RECISTEREO TRAOEMARK
PíScataway, New Jersey 08~4
• • 1 ~
THE COMPLEX VARIABLES PROBLEM SOLVER®
Copyright © 1987 by Research and Education Association. All rights reserved. No part of this book may be· reproduced in any form without permission of the publisher.
Printed in the United States of America
Library of Congress Catalog Card Number 87-60365
International Standard Book Number 0-87891-604-0
Revised Printing, 1989
reasons unc!erly
Students have generally found complex variables a difficult
subject to understand and learn. Despite the publication of
hundreds of textbooks in this fielci, each one intended to
provide an improvement over previous textbooks, students
continue to remain perplexed as a result of the numerous
conditions that often must be remembered and correlated in
solving a problem. Various possible interpretations of terms
use~ in complex variables hav.e also contributed to much of the
difficulties experienced by students.
In a study of the problem, REA found the following basic
reasons underlying students' diffic.\llties with complex v_ariables - .-;,- taught in schools: ~. /"::
(a) No systematic rules of analysis ha ve been developcd
which students may follow in a step-by-step manner to solve
the usual problems encountered. This results from the f act that
the numerous düferent conditions and principles which may be . . '
involved in a problem lead to many possible different methods
of solution. To prescribe a set of rules to be followed for each
of the possible variations, would involve an enormous number of
rules and steps to be searched through by students, and this
task would perhaps be more burdensome than solving the
problem directly with sorne accompanying trial and error_ to find
the correct solution route.
(b) Textbooks currently available will usually explain a
given principie in a few pages written by a professional who
has an insight in the subject · matter that is not shared by
students. Thé explanations ~re often written in an abstract '
manner which leaves the students confused as to the application
of the principie. The explanations given are not sufficiently
detailed and extensive to make the student aware of the wide
range of applications and different aspects of the principle
being studied. The numerous possible variations of principles
and their applications are usually not discussed, and it is Ieft
for the students to discover these for themselves while doing
iii
rediscover that which has been long known and practiced, but
not published or explained extensively.
(e) The examples usually following the explanation of a
topic are too few in number and too simple to enable the
student to obtain a thorough grasp of the principles involved.
The explanations do not provide sufficient basis to enable a
student to solve problems that may be subsequently assigned
for homework or given on examinations.
The examples are. presented in abbreviated forro which
le aves out mu ch material between steps, and requires · that
students derive the omitted material themselves. As· a result,
students find the examples difficult to understand--contrary to
the purpose of the examples.
Examples are, furthermore, often worded in a confusing
manner. They do not state the problem and then present the
solution. Instead, they pass through. a general discussion,
ne ver revealin g w hat is to be sol ved for.
Examples, also, do not always include diagrams/graphs,
wherever appropriate, and students do not obtain the training
to draw diagrams or graphs to simplify and organize their
thinking.
(d) Students can learn the subject only by doing the
exercises themselves and reviewi~g them in class, to obtain
experience in applying the principles with their different
ramification s.
they are required to devote considerably more time to complex
variables than to other subjects of comparable credits, beca use
they are uncertain with regard to the selection and application
. of the theorems and principies involved. H is also often
necessary for students to discover those "tricks" not revealed
in their texts ( or review books) :> that make it possible to solve
problerns easily. Students must usually resort to methods of
trial-and-error to discover these "tricks", and as a result they
find that they may sometimes spend several hours to solve a
iv --~· ... .__ .. -·-.
instructors usually request students to take turns in writing
solutions on the boards and e:xplaining them · to the class.
Students often find it difficult to explain in a manner that holds
the interest of the class, and enables the reinaining students to
follow -the material written on the boards. The remaining
students seated in the class are, furthermore, too occupied with
copyin g the material from the boards, to listen ·to the oral
explanations and concentrate on the methods of solution.
This book is intended to aid students in complex variables
in overcoming the difficulties described, by supplying detailed
illustrations of the solution methods which are usually not
apparent to students. The solution methods are illustrated by
problems selected from those that are most often assigned for
class work and given on examinations.. The problems are
arranged in order of complexity to enable students to learn and
understand a particular topic by reviewing the problems in
sequence. The problems are illustrated with detailed
step-by-step explanations, fo save the students the . large
amount of time that is often needed to fill in the gaps that are
usually found between steps of illustrations in textbooks or
review I outline books.
The staff of REA considers complex variables a subject
that is best learned by allowing students to view the methods of
analysis and solution techniques themselves. This approach to
learning the subject matter is similar to that practiced in
various scientific laboratories, particularly in the medica! fields.
In using this book, students may review and study the
illustrated . problems at their own pace; they are not limited to
the time allowed for explaining problems on the board in class.
When students want to look up a particular type of
problem and solution, they can readily locate it in the book by
referring to the index which has been extensively prepared. It
is also possible to locate a particular type of problem by
V
glancing at just the material within the boxed portions. To
facilitate rapid scanning of the probl~ms, each problem has a
heavy border around it. Furthermore, each problem is identified
with a number immediately above the problem- ~t the right-hand
margin.
familiarize themselves with the section, "How To Use This Book,"
located in the front pages.
To meet the objectives of this book, staff members of REA
have selected problems usually encountered in assignments and
examinatioñs, and ha ve sol ved each problem meticulously to
illustrate the steps which are difficult for students to
comprehend. Special gratitude is expressed to them for their
efforts in this area, as well as to the numerous contributors who
devoted brief periods of time to this work.
