1.complex_numbers_handout complex_numbers_handout.pdf

COMPLEX NUMBERS
ELECTRONIC VERSION OF LECTURE
Dr. Lê Xuân Đại
HoChiMinh City University of Technology
Faculty of Applied Science, Department of Applied Mathematics
Email: ytkadai@hcmut.edu.vn
HCMC — 2018.
Dr. Lê Xuân Đại (HCMUT-OISP) COMPLEX NUMBERS HCMC — 2018. 1 / 45

OUTLINE
1 PROBLEM
2 ALGEBRAIC FORM OF COMPLEX NUMBER
3 POLAR FORM OF A COMPLEX NUMBER
4 POWERS OF A COMPLEX NUMBER
5 COMPLEX EXPONENTS
6 FUNDAMENTAL THEOREM OF ALGEBRA
7 MATLAB

Problem
Complex numbers arise naturally in the
course of solving polynomial equations. For
example, the solutions of the quadratic
equation ax2
+bx+c = 0, which are given by
the quadratic formula
x =
−b±
p
b2 −4ac
2a
are complex numbers if the expression
inside the radical is negative (b2
−4ac < 0).

Algebraic form of complex number Definitions
To deal with the problem that the equation
x2
= −1 has no real solutions,
mathematicians invented the imaginary
number i =
p
−1, which is assumed to have
the property i2
= (
p
−1)2
= −1.
DEFINITION 2.1
An expression of the form z = a+bi; (a,b) ∈ R2
,
is called a complex number. The number a is
called the real part of z and is denoted by
Re (z), and the number b is called the
imaginary part of z and is denoted by Im (z).

DEFINITION 2.2
A complex number of the form z = 0+bi,b 6= 0
whose real part is zero is said to be pure
imaginary. For example i,3i,−i,... are pure
imaginary.
DEFINITION 2.3
A complex number of the form z = a+0i = a
whose imaginary part is zero is a real
number, so the real numbers can be viewed
as a subset of the complex numbers.

A complex number z = a+bi can be
associated with the ordered pair (a,b) and
represented geometrically by a point or a
vector in the xy−plane (complex plane).
Argand diagram

Algebraic form of complex number Operations
DEFINITION 2.4
Two complex numbers are considered equal
if and only if their real parts are equal and
their imaginary parts are equal; that is,
z1 = a1 +b1i = z2 = a2 +b2i
⇐⇒ a1 = a2 and b1 = b2.
(1)
Remark. Unlike the real numbers, the order
symbols <,É,> and Ê are not used with
complex numbers.

EXAMPLE 2.1
Find the real numbers x,y such that
(1+2i)x+(3−5i)y = 5−i
SOLUTION.
(1+2i)x+(3−5i)y = 1−3i
⇔ (x+3y)+(2x−5y)i = 5−i
⇔
(
x+3y = 5
2x−5y = −1
⇔
(
x = 2
y = 1

DEFINITION 2.5
Let z1 = a1 +b1i and z2 = a2 +b2i. Complex
numbers z1,z2 are added, subtracted by the
formulas
z1 +z2 = (a1 +a2)+(b1 +b2)i (2)
z1 −z2 = (a1 −a2)+(b1 −b2)i (3)

EXAMPLE 2.2
Find the real and imaginary parts of the
complex number z = (2+3i)+(−3+4i)−(6−5i)
SOLUTION.
z = (2−3−6)+(3+4+5)i = −7+12i
⇒ Re (z) = −7,Im (z) = 12.

DEFINITION 2.6
Let z1 = a1 +b1i and z2 = a2 +b2i. Complex
numbers z1,z2 are multiplied in accordance
with the standard rules of algebra but with
i2
= −1.
z1.z2 = (a1.a2 −b1.b2)+(a1.b2 +a2.b1)i (4)

EXAMPLE 2.3
Let z1 = 1+2i,z2 = 2+bi. Find all real
numbers b such that z1.z2 is a real number.
SOLUTION.
z1.z2 = (1×2−2×b)+(1×b+2×2)i =
= (2−2b)+(b+4)i.
In order that z1.z2 is a real number, then
b+4 = 0 ⇒ b = −4.

Algebraic form of complex number Complex conjugate
DEFINITION 2.7
If z = a+bi is a complex number, then the
complex conjugate of z is denoted by z (read
"z bar") and is defined by z = a−bi.

DEFINITION 2.8
The length of the vector z is called modulus
of z and is denoted by |z| or mod(z). The
modulus of z is defined by
|z| =
p
z.z =
p
(a+bi)(a−bi) =
p
a2 +b2
(5)

EXAMPLE 2.4
Find the modulus of complex number 1+i
p
3
SOLUTION.
|z| =
q
12 +
p
3
2
= 2.

