Download Chapter 1: Complex Numbers Why do we need complex numbers

Chapter 1: Complex Numbers
Why do we need complex numbers?
First of all, a simple algebraic equation like X 2 = −1 may not have a real solution.
Introducing complex numbers validates the so called fundamental theorem of algebra: every
polynomial with a positive degree has a root. Next, a basic problem in linear algebra is to
find eigenvalues and eigenvectors of a given matrix. Consider the following simple matrix
0 1
.
−1 0
Its characteristic equation is λ2 + 1 = 0 and hence we need complex numbers for its
eigenvalues. However, the usefulness of complex numbers is much beyond such simple
applications. Nowadays, complex numbers and complex functions have been developed
into a rich theory called complex analysis and become a power tool for answering many
extremely difficult questions in mathematics and theoretical physics, and also finds its
usefulness in many areas. For example, a famous result called the prime number theorem,
which was conjectured by Gauss in 1849, and defied efforts of many great mathematicians,
was finally proven by Hadamard and de la Vallée Poussin in 1896 by using the complex
theory developed at that time. Jacques Hadamard said: “The shortest path between two
truths in the real domain passes through the complex domain”.
We assume that the reader knows all basics about complex numbers. In this section
we give a quick review of them, emphasizing those which will be important for our future chapters. Also, we describe certain point of view which are suggestive for further
development outside the realm of complex numbers.
There are two basic ways to represent a complex number z algebraically:
Cartesian form: z = x + iy (or x + yi)
polar form: z = reiθ ≡ r(cos θ + i sin θ) with r ≥ 0.
Geometrically, z is represented as a vector with (x, y) as its coordinates in a Cartesian
plane, called complex plane. (This vector picture of a complex number is called an Argand
diagram. But we often only draw a point to simplify this picture.) The magnitude of this
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vector is called the absolute value or the modulus of z and is denoted by |z|. It is
equal to r given in its polar form:
|z| = r =
x2 + y 2 .
The angle between this vector and the x–axis is give by θ.
The value of r is uniquely determined by z, but the value of θ is not. For example, we
have eπi = cos π + i sin π = −1 as well as e−πi = cos(−π) + i sin(−π) = −1. Thus, both
π and −π are possible values of θ for z = −1.
When a complex number is represented in Cartesian form, say z = x + iy, the real
number x is called the real part of z and the real number y is called the imaginary part of
z. We write
x = Re z,
y = Im z.
Two complex numbers are equal if and only if both of their real parts and their imaginary
parts are equal. Thus For z = x + iy and z ′ = x′ + iy ′ , we have
z = z ′ ⇔ x = x′ and y = y ′ .
Using the Cartesian form, addition and multiplication of complex numbers are straightforward, as long as we keep in mind that i2 = −1:
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi)(c + di) = (ac − bd) + (ad + bc)i.
(1.1)
The usual algebraic identities still hold for complex numbers, such as (z + w)(z − w) =
z 2 − w2 and (z + w)2 = z 2 + 2zw + w2 .
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Example 1.1. Find the square roots of 5 − 12i.
Solution. We have to solve the equation z 2 = 5 + 12i. Write z = x + iy. Then
z 2 = x2 − y 2 + 2xyi. Thus we have x2 − y 2 = 5 and 2xy = 12. On the other hand,
√
x2 + y 2 = |z|2 = |z 2 | = |5 − 12i| = 52 + 122 = 169 = 13.
Together with x2 − y 2 = 5 we obtain 2x2 = 5 + 13 = 18 and hence x = ±3. On the other
hand, 2xy = 12 gives y = 6/x = ±2. Thus z = ±(3 + 2i) are the square roots of 5 + 12i.
The relation between the Cartesian form and the polar form is given by
x = r cos θ,
y = r sin θ.
If we write P for the point in the plane with Cartesian coordinates (x, y), then r is the
length of the line segment OP and θ is the angle between the line OP and the x–axis.