Gratitude is also expressed to the many persons involved in
the difficult task of typing the manuscript with its endless
change~, and to the REA art staff who prepared the numerous
detailed illustrations together with the layout and physical
f eatures of the book.
The difficult task of coordinating the efforts of all persons
was carried out by Carl Fuchs. His conscientious work deserves
much appreciation. He also trained and supervised art and
production personnel in the preparation of the book for printing.
Finally, special thanks are due to Helen Kaufmann for her
unique talents in rendering those difficult border-line decisions
and in making constructive suggestions related to the design and
organization of the book.
JI~ '· '· , I I' • ; :
t i HOWTO
2. Then l
to the "Table
HOW TO USE THIS BOOK
This ibook can be an invaluable aid to students in complex
variables as a supplement to their textbooks. The book is
subdivided into 24 chapters, each dealing with a separate topic.
The subject matter is developed beginning with complex
numbers, geometric representations, De Moivre's Theorem,
sequences and series, continuous functions, limits, Cauchy's
Theorem, power _series, and extending through radius of
convergence, residues, Taylor series, and Laurent's series.
Also included are problems on special kinds of integrals,
conforma! mappings, and symmetry principle. .An extensive
number of applications have been included, since these appear
to be more troublesome to students.
TO LEARN AND UNDERSTAND A TOPIC THOROUGHL Y
1. Re fer to your class text and re ad the. section pertainin g
to the topic. You should become acquainted with the principles
discussed there. These principies, however, may not be clear
to you at that time.
2. Then locate the topic you are looking for by referring
to the "Table of Contents" in front of this book, "The Complex
Variables Problem Solver."
3. Turn to the page where the topic begins and review the
problems under each topic, in the order given. For each topic,
the problems are arran ged in order of complexity, from the
simplest to the more difficult. Sorne problems may appear similar
to others, but each problem has been selected to illustrate a
different point or solution method.
To learn and understand a topic thoroughly and retain its
contents, it will be generally necessary for students to review
the problems several times. Repeated review is essential in
order to gain experience in recognizing the principles that
should be applied, and in selecting the best solution technique.
vii
J
To locate one or more problems related to a ·particular
subject matter, refer to the index. In using the index, be
certain to note that the numbers given there refer to problem
numbers, not page numbers. This arrangement of the index is
intended to facilitate finding a problem more rapidily, since two
or more problems may appear on a page.
If a particular type of problem cannot be found readily, it
is recommended that the student refer to the "Table of
Contents" in the front pages, and then turn to the chapter
which is applicable to the problem being sought. By scannin g
or glancing at the materhtl that is boxed, it will generally be
possible to find problems related to the one being sought,
without consuming considerable time. After the problems have
been located, the solutions can be reviewed and studied in
detail. For this purpose of locating problems rapidly, students
should acquaint themselves with the organization of the book as
found in the "Table of Contents" •
In preparing for an exam, locate the topics to be covered
on the exam in the "Table of Contents," and then review the
problems under those topics severa! times. This should equip
the student with what might be needed for the exam.
vfü
' ¡· "-t_":·.
,1
CONTEN
2 GEOME1 NUMBER
Reprt Appli1 Appfü
3 DE MOIV
a ·pa:rticular
1 COMPLEX NUMBERS, INTRODUCTORY REMARKS 1
lntegers, Rational and Real Numbers 1 Complex Numbers,, Imaginary Unit "in 6 Fundamental Operations with Complex Numbers 8 Field of Complex Numbers 16 Complex Conjugate 19 Absolute Value 21 Inequalities and Identities 26 Polynomials . 30
2 GEOMETRIC REPRESENT ATION OF COMPLEX NUMBERS 35
Representations of Complex Numbers 35 Applications of Complex Numbers in Geometry 39 Applications in Physics 65
3 DE MOIVRE'S THEOREM 67
Polar Form of Complex Numbers 68 Euler's Formula 75 Applications in Geometry 79 Dot and Cross Products 89 De Moivrc' s Theorem 94 Applications in Trigonometry, Identities 108 Series Polynomials 123
ix
4 COMPLEX NUMBERS AS A METRIC SPACE 135
Elements of Set Theory 136 Metric Spaces 139 Open Sets 145 Topological Spaces 148 Closure of a Set 154 Regions, Domains 157 Sequences 161 Complete and Compact Spaces 165 Theorems and Properties 169
5 SEQUENCES AND SERIES OF COMPLEX NUMBERS 176
Sequences 1 77 Series 191 Test for Convergence 197
6 CONTINUOUS MAPPINGS, CONTINUOUS CURVES, STEREOGRAPHIC PROJECTION 217
Mappings 218 Continuous Mappings, Homomorphisms 221 Properties of Topological Spaces 226 Curves 235 Convex Sets, Domains 240 . Continuous Curves 244 ~tereographic Projection, Riemann Sphere 246
