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Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9 Lecture Notes
The Complex Numbers
Winter 2017
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Limitations of the Real Numbers
Since x 2 ≥ 0 for every real number x, there is no real number that
satisfies the simple equation
x 2 + 1 = 0.
To this end, the number i denotes an imaginary solution to the
equation x 2 + 1 = 0, that is
i 2 = −1.
Surprisingly, it then turns out that any polynomial equation of the
form
an x n + an−1 x n−1 + · · · + a2 x 2 + a1 x + a0 = 0,
with ai ∈ R, will have solutions, either real, or in terms of i.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
The Quadratic Formula
For example, you should be familiar with the formula for solving
the quadratic equation:
√
b 2 − 4ac
−b
±
ax 2 + bx + c = 0 ⇒ x =
.
2a
If b 2 − 4ac < 0, then the roots to the quadratic equation involve
the square root of a negative number and i can be used to solve
the equation. For example,
√
−6
±
36 − 100
x 2 + 6x + 25 = 0 ⇒ x =
√2
−6 ± −64
⇒ x=
2
⇒ x = −3 ± 4 i
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
The Set of Complex Numbers
For any real numbers a, b the expression a + b i is called a complex
number. Let z = a + b i, then
I
a is called the real part of z, sometimes denoted by Re(z),
I
b is called the imaginary part of z, denoted by Im(z).
The set of all complex numbers is denoted by C. Complex numbers
are added and multiplied with the usual operations of addition and
multiplication, using the fact that i 2 = −1. Thus if a + b i and
c + d i are complex numbers, we define
I
(a + b i) + (c + d i) = (a + c) + (b + d) i, and
I
(a + b i)(c + d i) = ac − bd + (ad + bc) i,
as you can check. The real number a is equal to the complex
number a + 0 i, the imaginary number b i is equal to the complex
number 0 + b i.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 1
Let z = 3 − 2 i and w = −5 + 7 i. Then
1. z + w = (3 − 5) + (−2 + 7) i = −2 + 5 i
2. z − w = (3 + 5) + (−2 − 7) i = 8 − 9 i
3. zw = (3 − 2 i)(−5 + 7 i) = −15 + 10 i + 21 i − 14 i 2 = −1 + 31 i
4. z 2 = (3 − 2 i)(3 − 2 i) = 9 − 12 i + 4 i 2 = 5 − 12 i
5. w 2 = (−5 + 7 i)(−5 + 7 i) = 25 − 49 − 35 i − 35 i = −24 + 70 i
Division of complex numbers is a little more complicated.
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Complex Conjugates and Division of Complex Numbers
If z = a + b i is a complex number, define the complex conjugate
of z as
z̄ = a − b i,
and define the modulus of z, or the absolute value of z, as
p
|z| = a2 + b 2 .
Then
1. |z| = |z̄| =
√
a2 + b 2
2. |z|2 = z z̄ = a2 + b 2
a
bi
z̄
=
−
a2 + b 2 a2 + b 2
|z|2
is called the multiplicative inverse of z.
3. If z 6= 0, then z −1 =
If z 6= 0, then z −1
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 2
To divide by a non-zero complex number w use the fact that
|w |2 = w w ⇔
1
w
=
.
w
|w |2
With z and w as in Example 1:
z
w
3 − 2i
−5 + 7 i
(3 − 2 i)(−5 − 7 i)
=
(−5)2 + 72
−15 + 10 i − 21 i + 14 i 2
=
74
29 11
= − −
i
74 74
=
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Properties of Complex Conjugates and Absolute Values
Let z = a + b i and w = c + d i be complex numbers. Then
√
1. |z| = |z̄| = a2 + b 2
2. |z|2 = z z̄ = a2 + b 2
3. z ± w = z̄ ± w
4. zw = z̄ w
5. z ∈ R ⇔ z = z̄.
6. z = 0 ⇔ |z| = 0.
Proof of 4: z̄ w̄ = (a − bi)(c − di) = ac − bd − (bc + ad)i, and
zw = ac − bd + (bc + ad)i = ac − bd − (bc + ad)i.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 3
Consider again the quadratic equation and its roots:
√
−b
±
b 2 − 4ac
ax 2 + bx + c = 0 ⇒ x =
.