Algebraic form of complex number Reciprocals and Division
DEFINITION 2.9
If z 6= 0, then the reciprocal of z is denoted by
1
z
and is defined by the property
1
z
·z = 1 (6)
⇒
1
z
·z.z = z ⇔
1
z
·|z|2
= z ⇔
1
z
=
z
|z|2

DEFINITION 2.10
If z2 6= 0, then the quotient
z1
z2
is defined by
z1
z2
=
z2
|z2|2
·z1 =
z1z2
|z2|2
(7)
z1
z2
=
a1 +b1i
a2 +b2i
=
(a1 +b1i)(a2 −b2i)
(a2 +b2i)(a2 −b2i)
z1
z2
=
a1a2 +b1b2
a2
2 +b2
2
+
a2b1 −a1b2
a2
2 +b2
2
i.

EXAMPLE 2.5
Let z1 = 2+3i and z2 = 1+2i. Express z =
z1
z2
in
the form a+bi.
SOLUTION.
z =
2+3i
1+2i
=
2×1+3×2
12 +22
+
1×3−2×2
12 +22
i =
8
5
−
1
5
i.

SOME BASIC PROPERTIES OF CONJUGATE OPERATIONS
1
z+z = 2.Re (z), z−z = 2i.Im (z).
2
z.z = |z|2
.
3
z = z if and only if z is a real number.
4
z1 ±z2 = z1 ±z2.
5
z1.z2 = z1.z2.
6
z1
z2
=
z1
z2
·
7
z = z.
8
zn = (z)n
,∀n ∈ N

Polar Form of a Complex Number Definitions
If z = a+bi,z 6= 0, is a nonzero complex
number, and if ϕ is an angle from the real
axis to the vector z, then the real and
imaginary parts of z can be expressed as
a = |z|cosϕ, b = |z|sinϕ (8)

DEFINITION 3.1
The complex number z = a+bi can be
expressed as
z = r(cosϕ+isinϕ) (9)
which is called a polar form of z, where
r = |z| =
p
a2 +b2,tanϕ =
b
a
· The angle ϕ is
called the argument of z and denoted by
arg z.

REMARK
The argument arg z is not unique; any two
arguments of z differ by an integer multiple
of 2π. However, there is only one argument
whose radian measure satisfies
0 É ϕ < 2π or −π < ϕ É π.
This argument is called the principal
argument of z and denoted by Arg z

EXAMPLE 3.1
Express z = 1−
p
3i in polar form using the
principal argument.
SOLUTION. The modulus of z is
|z| =
q
12 +(
p
3)2 = 2
a = |z|cosϕ ⇒ 1 = 2cosϕ ⇒ cosϕ =
1
2
b = |z|sinϕ ⇒ −
p
3 = 2sinϕ ⇒ sinϕ =
−
p
3
2

The unique angle ϕ satisfies these
equations and whose radian measure
satisfies −π < ϕ É π, is −
π
3
· Thus, the
principal argument is −
π
3
· Therefore, a polar
form of z is
z = 2
h
cos
³
−
π
3
´
+isin
³
−
π
3
´i
= 2
³
cos
π
3
−isin
π
3
´

Polar Form of a Complex Number Multiplication of Complex Numbers
Let z1 = r1(cosϕ1 +isinϕ1) and
z2 = r2(cosϕ2 +isinϕ2). Multiplying, we
obtain
z1.z2 = r1.r2
£
cos(ϕ1 +ϕ2)+isin(ϕ1 +ϕ2)
¤
(10)

Polar Form of a Complex Number Multiplication of Complex Numbers
EXAMPLE 3.2
Use polar forms of the complex numbers to
compute z = (1+i
p
3)(2−2i).
SOLUTION.
z = 2
³
cos
π
3
+isin
π
3
´
.2
p
2
h
cos
³
−
π
4
´
+isin
³
−
π
4
´i
=
= 4
p
2
h
cos
³π
3
−
π
4
´
+isin
³π
3
−
π
4
´i
=
= 4
p
2
³
cos
π
12
+isin
π
12
´
.

Polar Form of a Complex Number Division of Complex Numbers
DIVISION OF COMPLEX NUMBERS
Let z1 = r1(cosϕ1 +isinϕ1) and
z2 = r2(cosϕ2 +isinϕ2). Dividing, we obtain
z1
z2
=
r1
r2
£
cos(ϕ1 −ϕ2)+isin(ϕ1 −ϕ2)
¤
, z2 6= 0
(11)

Polar Form of a Complex Number Division of Complex Numbers
EXAMPLE 3.3
Use polar forms of the complex numbers to
compute z =
1+i
p
3
1+i
SOLUTION.
z =
2
¡
cos π
3
+isin π
3
¢
p
2
¡
cos π
4
+isin π
4
¢ =
=
p
2
h
cos
³π
3
−
π
4
´
+isin
³π
3
−
π
4
´i
=
=
p
2
³
cos
π
12
+isin
π
12
´
.

Powers of a Complex Number Powers of i
POWERS OF i
i1
= i, i2
= −1, i3
= i2
.i = −i, i4
= (i2
)2
= 1,
i5
= i4
.i = i,i6
= i4
.i2
= −1,
i7
= i4
.i3
= −i,i8
= (i4
)2
= 1
THEOREM 4.1
Suppose n be a natural number, then in
= ir
,
where r is the remainder of division n by 4.

Powers of a Complex Number Powers of i
EXAMPLE 4.1
Evaluate i2011
.
SOLUTION.
We have 2011 = 4×502+3. So i2011
= i3
= −i.