7 FUNCTIONS, LIMITS AND CONTINUITY 261 .
Complex Functions 261 Limits 266 Continuity 270 Uniform Continuity 275 Functions Continuous or Connected, Compact, and Bounded Sets 281 ·
X
9 DIFFERENTIJ
1 O ELEMENTAi FUNCTION~
11 COMPLEX 11
Complex Li: Green's Th Cauchy's T Integrals h Application: Indefinite I
12 CAUCHY'S 1 THEOREMS
:E 135
9 DIFFERENTIAL OPERATORS HARMONIC FUNCTIONS 317
Differential Operators, Gradient, Divergence, and Curl Harmonic Functions 322 Diff erential Equations 328
1 O ELEMENT ARY FUNCTIONS, MUL TIPLE-VALUED FUNCTIONS 332
Exponential Functions 332 Trigonometric Function s 335 Hyperbolic Functions 337 Logarithmic Functions 340 Multiple-Valued Functions 346
11 COMPLE:>C INTEGRALS, CAUCHV'·s THEOREM 356
Complex Line Integrals 356 Green's Theorem 364 Cauchy's Theorem, The Cauchy-Gorsat Theorem 369 Integrals Independent of Path 374 Applications 377 Indefinite Integrals 384
12 CAUCHV'S INTEGRAL FORMULA AND RELATED THEOREMS 388
Cauchy' s Integral Formula 389 Applications 396 Theorems 401 Functions with Poles and Zeros 408
xi
Cauchy Convergence Criterion 420 Sorne Useful Theorems 425 Uniform Convergence 437 Power Series of Analytic Functions 442
14 TAYLOR SERIES 447
Taylor Theorem 447 Mac Laurin Series 451 Taylor Expansion 456 Applications 458 Taylor Series for Rational Functions More Applications 468
461
Laurent Series Expansion 474 Examples 482 Applications 490 Lagrange's Expansion 498 Analytic Continuation 502
16 SINGULARITIES, RA TIONAL ANO MEROMORPHIC FUNCTIONS 506
Singularities 506 Poles 512 Properties of Singularities 51 7 Removable Singularities, Singularities at Infinity 525 Theorems 533 Rational and Meromorphic Functions 536
17 RESIDUE CALCULUS 542
Residues 542 Residues at Pales 545
n ·a f F t• p(z) 5'':? .nes1 ues o une; IOP.s q(Z) >lu
Residue Theorem 555 Applications 560
xii
20 MAPPING E LINEAR FR,
Linear Tr~ Rotation ~ Functio: Linear rr:· Fixed Poin Transform; Sorne Spec
21 CONFORM' PROBLEM
Critica! an Con formal Theorems, Harmonic C Application Boundary Poisson's F
22 APPLICATlc
Heat Flow Elasticity Hydrodyna Flow Arour Electrostat
it'"' 18 APPLICATIONS OF RESIDUE CALCULUS, INTEGRATION OF REAL FUNCTIONS 574
19 APPLICATIONS OF RESIDUE CALCULUS.PART 11 611
20 MAPPING BY ELEMENTARY FUNCTIONS ANO LINEAR FRACTIONAL TRANSFORMA TIONS 662
Linear Transformation 662 Rotation, Translation 664 Function t 666 Linear Fractional Transformation 672 Fixed Points 675 Transformations of Regions 676 Sorne Special Functions 683
21 CONFORMAL MAPPINGS, BOUNDARY VALUE PROBLEM 686
Critica! and Non-Critica} Points 687 _ Conforma! Mapping 689 Theorems, Jacobian of the Transformation 699 Harmonic Conjugate, Dirichlet Problem 706
ORPHIC Applications 713 Boundary Value Problem, Neumann Problem, Poisson's Formula 722
finity 525 22. APPLICATIONS IN PHYSICS 727
Heat Flow 727 Elastici ty 7 46 Hydrodynamics and Aerodynamics 749 Flow Around an Object 766 Elecfrostatic Fields 776
xiii
SCHWARZ-CHRISTOFFEL TRANSFORMATION 801
Mapping of Polygons onto the Real Axis 801 Triangles,Rectangles and Degenerate Polygons 804 Various Polygons 821 Applications in Physics 824
24 SPECIAL TOPICS OF COMPLEX ANAL YSIS 828
Analytic Continuation 828 Principie of Reflection 837 Gamma Function 839 Beta Function 849 Infinite Products 856
. Differential Equations 863 Bessel Function 871 Legendre Polynomials 875 Hypergeometric Function 878 Zeta Function 879 Doubly Periodic Functions, Elliptic Functions 881 Asymptotic Expansions 890
APPENDIX 893
INDEX 900
1-16 ' 1-17
1-17
This chart is pr interrelationshi ter. Also shown subject matter.
INTEGERS, 1
2. the set Z o1 3. the set E o~
4. the set Q O'
5. the set R oJ
Using a diagrnm. sets of the typt
GS: THE
~TION 801
801 ygons
IS 828
Ions 881
Field of Complex Numbers, Isomorphism
1-16 , 1-17
1-17
Drill Problems 1-11, 1-12, 1-13, 1-14, 1-15
Modulus Absolute Val u e 1-21 1 1-22, 1-23, 1-24 , 1-26
Polynomials 1-5 J 1-31 , 1-32, 1-33 J
1-34, 1-35, 1-36
Sorne Inequalities 1-23 1 1-24 J 1-25 1-27, 1-28, 1-29, 1-30
This cbart is provided to facilitate rapid understanding of the interrelationships of the topics and subject matter in this chap­ ter. Also shown are the problem numbers associated with the subject matter.
INTEGERS, RATIONAL AND REAL NUMBERS • PROBLEM 1-f .
