University of Zimbabwe
SEMESTER 1: 2020
Lecturer
: Mr. T. M. Mazikana
Course title : Mathematical Methods for Engineering 1
Course code :
Purpose of the course
1. This course is designed to introduce students to mathematical methods routinely used in
Physics and Engineering. The applications of these mathematical methods are introduced
using Calculus of single variables and part of Calculus of several variables for undergraduate
students with some background of A-Level calculus.
2. The concentration is on motivating results and concepts geometrically rather than on providing
rigorous proofs.
3. Concepts are defined carefully and results stated precisely. Illustrations are through the use
of vivid and concrete examples.
AIMS
The aim of this course is to introduce gently the rigor of mathematical analysis and provide a good
background for applied mathematics.
1
Course Content
Section 1: Number systems, Basic sets and Basic Mathematical Notations.
1. Sets of numbers.
2. The Real Number System, Inequalities, Solution Sets and Geometrical Representations.
3. Absolute Value, Neighborhoods and Intervals.
Section 2: Sequences and series of functions and their limits.
1. Definitions and Notations, Limits of Sequences and their properties.
2. Bounded and Monotone Sequences.
3. Convergence and Divergence of Series.
4. Convergence and Divergence of a Power Series.
5. Definitions and Notation, Types of Functions and their Inverses.
6. Definition of a limit of a function and its application.
7. Left and Right hand Limits.
8. L’ Hospital’s Rule.
9. Continuous Functions.
Section 3: Differentiation and Integration
1. The concept of a derivative.
2. Methods of Differentiation.
3. Applications of the derivative.
4. Definite and Indefinite integrals.
5. Techniques of integration.
6. Reduction Formulas.
7. Improper integrals.
2
Section 4: Functions of several variables
1. Limits and continuity,
2. Partial differentiation,
3. The chain rule,
4. The extended chain rule,
5. maxima, Minima and Saddle points,
Section 5: Ordinary Differential Equations, Fourier Series and Laplace
Transforms.
Methods/Strategies to be used
1. Lecture method (Online and offline).
2. Group discussions and assignments.
3. Tests and tutorials.
Student Assessment
1. Students will write three tests.
2. The average of the tests will constitute the coursework mark where 25% of the coursework
mark will contribute to the final mark.
3. A 2 hour final examination will be written in the 14th or 15th week of the semester.
4. The examination will contribute 75% to the final mark.
5. The examination paper has two sections. Students answer ALL questions in section A and at
most TWO questions in section B.
Selected Resources(references)
Recommended reading
• S Lang, Calculus of Several Variables (Springer Science+Business Media New York).
3
• P D Lax, M S Terrell, Calculus with Applications (Springer Science+Business Media New
York).
• M R Spiegel, Advanced Calculus (Schaum’s Outline Series).
• J R Kirkwood, An Introduction To Analysis (PWS Publishing Company).
Additional reading
Any first year university Calculus text book.
November 30, 2020
4
Abstract
Calculus is one of the milestones of Western thought. Building on ideas of Archimedes, Fermat,
Newton, Leibniz, Cauchy, and many others, the calculus is arguably the cornerstone of modern
science. Any well-educated person should at least be acquainted with the ideas of calculus, and a
scientifically literate person must know calculus solidly.
Calculus has two main aspects: differential calculus and integral calculus. Differential calculus
concerns itself with rates of change. Various types of change, both mathematical and physical, are
described by a mathematical quantity called the derivative. Integral calculus is concerned with a
generalized type of addition, or amalgamation, of quantities. Many kinds of summation, both
mathematical and physical, are described by a mathematical quantity called the integral.
Calculus is one of the most important parts of mathematics. It is fundamental to all of modern
science. How could one part of mathematics be of such central importance? It is because calculus
gives us the tools to study rates of change and motion. All analytical subjects, from biology to
physics to chemistry to engineering to mathematics, involve studying quantities that are growing
or shrinking or moving, in other words, they are changing. Astronomers study the motions of the
planets, chemists study the interaction of substances, physicists study the interactions of physical
objects. All of these involve change and motion.