Powers of a Complex Number DeMoivre’s Formula
THEOREM 4.2 (DEMOIVRE’S FORMULA)
If r > 0 and n is a positive integer then
h
r(cosϕ+isinϕ)
in
= rn
(cosnϕ+isinnϕ)
(12)
THEOREM 4.3
If n is a positive integer then
(cosϕ+isinϕ)n
= cosnϕ+isinnϕ (13)

Powers of a Complex Number DeMoivre’s Formula
EXAMPLE 4.2
Find a minimum positive integer n such that
z = (−
p
3+i)n
is a pure imaginary.
SOLUTION.
z = (−
p
3+i)n
=
h
2(cos 5π
6
+isin 5π
6
)
in
=
= 2n
(cos 5nπ
6
+isin 5nπ
6
).
Therefore, in order to z is a pure imaginary
then cos
5nπ
6
= 0 ⇔
5nπ
6
=
π
2
+kπ ⇔ n =
3+6k
5
.
In order to 3+6k is divisible by 5, the
minimum number k is k = 2 ⇒ n = 3.

Powers of a Complex Number Roots of a complex number
THEOREM 4.4
Let z = a+bi = r(cosϕ+isinϕ) and let n be a
positive integer. Then z has the n distinct nth
roots
n
p
z = wk = n
p
r
µ
cos
ϕ+k2π
n
+isin
ϕ+k2π
n
¶
(14)
where k = 0,1,2,...,n−1.

Powers of a Complex Number Roots of a complex number
EXAMPLE 4.3
Let z = 1−i. Find 3
p
z.
SOLUTION.
z = 1−i =
p
2
³
cos
−π
4
+isin
−π
4
´
⇒ 3
p
z =
6
p
2
µ
cos
−π
4
+k2π
3
+isin
−π
4
+k2π
3
¶
,
(k = 0,1,2)

Complex Exponents Euler’s Formula
THEOREM 5.1 (EULER’S FORMULA)
eiϕ
= cosϕ+isinϕ (15)
If z = eiϕ
= cosϕ+isinϕ then |z| = 1 and z−1
=
=
1
cosϕ+isinϕ
=
cosϕ−isinϕ
(cosϕ+isinϕ)(cosϕ−isinϕ)
=
= cosϕ−isinϕ.
So z−1
= z.

Complex Exponents Complex Exponent
DEFINITION 5.1
If z = a+bi is any complex number, then z
can be expressed in the complex exponent
form
z = a+bi = r(cosϕ+isinϕ) = reiϕ
(16)

Remark. Complex exponents follow the
same laws as real exponents.
THEOREM 5.2
If z1 = r1eiϕ1
and z2 = r2eiϕ2
are nonzero
complex numbers, then
1
z1z2 = r1r2eiϕ1+iϕ2
= r1r2ei(ϕ1+ϕ2)
2
z1
z2
=
r1
r2
eiϕ1−iϕ2
=
r1
r2
ei(ϕ1−ϕ2)
3
z1 = r(cosϕ−isinϕ) =
r
h
cos(−ϕ)+isin(−ϕ)
i
= re−iϕ

EXAMPLE 5.1
Express the complex number z =
−1+i
p
3
1−i
in
the complex exponent form.
SOLUTION.
z =
−1+i
p
3
1−i
=
2ei2π
3
p
2ei−π
4
=
p
2ei2π
3 −i−π
4 =
=
p
2ei11π
12

EXAMPLE 5.2
Express complex numbers of the form
z = e2+iy
,y ∈ R on complex plane.
SOLUTION.
z = e2+iy
= e2
.eiy
= e2
(cosy +isiny).
Since y is a arbitrary real number then the
set of all complex numbers of the form
z = e2+iy
,y ∈ R is the circle with center at
origin O and radius of r = e2
.

Fundamental Theorem of Algebra
THEOREM 6.1
The equation
anxn
+an−1xn−1
+...+a1x+a0 = 0
(n ∈ N∗
,an 6= 0,ai ∈ C,i = 1,n) has exactly n
roots (real, complex and multiple roots).

THEOREM 6.2
If x = α is a root of the equation
anxn
+an−1xn−1
+...+a1x+a0 = 0,
(n ∈ N∗
,an 6= 0,ai ∈ R,i = 1,2,...,n),
then x = α is also a root of this equation.

EXAMPLE 6.1
Solve the equation z4
+z3
+3z2
+z+2 = 0 in
the set of complex numbers C if z = i is a root
of this equation.
SOLUTION. Since z = i is a root of the
equation then z = −i is also a root of this
equation. Therefore
z4
+z3
+3z2
+z+2 = 0 ⇔ (z2
+1)(z2
+z+2) = 0
⇔
"
z = ±i
z = −1±i
p
3
2

MatLab
MATLAB
1
To find a real part of complex number z :
real(z)
2
To find a imaginary part of complex
number z : imag(z)
3
To find modulus of complex number z :
abs(z)
4
To find the principle angle ϕ of complex
number z : angle(z)
5
To find complex conjugate of complex
number z : conj(z)

MatLab
THANK YOU FOR YOUR ATTENTION

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