Describe the following sets:
l. the set N of positive integers 2. tbe set z of all integers and zero 3. the set E of positive even integers 4. tbe set Q of rational numbers 5. the set R of real numbers
Using a diagram, illustrate the relationships between the sets of the type set - subset.
1
Solution: In this book we shall use the commonly accepted logical notation. A collection of things is called a set, a f ami ly, or a ·el ass. For our purposes, al l three terms mean the same thing. Sets are designated by upper-case letters, and their elements by the corresponding lower-case letters: a G A is read as "a belongs to the set A" or "a is an element of A". For example: Texas G USA. On the other hand, the notation b ~ A is read as "b is not an element of A".
A set of objects having a particular property is denoted by {x: P}. This statement is read "the set of all x which po­ ssess the property P". Sometimes, we simply list all of the elements of a set. In such cases, we explicitly show the con­ tents of the set. For example, {1,7,4} is the set consisting of the three elements 1, 4, and 7:
l. The set N of positive integers (natural numbers) is
N ;:·{1,2,3, ..• }
Note that "::" indica tes "equal by defini tion'!. N is closed under the operation of addition and multiplication. That is, i f a and b are natural numbers, then the sum a + b and the product ab are also natural numbers.
2. The set Z consists of z = { ... -4,-3,-2,-1,0,1,2,3, ..• } (2)
It is easy to see that Z is closed under the operation of addi­ tion, multiplication and subtraction. However, it is not closed under the operation of division because 4 G Z &3 G Z,but 4 3 'f, z.
3. The set of positive even integers is given by
E : {e : e = 2n, n G N} (3)
4. The set of rational numbers is
Q = { q : q = ii' m G Z, n G Z, n :f o} (4)
Q is closed under the four basic arithmetic operations: addi­ tion, subtraction, multiplication and division.
5. The set of real numbers R consists of all rational numbers and all irrational numbers. An irrational number is a number that cannot be expressed as a ratio E' where á and b are inte-
gers and b :f O. An example of an irrational number is /2' = 1.4142 ....
A one-to-one correspondence can be established between the set of real numbers and a line. Each number is represented by a point of the line called the real axis.
' 1 1 1 ,__.......___.__ ... _-'--___,__...___
-7 -6 -5 -4 -3 -2 -1-~ o 112' 2 3 4 s 6 7
Fig. 1
Any posi ti ve e
Any positive i
Any integer or
By defi::li tion,
1. a = 1 and
2. a {; z an\:
4. a eR an.
Svlution: 'fh
mly accepted ~alled a set, a ·ee terms mean the · letters, and · letters: a G A n element of A''. the notation
erty is clenoted all x which po­
st all of the ly show the con­ set consisting
nbers) is
1- b and the '
1 G Z & 3 G Z, bu t
>y
(3)
tional numbers r is a number 1d b are inte-
>er is 12'
t ,,
A set A is a subset of B, i.e. is included in B, when
X G A => X G J3 ( 5)
The notation =>is read ,as: then. We write A e: B to in- dicate that A is a subset of B. Let us note that A e B ctoes not exclude the possibility that A =B.
Any positive even integer is a positive integer, therefore
E e N (6)
N e Z (7)
Any integer or zero can be expressed in the form j, thus
Z e Q (8)
By definition, any rational number is a real number, thus
Q e R (9) We have
E e: N e Z e: Q e R ( 10)
Fig. 2
• PROBLEM 1-2
ax + b = O
Using the definitions given in Problem 1-1, find the smallest set containing the solutions of eq.(1) when
l. a = 1 and b G E
2. a G Z and b G N
3. a G Q and b G Q
4. a G R and b G Q.
Solution: The solution of eq.(1)
X = b a
.
.
l. If a = 1 and b (; E, then x = -b. Of course x ~ E because E is the set of positive even integers. The smallest set con- taining x is Z.
2. Si nce a b Z and b G N, b -- e Q. a
3. Since both a and b are rational numbers, b
b is a rational a number, hence, -- G Q. a
4. Here, a GR and b G Q. Let us note that ~ does not have
to be a rational number. For example, when a = /2 and b = 1,
b is not a rational number. a
Hence, b is a real number, a b G R. a
• PROBLEM 1-3
Imagine that due to sorne perfidious law, all irrational numbers have been banned from private, public and scientific life. As a law-abiding citizen and a good math­ ematician, what would you do to make this quandary bear­ able?
Solution: Certainly you know that there are severa! theories on real numbers. Axiomatic theories, even though convenient, are not very instructive. The theory of Cantor* (1845-1918), the founder of the theory of sets, is particularly handy. It takes as a starting point the set of rational numbers Q and considers sequences of rational numbers. These so-called Cauchy sequences play a key role. In the space a of Cauchy sequences, Cantor introduces the equivalence relation - The set of real numbers R is defined as a quotient space
R : ~ (1) The algebraic operations are carried over from the set of rational numbers Q to the space R. The reader will find a variety of books on Cantor's method, and it is not within our scope to present it here. An important conclusion of this theory (noted by Hausdorff) is that the space R is the comple­ mcnt of the set of rational numbers,
R = Q (2)
A practica! conclusion from this is that if we take any real number r e R and any positive rational number e b Q, regardless of how small, there exists a rational number q e Q such that
lr-qj <e (3)
Each real number can be approximated with any required accuracy by rational numbers. Computers, electronics, space technology etc. use rational numbers.
