 1, 2, 3. Thus,
(
3.
3 −i
)
1
4
⎛ π ⎞
⎛ 11π ⎞
⎛ 23π ⎞
⎛ 35π
4
4
= 2 4 cis ⎜ − ⎟ or 2 4 cis ⎜
⎟ or 2 cis ⎜
⎟ or 2 cis ⎜
⎝ 24 ⎠
⎝ 24 ⎠
⎝ 24 ⎠
⎝ 24
1
1
2n
1
cos 2nθ + i sin 2nθ = ( cisθ ) = ∑ Ck2 n ( cos θ ) ( i sin θ )
2n
k
2n−k
1
⎞
⎟.
⎠
.
k =0
n
cos 2nθ + i sin 2nθ = ∑ C22kn ( cos θ )
2k
( i sin θ )
2n−2k
k =0
n
= ∑ C22kn cos 2 k θ sin 2 n − 2 k θ ( −1)
k =0
n
+ ∑ C22kn−1 ( cos θ )
2 k −1
( i sin θ )
2 n − 2 k +1
k =1
n−k
n
+ i ∑ C22kn−1 cos 2 k −1 θ sin 2 n − 2 k +1 θ ( −1)
n−k
.
k =1
Equating the real parts on both sides,
n
cos 2nθ = ∑ ( −1)
k =0
n−k
n
C22kn cos 2 k θ sin 2 n − 2 kθ = ∑ ( −1) C22kn cos 2 n − 2 k θ sin 2 kθ .
k
k =0
Applications
1.
(a)
x 2 + y 2 − 4 x − 6 y − 23 = 0 .
(b)
2z
1
is purely imaginary iff z ≠ − , z ≠ 0 , and
3z + 1
3
4 ( 3x 2 + 3 y 2 + x )
2z
2 x + 2 yi
2 x − 2 yi
⎛ 2z ⎞
⎛ 2z ⎞
+⎜
+
=
0 = 2 Re ⎜
⎟=
⎟=
2
2
⎝ 3 z + 1 ⎠ 3z + 1 ⎝ 3z + 1 ⎠ ( 3x + 1) + 3 yi ( 3 x + 1) − 3 yi ( 3x + 1) + ( 3 y )
⎛ 1 ⎞
Thus 3x 2 + 3 y 2 + x = 0 excluding the points ( x, y ) = ⎜ − , 0 ⎟ or ( 0, 0 ) .
⎝ 3 ⎠
(c) From the definition of modulus, we have
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( x − 1) + yi
( x − 1)
2
( x − 1)
2
= ( x + 1) + yi + 1
( x + 1)
+ y2 =
2
+ y2 +1
+ y 2 = ( x + 1) + y 2 + 2
2
( x + 1)
2
+ y2 + 1
( x + 1) + y 2
2
16 x 2 + 8 x + 1 = 4 ( x + 1) + 4 y 2 .
4 x + 1 = −2
2
Thus, the locus of z is 12 x 2 − 4 y 2 − 3 = 0.
2.
Note that ∆ABC ∼ ∆XYZ ⇔
AB XY
=
and ∠ABC = ∠XYZ . However,
CB ZY
x− y
AB XY
a −b a −b
x− y
=
⇔
=
=
=
,
CB ZY
c−b
c−b
z−y
z− y
⎛ x− y⎞
⎛ a −b ⎞
∠ABC = ∠XYZ ⇔ arg ⎜
⎟.
⎟ = ± arg ⎜
⎝ c −b ⎠
⎝ z−y⎠
These two conditions together implies
a −b x− y
a −b x− y
=
or
=
.
c −b z − y
c −b z − y
Hence ∆ABC ∼ ∆XYZ iff
3.
a −b x− y
a −b x− y
=
=
or
.
c −b z − y
c −b z − y
The key to this problem is to use the formulae for geometric series to obtain
n
n −1
∑e
−
2 kπ
i
n
k =0
⎛ − 2π i ⎞
1− ⎜ e n ⎟
1 − e − 2π i
1−1
= ⎝ 2π ⎠ =
=
= 0.
2π
2π
− i
− i
− i
n
n
n
1− e
1− e
1− e
With this and the triangle inequality in mind, we have
n −1
∑ z−z
k =0
n −1
k
= ∑ z −e
2 kπ
i
n
k =0
n −1
= ∑ z −e
2 kπ
i
n
k =0
n −1
i1 = ∑ z − e
2 kπ
i
n
n −1
∑ ze
k =0
−
2 kπ
i
n
n −1
−n = z ∑ e
−
2 kπ
i
n
k =0
2 kπ
2 kπ
2 kπ
n −1 ⎛
n −1
i⎞ −
i
i
−
= ∑ ⎜ z − e n ⎟ie n = ∑ ze n − 1 ≥
k =0 ⎝
k =0
⎠
=
ie
−
2 kπ
i
n
⎛ − 2 knπ i ⎞
− 1⎟
⎜ ze
∑
k =0 ⎝
⎠
n −1
− n = 0 − n = n.
k =0
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