1
1
The true sign of intelligence is not knowledge but imagination—— Albert Einstein
Chapter 1
Number systems, Basic sets and Basic
Mathematical Notations.
Mathematics has its own language with numbers as the alphabet. The language is given structure
with the aid of connective symbols, rules of operation, and a rigorous mode of thought (logic). The
number systems that we use in calculus are the natural numbers, the integers, the rational numbers,
and the real numbers. Let us describe each of these :
1. The natural numbers are the system of positive counting numbers 1, 2, 3 . . . . We denote the
set of all natural numbers by N.
N = {1, 2, 3, 4, 5, 6, 7, 8, . . . }.
2. The integers are the positive and negative whole numbers and zero, . . . , −3, −2, −1, 0, 1, 2, 3, . . . .
We denote the set of all integers by Z.
Z = {. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . . }.
3. The rational numbers are quotients of integers or fractions, such as 32 , − 54 . Any number
p
of the form , with p, q ∈ Z and q 6= 0, is a rational number. We denote the set of all rational
q
numbers by Q.
p
Q=
p, q ∈ Z, q 6= 0 .
q
4. The real numbers are the set of all decimals, both terminating and non-terminating. We
denote the set of all real numbers by R. A decimal number of the form x = 3.16792 is actually
a rational number, for it represents
x = 3.16792 =
316792
.
100000
A decimal number of the form
m = 4.27519191919 . . . ,
1
with a group of digits that repeats itself interminably, is also a rational number. To see this,
notice that
100 · m = 427.519191919 . . .
and therefore we may subtract
100m = 427.519191919 . . .
m = 4.27519191919 . . .
Subtracting, we see that
99m = 423.244
or
423244
.
99000
So, as we asserted, m is a rational number or quotient of integers. To indicate recurring
decimals we sometimes place dots over the repeating cycle of digits, e.g., m = 4.2751̇9̇,
19
= 3.16̇.
6
m=
Another kind of decimal number is one which has a non-terminating decimal expansion that
does not keep repeating. An example is π = 3.14159265 . . . . Such a number is irrational, that
is, it cannot be expressed as the quotient of two integers.
In summary : There are three types of real numbers : (i) terminating decimals, (ii) nonterminating decimals that repeat, (iii) non-terminating decimals that do not repeat. Types
(i) and (ii) are rational numbers. Type (iii) are irrational numbers.
The geometric representation of real numbers as points on a line is called the real axis. Between
any two rational numbers on the line there are infinitely many rational numbers. This leads
us to call the set of rational numbers an everywhere dense set.
Real numbers are characterised by three fundamental properties :
(a) algebraic means formalisations of the rules of calculation (addition, subtraction, multiplication, division). Example : 2(3 + 5) = 2 · 3 + 2 · 5 = 6 + 10 = 16.
1
3
(b) order denote inequalities. Example : − < .
4
3
(c) completeness implies that there are “no gaps” on the real line.
Algebraic properties of the reals for addition (a, b, c ∈ R) are :
(A1) a + (b + c) = (a + b) + c. associativity
(A2) a + b = b + a. commutativity
(A3) There is a 0 such that a + 0 = a. identity
(A4) There is an x such that a + x = 0.
inverse
Order properties of the real numbers are :
(O1) for any a, b ∈ R, a ≤ b or b ≤ a. totality of ordering I
(O2) if a ≤ b and b ≤ a, then a = b. totality of ordering II
2
(O3) if a ≤ b and b ≤ c, then a ≤ c. transitivity
(O4) if a ≤ b, then a + c ≤ b + c. order under addition
(O5) if a ≤ b and c ≥ 0, then ac ≤ bc. order under multiplication
Some useful rules for calculations with inequalities are : If a, b, c are real numbers, then :
(a) if a < b and c < 0 ⇒ bc < ac.
(b) if a < b ⇒ −b < −a.