At sorne point in your life, you were II9 feet tall. Chances are that when measuring your height, you used the ap­ proximation 4 ft. 4 inches, and everybody was satisfied with the accuracy. Tbe answer to the question posed in the problem is to approximate all irrational nt~bers with rational numbers.
4
*Georg Car
l918 in Halle: the most prom1n( zÜrich, Gotti ng• in Halle. Can te theory of sets foundations of lished in the "
1. A linear eq·
2. Is the solu
Solution: l.
2. Dividing eq
b is a a rational
:> a does not have
12 and b = 1,
:lary bear-
~veral theories igh convenient, :-* ( 1845-1918), irly handy. It mmbers Q and ~ so-called ~ a of Cauchy ~lation The space
(1) the set of will find a not wi thin our
;ion of this l is the comple-
(2)
take any real ~ g Q, regardless G Q such that
(3)
in the problem ational numbers.
,, t'·
~-.
*Georg Cantor was born in 1845 in Petrograd and he died in 1918 in Halle. Cantor, the son of a Danish merchant, was one of the most prominent German mathematicians. Cantor studied at zÜrich, Gottingen, and Berlín. In 1872, he became a professor in Halle- Cantor introduced many original ideas into the tbeory of sets (theory of aggregates). He also worked on the foundations of the theory of numbers, these findings·were pub­ lished in the "Annalen" in 1379.
• PROBLEM 1-4
ax + b o
where a and b are rational numbers. Is the solution of this equation always a rational number?
2. Is the solution of the equation
ax 2 + bx + e = O
(1)
(2)
where a,b,c G R (i.e. all coefficients are real numbers) always a real number?
Solution: l. The solution of eq.(1) is
b X = -a' a f. O
Since a and b are rational numbers, number.
b is always a rational a
2. Dividing eq.(2) by a~ O, we obtain the equation
2 b e X + - X + - 0 a a
which can be transformed to
so that
+ 2a = 4a 2
-Jb 2 -4ac b -X¡ - 4a 2 - 2a -
-jb 2 -4ac b
X2 - 2a 4a 2
-b-lb 2 -4ac 2a
5
(3)
(4)
(5)
(6)
(7)
(8)
b X :: -2a (9)
The situation becomes more complicated when b 2 -4ac <O. The
left-hand side of eq.(6) is then positive, while the right-hand side is negative. Such equations cannot be solved in terms of real numbers.
COMPLEX NUMBERS, IMAGINARY UNIT "i" • PROBlEM 1-5
A complex number z can be expressed in the form
z = a + bi
wbere a and b are real numbers and i has the property
i 2 = -1
(1)
(2)
Find the solutions of the following equations using com­ plex numbers
l. x 2 + 1 = O
2. 2x 2 + X + 3 = 0
(3)
(4)
Solution: The representation of complex numbers in the form z = a + bi is probably the most popular. The mathematical op­ erations of addition, subtraction and multiplication of real numbers carry on into the complex number system. However, one must remember that i 2 = -1. Thus,
1. x 2 + 1 = O (5)
x 2 = -1 (6)
X¡ i' (8)
Let us now verify our results. Substituting x 1 we f ind
i in to eq.(3)
Substituting x2 = -i into eq.(3) results in
x 2 + 1 = (-i) 2 + 1 = i 2 + 1 = O
(9)
(10)
2. To solve eq.(4), we use the quadratic formula to obtain the two solutions.
X 1 =
Substi tutii
[ 1 . 12:
substituting ;-{2
2 (1~ + i 12:
in the domain oi the equation
was introduced t notation of com~
One should following fallac
-1 i 2 = i
thus -1 l. E:x
Solution: The s numbers a+ib whe addition and mul
(a+ib)
(a+ib)
(9)
- 4ac < O. The le the right-hand lved in terms of
lT "i" • PROBLEM 1-5
(3}
(4)
~s in the form riathematical op­ ~ation of real '.l. However, one
(5}
(6)
(7)
(8)
4 1 i 123 -4 - -4- (12)
Let·us verify that x1 and x2 are indeed the roots of eq.{4).
Substituting x1 in to eq. ( 4) results in
2[-~+ i ~r + ( 1 . -4+ 1 ~) + 3
2[1 . 123 23) 1 i 123 + 3 o = 16- 1 -a-16 - 4 + 4
(13)
2 (-~ - i mr -- + 4 . (-~- i ~) + 3
- 2 [ 1 + . 123 23) + ( 1 . ~) + 3 = o - 16 1 -g- 16 -4- 1
(14)
Complex numbers were introduced into mathematics in connection with the solution of algebraic equations. Since it is impossible to solve the polynomial equation
x 2 + 1 = O (15)
in the domain of real numbers, the imaginary unit i, defined by the equation
i 2 = -1 (16)
was introduced by Euler in 1777 and fully incorporated into the notation of complex numbers by Gauss.
One should execute caution in handl irrg i,. r-T etc. The following fallacy indicates why.
-1 i 2 = i • 1 = r-t r-r = l<-1 )(-1) = /1 = 1
thus -1 = l. Explain what went wrong.
• PROBLEM 1-6
Define the system of complex numbers. When are two num­ bers equal?