The angle θ in the polar form z = reiθ is called the argument of z. It is not uniquely
determined by z. It depends on our choice: when r ≡ |z| = 0, any real number would
do; when r = 0, any θ satisfying cos θ = x/r and sin θ = y/r would do. You may
see θ = arg z in some old–fashioned textbooks. Strictly speaking, this is incorrect.
(Sometimes, for convenience or merely following tradition, we use an incorrect expression
√
−1 for i so that we can
with correct understanding. A typical example is writing
reserve the letter i for other purposes. But we try to avoid incorrect usage as much as
possible.) Notice that, in the polar form z = reiθ , we use the identity
eiθ = cos θ + i sin θ.
(1.2)
Here we consider cos θ + i sin θ as the definition of the expression eiθ . Anyway, there is
a name for this identity: Euler’s formula. One of the most amazing things in complex
numbers is the following so–called addition formula:
ei(α+
β)
= eiα eiβ .
(1.3)
Using Euler’s formula, we can rewrite it as
cos(α + β) + i sin(α + β) = (cos α + i sin α)(cos β + i sin β).
Multiplying out the right hand side and comparing the real and imaginary parts of both
sides, we have arrived at
cos(α + β) = cos α cos β − sin α sin β,
sin(α + β) = cos α sin β + sin α cos β,
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which are well-known (but by no means obvious) identities in trigonometry. At present,
identity (1.3) is quite mystifying. We will give a thorough explanation in the future.
By induction, we can deduce form (1.3) that, for all integers n,
eint = (eit )n ,
or
cos nt + i sin nt = (cos t + i sin t)n
(1.4)
which is usually called de Moivre’s theorem. Though pretty well–known among school
teachers, this theorem does not look too impressive. In the future we will give some
applications to re-assure its importance.
Consider the special case n = 3 of (1.4). By means of the well knonw identity (a+b)3 =
a3 + 3a2 b + 3ab2 + b3 , we have
cos 3t + i sin 3t = (cos t + i sin t)3
= cos3 t + 3 cos2 t(i sin t) + 3 cos t(i sin t)2 + (i sin t)3
= cos3 t − 3 cos t sin2 t + i(3 cos2 t sin t − sin3 t).
By comparing the real parts and the imaginary parts from both sides, we get
cos 3t = cos3 t − 3 cos t sin2 t = cos3 t − 3 cos t(1 − cos2 t) = 4 cos2 t − 3 cos2 t;
sin 3t = 3 cos2 t − sin3 t = (3 cos2 t − sin2 t) sin t = (4 cos2 t − 1) sin t.
which are rewritten as cos 3t = T3 (cos t) and sin 3t = U2 (cos t) sin t, where T3 (x) =
4x3 − 3x2 , a polynomial of degree 3, and U2 (x) = 4x2 − 1, a polynomial of degree 2.
This suggests that, for any positive integer n, we can write cos nt = Tn (cos t) for some
polynomial Tn (x) of degree n and sin nt = Un−1 (cos t) sin t for some polynomial Un−1 (x)
of degree n − 1. We prove this by induction. We have already checked the case for n = 3.
The reader should check the cases for n = 1 and n = 2. Now make the induction hypothesis
that
eint = Tn (x) + iUn−1 (x) sin t
with x = cos t,
where Tn and Un−1 are polynomials of degrees n and n − 1 respectively. So
ei(n+
1)t
= eint eit = (Tn (x) + iUn−1 (x) sin t)(cos t + i sin t)
= Tn (x) cos t − Un−1 (x) sin2 t + i(Tn (x) sin t + Un−1 (x) sin t cos t)
= Tn (x)x + Un−1 (x)(1 − x2 ) + i(Tn (x) + Un−1 (x)x) sin t.
Thus we have ei(n+
Tn+
1 (x)
1)t
= Pn+
1 (x)
+ iQn (x) sin t, where
= Tn (x)x + (1 − x2 )Un−1 (x)
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Un (x) = Tn (x) + xUn−1 (x).