1
(c) if a > 0 ⇒ > 0.
a
(d) if a and b are both positive or negative, then a < b ⇒
1
1
< .
b
a
The completeness property can be understood by the following construction of the real
numbers : Start with the counting numbers 1, 2, 3, . . . .
• N = {1, 2, 3, 4, . . . } natural numbers ⇒ Can we solve a + x = b for x?
• Z = {. . . , −2, −1, 0, 1, 2, . . . } integers ⇒ Can we solve ax = b for x?
• Q = { pq |p, q ∈ Z, q 6= 0} rational numbers ⇒ Can we solve x2 = 2 for x?
√
• R real numbers, for example : The positive solution to the equation x2 = 2 is 2. This
is an irrational number whose decimal representation is not eventually repeating.
⇒ N⊂Z⊂Q⊂R
In summary, the real numbers R are complete in the sense that they correspond to all points on
the real line, i.e., there are no “holes” or “gaps”, whereas the rationals have “holes” (namely
the irrationals).
1.1
Intervals
Definition 1.1.1. A subset of the real line is called an interval if it contains at least two numbers
and all the real numbers between any of its elements.
Examples :
1. x > −2 defines an infinite interval. Geometrically, it corresponds to a ray on the real line.
2. 3 ≤ x ≤ 6 defines a finite interval. Geometrically, it corresponds to a line segment on the real
line.
3
Finite Intervals. Let a and b be two points such that a < b. By the open interval (a, b) we mean
the set of all points between a and b, that is, the set of all x such that a < x < b. By the closed
interval [a, b] we mean the set of all points between a and b or equal to a or b, that is, the set of all
x such that a ≤ x ≤ b. The points a and b are called the endpoints of the intervals (a, b) and [a, b].
By a half-open interval we mean an open interval (a, b) together with one of its endpoints. There
are two such intervals : [a, b) is the set of all x such that a ≤ x < b and (a, b] is the set of all x such
that a < x ≤ b.
Infinite Intervals. Let a be any number. The set of all points x such that a < x is denoted by
(a, ∞), the set of all points x such that a ≤ x is denoted by [a, ∞). Similarly, (−∞, b) denotes the
set of all points x such that x < b and (−∞, b] denotes the set of all x such that x ≤ b.
1.2
Solving Inequalities
Solve inequalities to find intervals of x ∈ R. Set of all solutions is the solution set of the inequality.
Examples:
1.
2x − 1 < x + 3
2x < x + 4
x < 4.
2. For what values of x is x + 3(2 − x) ≥ 4 − x?
x + 3(2 − x)
x + 6 − 3x
6 − 2x
2
≥
≥
≥
≥
4 − x when
4−x
4−x
x ⇒ x ≤ 2.
3. For what values of x is (x − 4)(x + 3) < 0?
Case 1: (x − 4) > 0 and (x + 3) < 0, =⇒ x > 4 and x < −3.
Impossible since x cannot be both greater than 4 and less than −3.
Case 2: (x − 4) < 0 and (x + 3) > 0, =⇒ x < 4 and x > −3 =⇒ −3 < x < 4.
4
You Try It: Solve the inequality
1.3
2
3
<
.
x−1
2x + 1
The Absolute Value
It is a quantity that gives the magnitude or size of a real number. The absolute value or modulus
of a real number x, denoted by |x|, is given by
x,
if x ≥ 0
|x| =
−x,
if x < 0.
Geometrically, |x| is the distance between x and 0. For example, | − 6| = 6, |5| = 5, |0| = 0.
1.3.1
Properties of the Absolute Value
1. The absolute value of a real number x is non-negative, that is, |x| ≥ 0.
2. The absolute value of a real number x is zero if and only if x = 0, that is, |x| = 0 ⇐⇒ x = 0.
3. In general, if x and y are any two numbers, then
(a) −|x| ≤ x ≤ |x|.
(b) | − x| = |x| and |x − y| = |y − x|.
(c) |x| = |y| implies x = ±y.