Solution: The system of complex numbers is the set of all numbers a+ib where a and b are real, with two binary operations, addition and multiplication, defined as follows:
(a+ib) + (c+id) - a+c+i(b+d)
{a+ib) (c+id) _ (ac-bd) + i(ad+bc)
(1)
- ( 2)
The system of complex numbers, denoted (C,+,•) is a field whose multiplicative neutral element is 1,
1 = 1 + iO (3)
and in which i 2 = -1.
( 4)
The f ield of real numbers R is a subset of C. Strictly speaking, the elements of C of the form a = a + iO form a sub­ field of C, which is isomorphic as a field to the field of real numbers. The function a~ a• 1 + iO from R into e maps R one­ to-one onto this subfield, preserving products and sums, so it is an isomorphism.
Two complex numbers
Zz = a2 + ib2
(5)
and (6) bi b2
The student should keep in mind that the statement z 1 > z 2
is meaningless unl~ss both z 1 and z 2 are real.
FUNDAMENTA~ OPERATIONS WITH COMPLEX NUMBERS
l. Prove the theorem:
Zt + Z2 = Z2 + Z1
(a + ib) +
(4 + 2i) +
(2 + 4i) -
l. Let
Z1 + Z2
(-a+ ib)
(3 - 3i)
(1)
(2)
(3)
(4)
Here we used the fact that the addition of real numbers is com­ mutative. We have,
8
+ i 2 b1b2
(4)
of C. Strictly t + iO form a sub­ > the f ield of real .nto e maps R one­ :s and surns, so i t
(5)
(6)
• PROBLEM 1-7
[a1 + (a2+a3)] + i[b1 + (b2+b3)] (6)
f(a1+a2) + a3] + i[(b1+b2) + b 3 ]
(z1+z2) + Z3
We used the fact that addition of real nurnbers is associative.
2. (a+ib) + (-a+ib) = ib + ib = 2bi (7)
(4+2i) + (3-3i) 7 - i
(2+4i) - (6-3i) + (4+7i)
2 + 4i 6 + 3i + 4 + 7i 14i
(8)
(9)
(10)
Z¡ • Z2 = Z2 • Z1
(b) Multiplication is associative
2. Reduce the following numbers to the form a + ib.
(a) (3-2i) • (4+5i) + (3-2i) • (4-5i)
(b) (2-i). (4+31). (5+2i)
Solution: Let Z1 = a1 + ib l
Z2 = a2 + ib2
(1)
(2)
( 3)
(4)
(5)
(7)
(a) Since multiplication of real numbers is commutative, we obtain z1 • z2 = (a 1+ib 1 ) • (a 2 +ib 2 ) = a 1 a 2 + i(a
1 b
2 +b
1 a
2 )
+ i 2 b 1 b 2 = a 1 a 2 - b 1b 2 + i(a 1 b 2+a 2b 1 ) (8)
(b) Multiplication of real numbers is associative, tbus z 1 • (z2 • z 3 ) = (a 1 +ib 1 ) • [ (a2+ib 2 ) • (a 3 +ib 3 )]
9
- ·- --··---·-- - . ·--------- . ..i
= (a1+ib1). r (a2a3-b2b3) + i(b2a3+a2b3 )l a1(a2a3-b2b3) - b1(b2a3+a2b3)
+ i{b1(a2a3-b2b3) + a1(b2a3+a2b3)] (9)
= (a1a2-b1b2)a3 - (a 1b2+a2b1)b3 + i[(a1b2+b1a2)a 3 + (a1a2-b1b2)b3]
[(a1+ib1) • (a2+ib2 )] • (a3+ib3)
= ( Z 1 • Z2 ) • Z 3
(e) Now, to prove eq.(3), we write z
1 • (z2 +z 3 ) = (a 1 +ib 1 ) • [(a 2 +ib 2 ) + (a 3 +ib 3 )]
= (a 1 +ib 1 ) • [(a 2 +a 3 ) + i(b 2 +b 3 )]
a 1 (a 2 +a 3 ) - b 1 (b 2 +b 3 ) + ib 1 (a 2 +a 3 ) + ia 1 (b2 +b 3 )
= (a 1 a 2 -b 1b 2 ) + i(b 1 a 2 +a 1 b 2 ) (10)
+ (a 1 a 3 -b 1b 3 ) + i(a1 b 3 +b 1 a 3 )
z i • Z2 + z 1 • z 3
Instead of wri ting z 1 • (z 2 • z 3 ), we skip the brackets and wri te Z 1 • Z2 • Z 3.
2.(a) (3-2i) • (4+5i) + (3-2i) • (4-Si)
= (3-.2i). [(4+5i) + (4-5i)] (li)
= ( 3-2i) • 8 :::: 24 - 16i
( 11+2i )( 5+2i) (b) (2-i). (4+3i). (5+2i)
= 51 + 32i (12)
• PROBLEM 1-9
l. Prove that if z1 = a 1 + ib 1 and z 2 = a 2 + ib 2 are com­ plex numbers, then there exists a unique complex number z such that
Z l + Z = Z2 (1)
2. Given two complex numbers z 1 and z 2 , we define the dif­ ference z, denoted by z 2 - z 1 , to be a complex number sucb that
Find
is true if and only if
and
10
(2)
(3)
(4)
(5)
(6)
(7)
1. Prove the :
The quotient. o.
2. Reduce the
Eq • ( 5) is true
)J ) ]
(9)
+b 3
(1)
(2)
(3)
(4)
(5)
(6)
(7)
r ~ ~-
a = a 2 - ª1 b = b2 - b1
Thus z exists and is unique.