2a
If b 2 − 4ac < 0, then the roots to the quadratic equation are
complex numbers; in fact they will be complex conjugates if a, b, c
are all real numbers. Why?
az 2 + bz + c = 0 ⇒ az 2 + bz + c = 0̄ = 0
⇒ ā(z̄)2 + b̄ z̄ + c̄ = 0
⇒ a(z̄)2 + b z̄ + c = 0
So if z is a solution to the quadratic, so is z̄.
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Complex Numbers and the Complex Plane
√
Let i = −1; let z = a + b i, with a, b ∈ R, be a complex number.
The real part of z, denoted by Re(z), is a; the imaginary part of z,
denoted by Im(z), is b. Every z ∈ C can be represented by a point
in the complex plane. The horizontal axis is called the real axis;
6
b
z =r a + b i
|z|
the vertical axis is called the imaginary axis.
z̄ = a − b i
real axis
a
@
-
@
|z̄| @
@r
z̄ = a − b i
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
is called the conjugate of z; it is
the reflection
of z in the real axis.
√
2
|z| = a + b 2 is called the absolute value, or modulus, of z.
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Polar Form of a Complex Number
The complex number z = a + b i
in the complex plane, can be represented another way, in terms of its
‘polar coordinates.’ Let r = |z| be
the distance from the origin to the
point (a, b). Let θ be the angle, measured in radians, that the line from
the origin to the point (a, b) makes
with respect to the positive real axis,
√
θ ∈ [0, 2π). θ is called the argument of z. r = |z| = a2 + b 2 , and
z = r (cos θ + i sin θ)
is called the polar form of the complex number z.
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Chapter 9: The Complex Numbers
Example 1
Find the polar form of z = −3 + 3 i.
p
√
Solution: r = |z| = (−3)2 + 32 = 3 2. For the argument, use
tan θ =
3
= −1;
−3
take
θ=
Thus
3π
.
4
√
3π
3π
−3 + 3 i = 3 2 cos
+ i sin
.
4
4
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Properties of Complex Numbers in Polar Form
If z = |z|(cos α + i sin α), w = |w |(cos β + i, sin β), then
z w = |z||w |(cos(α + β) + i sin(α + β)).
and
z/w = |z|/|w |(cos(α − β) + i sin(α − β)).
In particular, you multiply complex numbers by multiplying their
moduli and adding their arguments; you divide complex numbers
by dividing their moduli and subtracting their arguments.
Proof:
zw
= |z|(cos α + i sin α)|w |(cos β + i sin β)
= |z||w |((cos α cos β − sin α sin β) + i(sin α cos β + sin β cos α))
= |z||w |(cos(α + β) + i sin(α + β))
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
DeMoivre’s Theorem
Theorem 9.2.6: (r (cos θ + i sin θ))n = r n (cos(nθ) + i sin(nθ)).
Proof: by induction. Its obviously true if n = 1. Now suppose the
formula is true for n and calculate (r (cos θ + i sin θ))n+1 , making
use of appropriate trig identities, again:
(r (cos θ + i sin θ))n+1
= (r (cos θ + i sin θ))n r (cos θ + i sin θ)
= r n ((cos(nθ) + i sin(nθ)) r (cos θ + i sin θ)
= r n+1 (cos(nθ) cos θ − sin(nθ) sin θ + i (cos(nθ) sin θ + sin(nθ) sin θ))
= r n+1 (cos((n + 1) θ) + i sin((n + 1) θ))
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 2
Let z = 1 + i. Find z 8 , and a square root of z.
√
Solution: put z into polar form. |z| = 2; take θ = π/4. Thus
π √ π z = 2 cos
+ i sin
.
4
4
A direct application of DeMoivre’s Theorem gives
√
8π
8π
z 8 = ( 2)8 cos
+ i sin
= 16(cos(2π)+sin(2π)) = 16.