(d) |xy| = |x| · |y| and
x
|x|
=
if y 6= 0.
y
|y|
(e) |x + y| ≤ |x| + |y|. (Triangle inequality)
4. If a is any positive number, then
(a) |x| = a if and only if x = ±a.
(b) |x| < a if and only if −a < x < a.
(c) |x| > a if and only if x > a or x < −a.
(d) |x| ≤ a if and only if −a ≤ x ≤ a.
(e) |x| ≥ a if and only if x ≥ a or x ≤ −a.
5
Example: Show that for all real numbers x, | − x| = |x|.
Solution: If x ∈ R, then either x > 0, x = 0 or x < 0. If x > 0, then −x < 0. Thus,
| − x| = −(−x) = x = |x|, that is, | − x| = |x|.
If x = 0, then | − x| = | − 0| = |0| = 0, that is, | − x| = |x|.
If x < 0, then −x > 0. Now |x| = −x = | − x| since −x > 0.
Therefore in all cases | − x| = |x|.
Solving an Equation with Absolute Values: Solve the equation |2x − 3| = 7.
Solution: Hence 2x − 3 = ±7, so there are two possibilities,
2x − 3 = 7
2x = 10
x = 5
2x − 3 = −7
2x = −4
x = −2
The solutions of |2x − 3| = 7 are x = 5 and x = −2.
Solving Inequalities Involving Absolute values: Sole the inequality 5 −
2
< 1.
x
Solution: We have
2
2
< 1 ⇐⇒ −1 < 5 − < 1
x
x
2
⇐⇒ −6 < − < −4
x
1
⇐⇒ 3 > > 2
x
1
1
⇐⇒
<x< .
3
2
Solve the inequalities and show the solution set on the real line. (a) |2x − 3| ≤ 1
5−
Solution: (a)
|2x − 3| ≤ 1 ⇐⇒ −1 ≤ 2x − 3 ≤ 1
⇐⇒ 2 ≤ 2x ≤ 4
⇐⇒ 1 ≤ x ≤ 2.
The solution set is the closed interval [1, 2].
(b)
|2x − 3| ≥ 1 ⇐⇒ 2x − 3 ≥ 1 or 2x − 3 ≤ −1
⇐⇒ x ≥ 2 or x ≤ 1.
The solution set is (−∞, 1] ∪ [2, ∞).
You Try It: Solve the inequality 4|x| < 7x − 6.
6
(b) |2x − 3| ≥ 1.
Chapter 2
Complex Numbers
2
3
A drunkard may not know which number is larger, or , but he knows that 2 bottles of vodka for 3 people
3
5
is better than 3 bottles of vodka for 5 people.
—Sir veMaths
2.1
Introduction
No one person invented complex numbers, but controversies surrounding the use of these numbers
existed in the sixteenth century. In their quest to solve polynomial equations by formulas involving
radicals, early dabblers in mathematics were forced to admit that there were other kinds of numbers
besides positivepintegers. Equations
such as x2 + 2x + 2 = 0 and x3 = 6x + 4 that yielded solutions
p
√
√
√
3
3
1 + −1 and 2 + −2 + 2 − −2 caused particular consternation within the community
√ of
−1
fledgling
mathematical
scholars
because
everyone
knew
that
there
are
no
numbers
such
as
√
and −2, numbers whose square is negative. Such numbers exist only in one’s imagination, or
as one philosopher opined, “the imaginary, (the) bosom child of complex mysticism.” Over time
these imaginary numbers did not go away, mainly because mathematicians as a group are tenacious
and some are even practical. A famous mathematician held that even though they exist in our
imagination, nothing prevents us from employing them in calculations. Mathematicians also hate
to throw anything away. After all, a memory still lingered that negative numbers at first were
branded fictitious. The concept of number evolved over centuries; gradually the set of numbers grew
from just positive integers to include rational numbers, negative numbers, and irrational numbers.
But in the eighteenth century the number concept took a gigantic evolutionary step forward when
the German mathematician Carl Friedrich Gauss put the so-called imaginary numbers or complex
numbers, as they were now beginning to be called on a logical and consistent footing by treating
them as an extension of the real number system.