2. (3+4i)
(9)
(10)
( 11)
It immediately follows from the above theorem that the differ­ ence of two complex numbers exists and is unique.
• PROBLEM 1-10
l. Prove the following tbeorem:
Given two complex numbers z 1 = a 1 + ib 1 and z2 =,a2 + ib2, z2 f O, there exists a unique complex number z, such that
ZZ2 = Z1
The quotient of z1 and z2 is denoted by ~. Z2
2. Reduce the following nwiibers to the form a + ib.
(a)
(b)
Solution: l. Let z = a + ib.
We shall show that eq.(1) determines uniquely a and b. We have
ZZ2 (a+ib)(a2+ib2)
Eq.(5) is true if, and only if,
aa2 bb2 a1
ba2 + ab2 b1
a1a2 + b1b2 a = a2 + b2
2 2
Hence, z = a + ib is uniquely determined.
11
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
_,.l
12 - 9i - Si + 6i 2
16 - 12i + 12i - 9i 2
(10) ~...._
7+i 2 - 3i (7+i)(7-i)
6 8 . 25 - 25 1 -
+
+ 17. 13 l (11)
l. Reduce the followíng numbers to the form a + ib
(a) i .. - 3i 3 + 4i 2 + 2i - 6 (1)
(b) (2!_) .. l+i
(2)
2. Prove that z 1z2 = O if and only if at least one of the numbers z 1 , z2 is zero.
Solution: 1.(a) i 4 - 3i 3 + 4i 2 + 2i - 6
(-1)(-1) 3. (-l)i + 4 • (-1) + 2i - 6
1 + 3i - 4 + 2i - 6 -9 + Si
(b) [2!_) lf
_ [-4 J 2 Lrr+T J - 2i = - 4
2. Let z1 = a1 + ib1 and z 2 = a 2 + ib 2 , then
(a1a2-b1b2) + i(a1b2+a2b1)
Eq.(5) implies that
(a~ + b~)(a~ + b~)
o
(3)
(4)
(5)
(6)
(7)
(8)
If both a~ + b~ and a~ + b~ are different from zero, eq.(8) is not true.
12
Let us not itive integer,
1. Show that :
(a 2 -ia:
Derive the equ
from eq. ( 3).
1 + .
sino.+ sir
i - icosa. ·
Eqs. (8) and (f Thus, we have
2. The h~:f '!'.-í
) - 17i ·16 + 9
7 • 31
¡ ~ ¡s .. !
:
Henc·e, z 1 z2 = O i f and only i f a t least one of the numbers z 1 , z 2 is zero.
Let us note tbat if z is a complex number and n is a pos­ i tive integer, then
n z z·z· ..• •z
n times
• PROBLEM 1-12
then l+z . a -- = 1 cot -1- z 2
2. If a and b are real numbers, show that
(a 2 -iab-b 2 ){a+ib) = a 3 - ib 3
Derive the equation
a 6 + b6
cot ~ = 1 +.cosa 2 sin a
and eq.(1) into eq.(2), we find
1+ cosa+ i sina = i(l+cosa) 1 - cosa - i sina sin a
Eq.(6) can be transformed to sina + sinacosa + isin 2 a = (i+icosa)(l-cosa-isina)
i -icosa+sinn+ icoscx- icos 2 cx+ cosa.sincx
Eq. ( 7) leads to
(1)
(2)
(3)
( 4)
Eqs. (8) and (9) are obviously true, since sin 2 cx + cos 2 cx l. Thus, we have preve eq.(2).
2. The left-hand side of eq.(3) yields
(5)
(6)
(7)
(8)
(9)
(10)
1J
Let us replace, in eq.(10), a by a 2 and b by -ib2 •
a + a 2
b + -ib 2
[a i. - a 2 b2 + b 4] [a 2 + b2]
= [(a2-b2)2 + a2b2](a2+b2)
(a2)3 _ i(-ib2)3 =as+ bs
( 11)
(12)
(13)
Thus,
if n =
.n 1
(5)
In general i i.m = 1
Combining eqs.(5) and (7), we find
Hence,
4x + 2ixy - iy = 5 + 3i
14
(7)
(8)
(9)
From eq. ( 1)
You would cer1 tbe modern al~ wri tten in 1:hE
"Ex irra1 imaginariae, e utili tas; ets ·: c.ent, tamen r:.c quomoolo quae:: etiam interver See Werke, Gc1
-ib2 •
( 11)
(12)
(13)
ne.