The last two identities tells us how to generate polynomials Tn (x) and Un (x) recursively.
Polynomials Tn (x) are called Chebyshev’s polynomials and Un (x) are known as
Chebyshev’s polynomials of the second kind. They have extensive applications in umerical
analysis and some extremal problems arising from electrical engineering.
Every complex number z has a “twin sister” z, called the complex conjugate
of z. The twins z and z do not quite have exactly the same look. They are more like
mirror images to each other. We put the pair z and z together in the Cartesian form
and the polar form as follows:
z = x + iy
z = x − iy.
z = reiθ = r(cos θ + i sin θ)
(1.5)
z = re−iθ = r(cos(−θ) + i sin(−θ)).
It is clear from the above identities that the complex conjugate of z is z. Also, z is a
real number if and only if z = z.
Example 1.2. We are asked to prove that if |z| = 1 and z = −1, then
w=i
z−1
z+1
is a real number. To do this, it is enough to check that w = w. Notice that |z| = 1 gives
zz = |z|2 = 1. Since that complex congugates of i and z are −i and ovz respectively, we
have
1−z z
z − zz
z−1
1−z
= −i
= −i
=i
= w.
1+z z
z + zz
z+1
1+z
Hence w is a real number.
w = (−i)
One of the most useful identities about complex numbers is the following
|z|2 = zz
The proof of this is simple: writing z = x + iy, we have
zz = (x + iy)(x − iy) = x2 − i2 y 2 = x2 + y 2 = |z|2 .
The following example illustarte its usefulness.
Example 1.3. Let a be a complex number with |a| < 1. We are asked to verify that
if z is unit modulus, that is, |z| = 1, then so is
w=
z−a
.
1 − az
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We need to check that |w| = 1. It is enough to check that |w|2 = 1, or equivalently,
ww = 1. Now
z−a z−a
zz − az − az − aa
ww =
=
.
1 − az 1 − az
1 + az − az + aazz
Notice that |z| = 1 gives zz = |z|2 = 1. Thus, replacing zz = 1 in the last fraction, we see
that its numerator and denominator are the same. Hence ww = 1.
(The rest of the material in the present chapter is optional.) Here we mention some
formal mathematical definitions of complex numbers. In some textbooks, complex numbers
are defined as planar vectors of the form (x, y) at the beginning. After identifying (1, 0)
with 1 and (0, 1) with i, the expression (x, y) is converted into a more preferable form:
(x, y) = (x, 0) + (0, y) = x(1, 0) + y(0, 1) = x + yi.
This definition provides a rigorous way to build complex numbers, but somehow it is too
artificial. A better way is defined the complex field C as the splitting field of X 2 +1 over
the real field. This is a highly motivated definition since, after all, historically, complex
numbers were introduced in order to give a solution to X 2 + 1 = 0. But this approach
needs too much background in field theory, which is an advanced course. Here we introduce
an interesting approach called “the doubling construction”, using some tools from linear
algebra. Recall that the matrix
cos θ
Aθ ≡
sin θ
− sin θ
.
cos θ
(†)
represents the rotation of the Cartesian plane (about the origin) through angle θ. It is a
priori clear that if we rotate a vector by angle α, followed by a rotation through angle β,
the result is the same as a single rotation with angle α + β. Putting this in mathematical
symbols, we have Aα Aβ = Aα+ β . Multiplying out the left hand side matrices of the identity
and comparing both sides, we recapture the well known identities in trigonometry again:
cos(α + β) = cos α cos β − sin α sin β, sin(α + β) = cos α sin β + sin α cos β. Expression (†)
suggests the following approach: consider 2 × 2 matrices of the form
a −b
Z=
,
b
a
where a, b ar some real numbers. We can rewrite it in the form
1 0
0 −1
Z = a1 + bi,
where 1 =
and i =
.