7
2.2
Complex Numbers
The set of all complex numbers is usually denoted by C. Since x2 ≥ 0 for every real number, x, the
equation
x2 + 1 = 0
has no real solutions.
Introduce the imaginary number 1 ,
i=
which is assumed to have the property
√
−1
√
i2 = ( −1)2 = −1.
Complex numbers are usually written in the form a + bi where a and b are real numbers or can be
regarded as the ordered pair (a, b).
Ordered Pair Equivalent Notation
(3, 4)
3 + 4i
(0, 1)
0+i
(2, 0)
2 + 0i
(4, −2)
4 + (−2)i
Geometrically, a complex number can be viewed either as a point or vector in the xy−plane.
Let us denote
z = a + bi.
The real number a is called the real part of z and the real number b is called the imaginary part
of z.
These numbers are denoted Re(z) and Im(z) respectively.
Example 2.2.1. Re(4 − 3i) = 4
and
Im(4 − 3i) = −3.
When complex numbers are represented geometrically in the xy-coordinate system, the x-axis is
called the real axis, the y-axis, the imaginary axis, and the plane is called the complex plane.
Definition 2.2.1. Two complex numbers a + bi and c + di are defined to be equal, when
a + bi = c + di if
a = c and
b = d.
Numbers of the form where a = 0, then a + bi reduces to 0 + bi = bi, these complex numbers which
correspond to points on the imaginary axis, are called purely imaginary numbers. For example
z = 8i is a purly imaginary number.
1
was first used by the Swiss mathematician Leonhard Euler in 1777.
8
2.2.1
Operations
Complex numbers can be added, subtracted, multiplied and divided.
(a + bi) + (c + di) = (a + c) + (b + d)i.
(a + bi) − (c + di) = (a − c) + (b − d)i.
k(a + bi) = (ka) + (kb)i, k ∈ R. (multiplication
by
a
real
Because (−1)z + z = 0, we denote (−1)z as −z and call it the negative of z.
Example 2.2.2. If z1 = 4 − 5i, z2 = −1 + 6i, find z1 + z2 , z1 − z2 , 3z1 and −z2 .
Solution:
z1 + z2
z1 − z2
3z1
−z2
=
=
=
=
(4 − 5i) + (−1 + 6i) = (4 − 1) + (−5 + 6)i = 3 + i.
(4 − 5i) − (−1 + 6i) = (4 + 1) + (−5 − 6)i = 5 − 11i.
3(4 − 5i) = 12 − 15i.
−1(z2 ) = (−1)(−1 + 6i) = 1 − 6i.
Multiplying two complex numbers as (a + bi)(c + di), treating i2 = −1, this yields
(a + bi)(c + di) = ac + bdi2 + adi + bci = (ac − bd) + (ad + bc)i.
Example 2.2.3.
1. (3 + 2i)(4 + 5i) = (3 · 4 − 2 · 5) + (3 · 5 + 2 · 4)i = 2 + 23i.
2. i2 = (0 + i)(0 + i) = (0 · 0 − 1 · 1) + (0 · 1 + 1 · 0)i = −1.
2.2.2
Axioms of Complex Arithmetic
Given that z1 , z2 , z2 ∈ C, then
1. z1 + z2 = z2 + z1 .
2. z1 z2 = z2 z1 .
3. z1 + (z2 + z3 ) = (z1 + z2 ) + z3 .
4. z1 (z2 z3 ) = (z1 z2 )z3 .
5. z1 (z2 + z3 ) = z1 z2 + z1 z3 .
6. 0 + z = z.
7. z + (−z) = 0.
8. 1 · z = z
9
number)
2.3
2.3.1
Modulus, Complex Conjugate and Division
Complex Conjugate
If z = a + bi, is any complex number, then the conjugate of z denoted by z is defined as
z = a − bi.
Geometrically, z is the reflection of z about the axis.
116/cj.png 116/figures/../../MT 116/cj.png
Figure 2.1: z and its conjugate on an argand diagram
Example 2.3.1.