4
4
For a square root, we go in ‘reverse:’
π π √ 1/2
1π
1π
1/4
( 2)
cos
+ i sin
=2
cos
+ i sin
.
24
24
8
8
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 3
Find the cube roots and fourth roots of unity; that is solve
(a) z 3 = 1, and (b) z 4 = 1.
Solution: note that the argument of the real number 1 is θ = 0.
Let z = cos θ + i sin θ. Then z n = cos(nθ) + i sin(nθ). So
I (a) z 3 = 1 ⇒ 3θ = 0 + 2kπ ⇒ θ = 2kπ/3. Take
θ = 0, 2π/3, 4π/3, and consequently the three distinct
solutions are
√
√
1
3
1
3
z = 1, − +
i, or − −
i.
2
2
2
2
I
(b) z 4 = 1 ⇒ 4θ = 0 + 2kπ ⇒ θ = 2kπ/4. Take
θ = 0, π/2, π, 3π/4, so the four distinct solutions are
z = 1, i, −1, −i.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Alternate Calculations for Example 3
We could have just factored:
I
(a) z 3 = 1 ⇔ z 3 − 1 = 0 ⇔
(z − 1)(z 2 + z + 1) = 0. Thus
z = 1 or
√
√
1
−1 + 1 − 4
3
=− ±
z=
i.
2
2 2
I
(b) z 4 = 1 ⇔ z 4 − 1 = 0 ⇔
(z 2 − 1)(z 2 + 1) = 0. Thus
z = ±1 or ± i.
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Chapter 9: The Complex Numbers
Example 4
√
Find the all the solutions of z 3 = 1 + 3 i.
√
Solution: first put 1 + 3 i into polar form.
√
√
|1 + 3 i| = 1 + 4 = 2,
and so
1+
√
√
3 =2
1
3
+
i
2
2
!
π
π
= 2 cos + i sin
.
3
3
√
Let z = r (cos θ + i sin θ) and suppose z 3 = 1 + 3 i. Then
π
π
3
r (cos(3θ) + i sin(3θ)) = 2 cos + i sin
.
3
3
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Therefore,
r 3 = 2 and 3θ =
π
+ 2kπ.
3
Thus r = 21/3 and
π 2kπ
+
,
9
3
for which the three distinct solutions in the interval [0, 2π) are
θ=
θ=
13π
π 7π
.
, , or
9
9 9
Thus the three values of z are
π
π
1/3
z1 = 2
cos + i sin
,
9
9
7π
7π
13π
13π
z2 = 21/3 cos
+ i sin
, or z3 = 21/3 cos
+ i sin
.
9
9
9
9
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra
Having defined the number i to be a solution to the simple
equation x 2 + 1 = 0, it may come as a surprise to find that any
non-constant polynomial will always have solutions in terms of
complex numbers. This is known as the Fundamental Theorem of
Algebra.
Theorem 9.3.1: every non-constant polynomial with complex
coefficients has a complex root.
Proof: the proof of this theorem requires results which are beyond
the scope of this course.
So we will just look at some implications of this result. In
particular, we shall investigate how many roots a polynomial can
have.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Some Definitions
1. An expression of the form
an z n + an−1 z n−1 + · · · + a2 z 2 + a1 z + a0 ,
with coefficients ai ∈ C, and an 6= 0, is called a polynomial of
degree n. We will often write it as p(z), or q(z).
2. The complex number r is a root of the polynomial p(z) if
p(r ) = 0.
3. The polynomial f (z) is a factor of the polynomial p(z) if
there is a polynomial q(z) such that p(z) = f (z)q(z).
4. The root r of the polynomial p(z) has multiplicity k if
f (z) = (z − r )k is a factor of p(z), but (z − r )k+1 is not a
factor of p(z).