0 1
1
0
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When a2 + b2 = 1, P = (a, b) becomes a point on the circle x2 + y 2 = 1 and hence we
can write a = cos θ and b = sin θ (here θ is the angle between OP and the x–axis)
√
and hence Z becomes Aθ in (†). In general, when Z = O, we have r = a2 + b2 = 0
and the point (a/r, b/r) is a point on he circle x2 + y 2 = 1. Consequently we can write
a = cos θ and b = sin θ for some θ and hence
a −b
a/r −b/r
cos θ − sin θ
Z=
=r
=r
= r(cos θ 1 + sin θ i).
b
a
b/r
a/r
sin θ
cos θ
It is easy to check that i2 = −1. Also, we we denote by Z ∗ the transpose of Z, then
Z ∗ Z = (a2 + b2 ) 1 and Z ∗ + Z = 2a 1. Also, the determinant and the trace of Z are
given by det Z = a2 + b2 and tr Z = 2a.
Exercise 1.1. Do all of the checking mentioned above.
Thus, if we identify the matrix Z = a1 + bi, then:
1. addition of matrices corresponds to addition of complex numbers;
2. multiplication of matrices corresponds to multiplication of complex numbers;
3. the transpose Z ∗ of Z corresponds to the complex conjugate z of z;
4. the determinant det Z is equal to |z|2 and the identity Z ∗ Z = (a2 + b2 )1
corresponds to zz = |z|2 ;
5. the trace tr Z is equal to 2Re z; the identity Z ∗ + Z = 2a1 corresponds to
z + z = 2Re z, or Re z = (z + z)/2; and
6. the rotation Aθ corresponds to the Eulerian expression eiθ .
Drills
1. Find the complex conjugate of each of the following numbers:
2 + 3i, −4, 1 − i, 4i − 1, 3i, (2 + 5i)99 ,
1
√
3−i 2
2. Express each of the following complex numbers in the polar form reiθ :
√
2i, −2, 1 − i, i 3 − 1, 2 cos π7 − 2i sin π7 , 2 sin π7 + 2i cos π7
3. Find the value of each of the following products
√
√ √
√
(a) (2 + 3i)(1 − i) (b) (1 + i)4 (c) ( 3 + i 2)( 2 + i 3)
7
(d) (10000 + 9999i)2 .
4. Find the value of each of the following quotients
√
√
1−i
2 + 3i
3−i 2
cos(π/8) − i sin(π/8)
√
(a)
(b)
(c) √
(d)
.
1−i
1+i
cos(π/8) + i sin(π/8)
2+i 3
5. Find the square roots of each of the following complex numbers
(a) 3 + 4i (b) 16 + 30i (c) 4 − 3i (d) 4 + 3i; (notice the relation to part (c)).
6. Let ∆ABC be a triangle in the complex plane, where vertices A and B are located at
i and 2 respectively. Find the complex numbers which give possible locations of C.
7. If you walk from the origin of the complex plane, arrive at a point A represented by the
complex number z, and then turn left (i.e. turning 90o to the left) and walk the same
distance to a point B, what makes you think that the complex number representing
B is (1 + i)z?
Exercises
1.2. Find the roots of z 2 − (3 + i)z + 4 + 3i = 0
1.3. Prove the identity |z − w|2 + |z + w|2 = 2|z|2 + 2|w|2 . Give a geometric interpretation
of this identity. (Hint: Use |z|2 = zz several times.)
|z + w|2 − |z − w|2
1.4. Prove the identity Re zw =
.
4
z−w
is a real number.
1.5. Prove that, if |z| = 1, |w| = 1 and z = w, then
1 − zw
1.6*. (This is a hard problem) Prove the following so–called Hlawka’s inequality
|z1 + z2 | + |z2 + z3 | + |z3 + z1 | ≤ |z1 | + |z2 | + |z3 | + |z1 + z2 + z3 |.
(Hint: Square both sides.)
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