1. z = 3 + 2i, then z = 3 − 2i.
2. z = −4 − 2i, then z = −4 + 2i.
3. z = 4, then z = 4.
So z = z if and only if z is a real number.
2.3.2
Modulus of a Complex Number
Definition 2.3.1. The modulus of a complex number z = a + bi, denoted |z|, is defined by
√
|z| = a2 + b2 .
10
If b = 0, then z = a is a real number, and
√
√
|z| = a2 + 02 = a2 = |a|.
So the modulus of a real number is simply its modulus value.
Example 2.3.2. Find |z| if z = 3 − 4i.
Solution: |z| =
p
√
32 + (−4)2 = 25 = 5.
Theorem 2.3.1. For any complex number
zz = |z|2
or
|z| =
√
zz.
Proof. If z = a + bi, then
zz = (a + bi)(a − bi) = a2 − abi + bai − b2 i2
= a2 + b 2
= |z|2 .
The modulus of a complex number z has the additional properties |z1 z2 | = |z1 ||z2 | and
2.3.3
Division of Complex Numbers
For division
Example 2.3.3. Express
z1 z 2
z1
=
.
z2
|z2 |2
3 + 4i
in the form a + bi.
1 − 2i
Solution:
3 + 4i
(3 + 4i)(1 + 2i)
=
1 − 2i
(1 − 2i)(1 + 2i)
3 + 6i + 4i + 8i2
=
1 + 2i − 2i − 4i2
−5 + 10i
=
5
= −1 + 2i.
11
|z1 |
z1
=
.
z2
|z2 |
2.3.4
Properties of the Conjugate
Theorem 2.3.2. For any complex numbers z, z1 and z2 , then
(a) z1 + z2 = z1 + z2 .
(b) z1 − z2 = z1 − z2 .
(c) z1 · z2 = z1 · z2 .
z1
z1
(d)
= .
z2
z2
(e) z = z.
Proof. (a) Let z1 = a1 + b1 i and z2 = a2 + b2 i, then
z1 + z2 =
=
=
=
1
(a1 + a2 ) + (b1 + b2 )i
(a1 + a2 ) − (b1 + b2 )i
(a1 − b1 i) + (a2 − b2 i)
z1 + z2 .
√
1
a2 + b2 = ((Re(z))2 + (Im(z)2 )) 2 , then
p
p
Re(z) ≤ |Re(z)| = (Re(z))2 ≤ (Re(z))2 + (Im(z))2 = |z|.
Since |z| = (zz) 2 =
Similarly,
Im(z) ≤ |Im(z)| ≤ |z|.
For any two complex numbers, z1 and z2 , we have that
|z1 + z2 | ≤ |z1 | + |z2 |.
This is called the triangle inequality.
Proof.
|z1 + z2 |2 = (z1 + z2 )(z1 + z2 ) = (z1 + z2 )(z1 + z2 )
= z1 z1 + 2Re(z1 z2 ) + z2 z2 .
Using the fact that 2Re(z1 z2 ) ≤ 2|z1 z2 | = 2|z1 ||z2 |, we get
|z1 + z2 |2 ≤ |z1 |2 + 2|z1 ||z2 | + |z2 |2 = (|z1 | + |z2 |)2 .
Taking square roots the result follows, that is
|z1 + z2 | ≤ |z1 | + |z2 |.
12
2.4
Polar Representation of Complex Numbers
If z = x + iy is a non-zero complex number, r = |z| and θ measures the angle from the positive real
axis to the vector z,
P = (r, θ)
r = directed distance
θ = directed angle
Figure 2.2: Polar form
then
x = r cos θ
and y = r sin θ,
so that z = x + iy can be written as
z = r cos θ + ir sin θ = r(cos θ + i sin θ).
This is called a polar form of z. The angle θ is called an argument of z and is denoted by
θ = arg z. The argument of z is not uniquely determined because we can add or subtract any
multiple of 2π from θ to produce another value of the argument.