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Long Division by a Linear Factor
The following type of ‘long division’ should be familiar to you:
z + 2 ) z4 −
z4 +
−
−
z3
z3
2z 3
3z 3
3z 3
−
+
3z 2 + 11z
5z 2
− 22
+ 10
+
−
5z 2
6z 2
11z 2
11z 2 + 22z
− 22z
− 22z
+ 10
+ 10
+ 10
− 44
54
Thus z 4 − z 3 + 5z 2 + 10 = (z + 2)(z 3 − 3z 2 + 11z − 22) + 54.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Theorem 9.3.4: if r ∈ C and p(z) is a non-constant polynomial
then there exists a polynomial q(z) and a number c such that
p(z) = (z − r )q(z) + c.
Proof: by complete mathematical induction of the degree of p(z).
n = 1 : if p(z) = az + b then check that p(z) = a(z − r ) + ar + b.
Now suppose p(z) = an+1 z n+1 + an z n + · · · + a1 z + a0 has degree
n + 1. Check that
p(z) = (z − r )an+1 z n + (an + ran+1 )z n + an−1 z n−1 + · · · a1 z + a0 .
|
{z
}
g (z), deg g (z) ≤n
By induction, g (z) = (z − r )h(z) + c, for some polynomial h(z)
and constant c. Then
p(z) = (z−r )an+1 z n+1 +(z−r )h(z)+c = (z−r )(an+1 z n+1 + h(z))+c.
|
{z
}
q(z)
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
The Factor Theorem
Theorem 9.3.6: the complex number r is a root of the polynomial
p(z) if and only if z − r is a factor of p(z).
Proof: if z − r is a factor of p(z), then p(z) = (z − r )q(z), for
some polynomial q(z), and p(r ) = (r − r )q(r ) = 0. Now suppose r
is a root of p(z). By Theorem 9.3.4, we can write
p(z) = (z − r )q(z) + c,
for some constant c. Then
p(r ) = 0 ⇔ c = 0,
implying that z − r is a factor of p(z).
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 1
Consider the polynomial p(z) = i z 3 + z 2 − 4. Since
p(2i) = i(2i)3 + (2i)2 + 4 = i(−8i) + (−4) − 4 = 8 − 4 − 4 = 0,
(z − 2i) is a factor of p(z). Use long division to check that
p(z) = (z − 2i)(i z 2 − z − 2i).
Thus p(z) has two more roots:
i z 2 − z − 2i = 0 ⇒ z 2 + i z − 2 = 0
√
−i ± i 2 + 8
⇒ z=
√ 2
7
i
⇒ z =±
−
2
2
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla
Chapter 9: The Complex Numbers
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
How Many Roots Can a Polynomial Have?
Theorem 9.3.8: a polynomial p(z) of degree n has at most n
roots; if multiplicities are counted, it has exactly n roots.
Proof: p(z) has at least one root, by the Fundamental Theorem
of Algebra; call it r1 . By the factor theorem, p(z) = (z − r1 )q(z),
where q(z) is a polynomial with degree n − 1. By the Fundamental
Theorem of Algebra, q(z) has a root, call it r2 . Thus
p(z) = (z − r1 )(z − r2 )r (z),
for some polynomial r (z) with degree n − 2. Continuing in this
fashion, n times, there will be roots r1 , r2 , . . . , rn , not all necessarily
distinct, such that
p(z) = k(z − r1 )(z − r2 ) · · · (z − rn ),
for some constant k.
Chapter 9 Lecture Notes The Complex Numbers
Chapter 9: The Complex Numbers
MAT246H1S Lec0101 Burbulla
9.1: What Is A Complex Number?
9.2: The Complex Plane
9.3: The Fundamental Theorem of Algebra
Example 2
In Example 1, we saw that the cubic polynomial
p(z) = i z 3 + z 2 − 4
has three distinct roots. Now consider the fifth degree polynomial
q(z) = z 5 + 2z 4 + 2z 3 + 4z 2 + z + 2.
Since q(−2) = 0, as you may check,
q(z) = (z + 2)(z 4 + 2z 2 + 1) = (z + 2)(z 2 + 1)2 .
Thus the distinct roots of q(z) are −2, i and −i. Each of the roots
±i has multiplicity two. So counting multiplicities, q(z) has five
roots.
Chapter 9 Lecture Notes The Complex Numbers
MAT246H1S Lec0101 Burbulla