One value of the argument in radians that satisfies −π < θ ≤ π is called the principal argument
of z and is denoted by θ = Arg z.
√
Example 2.4.1. Express z = 1 + 3i in polar form using the principal argument.
√
√
√
Solution: The value of r is r = |z| = (1)2 + ( 3)2 = 4 = 2. Since x = 1 and y = 3, it
√
√
follows that 1 = 2 cos θ and 3 = 2 sin θ. So cos θ = 12 and sin θ = 23 . The only value of θ that
satisfies these relations and meets the requirement −π, θ ≤ π is θ = π3 . The polar form of z is
π
π
.
z = 2 cos + i sin
3
3
q
13
We now show how polar forms can be used to give geometric interpretations of multiplication and
division of complex numbers.
Let z1 = r1 (cos θ1 + sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ). Multiplying, we obtain
z1 z2 = r1 r2 [(cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(sin θ1 cos θ2 + cos θ1 sin θ2 )].
Recall:
cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 .
sin(θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 .
We obtain
z1 z2 = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]
which is a polar form of the complex number with modulus r1 r2 and argument θ1 + θ2 . Thus, we
have shown that
|z1 z2 | = |z1 ||z2 | and arg(z1 z2 ) = arg z1 + arg z2 .
Also
z1
r1
= [cos(θ1 − θ2 ) + i sin(θ1 − θ2 )] ,
z2
r2
from which, it follows that
z1
|z1 |
=
,
z2
|z2 |
and
arg
2.5
z1
z2
if z2 6= 0
= arg z1 − arg z2 .
De Moivre’s Formula
If n is a positive integer and z = r(cos θ + i sin θ), then
z n = z · z · z · · · z = rn [cos (θ + θ + · · · + θ) +i sin (θ + θ + · · · + θ)]
|
{z
}
{z
}
|
n terms
n terms
or
z n = rn (cos nθ + i sin nθ).
In the special case, if r = 1, we have for z = (cosθ + i sin θ), so that (2.1) becomes
(cos θ + i sin θ)n = cos nθ + i sin nθ
which is called the De Moivre’s formula.
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(2.1)
2.5.1
Application of De Moivre’s Formula
It is used to obtain roots of complex numbers
Recall from algebra that −2 and 2 are said to be square roots of the number 4 because (−2)2 = 4
and (2)2 = 4. In other words, the two square roots of 4 are distinct solutions of the equation w2 = 4.
If n is a positive integer and z is any complex number, then we define the nth root of z to be any
complex number that satisfies the equation
wn = z
(2.2)
1
and denote the nth root of z by z n .
Let w = ρ(cos α + i sin α) and z = r(cos θ + i sin θ), then
ρn (cos nα + i sin nα) = r(cos θ + i sin θ).
Comparing the moduli of the two sides, we see that
ρn = r
or ρ =
√
n
r
√
where n r denotes the real positive nth root of r. In order to have cos nα = cos θ and sin nα = sin θ,
the angles nα and θ must be either equal or differ by a multiple of 2π, that is
nα = θ + 2kπ, k = 0, ±1, ±2, . . .
θ 2kπ
+
, k = 0, ±1, ±2, . . .
α =
n
n
Thus, the values of w = ρ(cos α + i sin α) that satisfy (2.2) are given by
√
θ 2kπ
θ 2kπ
n
w = r cos
+
+
+ i sin
, k = 0, ±1, ±2, . . .
n
n
n
n
Although there are infinitely many values of k, it can be shown that k = 0, 1, 2, . . . , n − 1 produces
distinct values of w satisfying (2.2), but all other choices of k yield duplicates of these.
Example 2.5.1. Find all the cube roots of −8.
Solution: Since −8 lies on the negative real axis, we can use π as an argument.
Here r = |z| = | − 8| = 8, so a polar form of −8 is
−8 = 8(cos π + i sin π).
Here n = 3, hence
1
3
(−8) =
√
3
8 cos
π 2kπ
+
3
3
+ i sin
15
π 2kπ
+
3
3
,
k = 0, 1, 2.
Thus, the cube roots of −8 are
π
π
= 2
2 cos + i sin
3
3
k = 0,
√ !
√
3
1
+
i = 1 + 3i.
2
2
k = 1, 2(cos π + i sin π) = 2(−1) = −2.
√ !
√
5π
5π
1
3
+ i sin
= 2
−
i = 1 − 3i.
k = 2, 2 cos
3
3
2
2
2.6
2.6.1
Applications of Complex Numbers
The Quadratic Formula
Example 2.6.1. Solve the quadratic equation z 2 + (1 − i)z − 3i = 0.
Solution: From the quadratic formula, we have
1
−(1 − i) + [(1 − i)2 − 4(−3i)] 2
z =
2
i
1
1h
=
−1 + i + (10i) 2 .
2
√
1
We compute (10i) 2 with r = 10 and θ = π2 and n = 2 for k = 0 and k = 1. The two square roots
of 10i are
√ √
√
π
π √
1
1
w0 =
10 cos + i sin
= 10 √ + √ i = 5 + 5i
4
4
2
2
√
√
√
√
1
5π
1
5π
+ i sin
w1 =
10 cos
= 10 − √ − √ i = − 5 − 5i.
4
4
2
2
√
√
√
√
Therefore the two values are z1 = 21 [−1 + i + ( 5 + 5i)] and z2 = 12 [−1 + i + (− 5 − 5i)]. These
solutions written in the form z = a + bi, are
1 √
1 √
z1 = ( 5 − 1) + ( 5 + 1)i
2
2
2.6.2
and
1 √
1 √
z2 = − ( 5 + 1) − ( 5 − 1)i.
2
2
Roots of Polynomials
A polynomial in x is a function of the form
p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 .
Example 2.6.2. x3 − 2x + 4.
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A number (real or complex) a is said to be a root of the polynomial p(x) if p(a) = 0.
Example 2.6.3. x = 1 is a root of x2 − 2x + 1, since 12 − 2 + 1 = 0.
A number a (real or complex) is a root of the polynomial p(x) if and only if (x − a) is a factor of
p(x). It may be the case that you pull more than one factor (x − a) out of the polynomial. In such
cases a is said to be a multiple root of p(x).
A root is called a simple root if it produces one factor.
2.6.3
The Fundamental Theorem of Algebra
Theorem 2.6.1 (The Fundamental Theorem of Algebra). Let p(x) be any polynomial of degree n.
Then p(x) can be factorized into a product of a constant and n factors of the form (x − a), where
a may be real or complex.
Suppose the complex number z is a root of the polynomial, then the complex conjugate z is also a
root.
Example 2.6.4. Let p(z) = z 4 − 4z 3 + 9z 2 − 16z + 20. Given that 2 + i is a root, express p(z) as
a product of real quadratic factors.
Solution: Given that 2 + i is a root, it follows that 2 − i must also be a root and so the quadratic
(z − (2 + i))(z − (2 − i)) = z 2 − 4z + 5
must be a factor. Dividing the given polynomial by this factor gives
p(z) = z 4 − 4z 3 + 9z 2 − 16z + 20 = (z 2 − 4z + 5)(z 2 + 4).
Example 2.6.5. Solve z 3 +3z 2 +2z −6 = 0 and express the left hand side as a product of irreducible
factors.
Solution: Since the equation is a polynomial equation of odd degree there is at least one real
solution. To find that solution by trial and error the factors of the constant terms are substituted
into the polynomial. The factors of 6 are ±1, ±2, ±3, ±6.
Substituting z = 1 gives
1+3+2−6=0
z 3 + 3z 2 + 2z + 6
= z 2 + 4z + 6 and the other
so z = 1 is a solution and (z − 1) is a factor. So
z
−
1
√
solutions are z = −2 ± 2i and so
z 3 + 3z 2 + 2z − 6 = (z − 1)(z 2 + 4z + 6)
as a product of irreducible real factors.
Exercise 2.6.1. Express z 5 − 1 as a product of real linear and quadratic factors.
17