(7)
(8)
(9)
~---
Solution: Two complex numbers z 1 are equai if and only if
and
we obtain
(2)
(3)
( 4)
(5)
x 4 + a 4 = (x4a/-r-1)(x-al-r-T)(x+a/r-I)(x-a/r-T) (1)
Prove that he was right_
Solution: Multiplying the first two terms of eq.(l), we find
(x+a/-r-T) (x-a/-r-I)
(x+a/r-T)( x-a/r-1) = x 2 - a 2 r-1
Thus,
(2)
(3)
(4)
You would certainly appreciate the clarity and sirnplicity of the modern algebraic notation if you would know that eq.(1) was written in the form
"Ex irrationalibus oriuntur quantitates impossibiles seu imaginariae, quarum mira est natura, et tamen non contemnenda utilitas; etsi enim ipsae per se aliquid impossibile signifi­ cent, tamen non tantum ostendunt fontem impossibilitatis, et quomoolo quaestio corrigi potuerit, ne esset impossibilis, sed etiam interventu ipsarum exprimí possunt quantitates reales." See Werke, Gerhardt ed. (Berlín, 1850), VII, 69. "-
15
a PROBLEM 1-16
An important concept in mathernatics is the notion of a field. A set S on which two operations of addition and multiplication are defined is called a field if the opera­ tions satisfy the following axioms:
I. If a G S, b G S, then a+b G S and ab G S.
II. Commutative Laws.
a+b=b+a
a(bc) ( ab)c
IV. Distributive Law.
If a e S, b G S, e G S, then
a(b+c) = ab + ac
V. There exists an element O G S such that for every a G S
a + O = O + a = a
VI. There exists an element 1 G S such that for every a G S
Show that the zero element, O G S, and the unit element, 1 G S are unique.
Show that the set of all complex numbers, that is numbers a + ib, where a and b are real numbers, with addition and multiplication, as defined in Problem 1-6,. forms a field.
Solution: To show that O G S is unique, let us assume that it isn't, that is that there exist two zeros O and O*. Then by axiom V
and hence,
and hence
The set of comJ operations of :: Problem 1-6. ~ a e e and b G e and 1-8, we shc mutative, assoc O + iO = O, be<
For all a + i6 a + i 8 is any <
Thus, the set · plication as d properties and
Consider the s wi th two binar
defined as
(
(a,b) ·
between all · bers a
or every
or every
as sume tha t i t O*. Then by
(1)
(2)
(3)
IL
Now, if there are two uni t elements 1 and 1*, then by axiom VI
1* • 1
l* • 1
and hence
1 = 1*
l*
1
(4)
(5)
(6)
The set of complex numbers is usually denoted by C. The operations of addition and multiplication were defined in Problem 1-6. Since C is the set of all numbers a + iB, then if a G C and b G C, then a + b G C and ab G C. In Problems 1-7 and 1-8, we showed that multiplication and addition are com­ mutative, associative and distributive. The zero element is O + iO = O, because
(a+iB) + (O+iO) a + iB
For all a + iB G C. The unit element is 1 + iO. Indeed if a + iB is any complex nurnber, then
(7)
(a+iS)(l+iO) := a + iB (8)
Thus, the set C of all complex numbers with addition,and multi­ plication as defined in Problem 1-6 is a field. It has all the properties and satisfies all the theorems concerning fields.
• PROBLEM 1-17
Consider the set of all ordered pairs (a,b) of real numbers, with two binary operations, addition, +, and multiplication,
defined as follows:
(a,b) • (c,d) = (ac-bd, ad+bc)
(1)
(2)
It is very easy to show that this set is a field, with the zero element (0,0), indeed
(a,b) + (0,0) (a, b) (3)
and the unit element (1,0)
(a,b) • (1,0) = (a-O, O+b) = (a,b). (4)
Two pairs (a,b) and (c,d.) are equal if, and only if, a= e and b = d. A one-to-one correspondence can be established between all the pairs of the form (a,0) and the real nu~­ bers a
(a,O) ~ a (5)
The set of real numbers
The set of all ordered pairs (a,b)
Fiq. 1
From eqs.(1) and (2), we can deduce the following proper­ ties of the one-to-one correspondence
(a,O) ++ a
(a,0) • (b,O) = (ab, O) ++ ab
A one-to-one function which satisfies eqs.(6), (7) and
(6)
( 7)
(5) is called an isomorphism. We say that the set of all p'irs of the type (a,O) is isomorphic with the set af real numbers R. In practice, it is convenient to write
(a,O) = a (8)
Denote (0,1) by i' (0,1) i (9)
It is not without a reason that we use the same "i" as in the complex number a + ib. Compute
and
(a,O) + (0,l)(b,0)
Can you draw any conclusions based on the statement (a,0) = a?
Solution: Since (0,1) = i we get
1 2 = (0,1) • (0,1) = (-1,0) = -1
(a,b) = (a,O) + (O,b) = (a,O) + (0,l)(b,O)
Substituting (a,0) = a
The set o: the set of com1
Fig. 2 oi of complex numl
positive e'\·en
COMPLEX C
I. Definitio
The conju
The imaginary
.ng proper-
(7) and
:-ite
(8)
·4 ..•
(a, b) = a + ib (15)
That leads us to the conclusion tbat a complex number can be represented by a+ ib, or by an ordered pair (a,b).
The set of real numbers R can be regarded as a subset of the set of complex numbers C.
Fig. 2 of Problem 1-1 can be modified to include the set of complex numbers, as shown.
positive even inte(Jers rational numbers
complex nwnbers
real numbers
Fig. 2
In this book we will use the representation a + ib rather than (a,b).
COMPLEX CONJUGATE
• PROBLEM 1-18
I. Definition
::! "i" as in
Show that
-z = z
II. Definition
The real part of the complex number z noted by Re z, or Re(z},. hence
Re(z) = a
The imaginary part of z is denoted by Im(z), hence,
19
(7)
(8)
(9)
(10)
Solution: l. If z1 = a1 + ib 1 and z 2 = a 2 + ib2, then
Z = (a+ib) = (a-ib) = a + ib = z
II. !(z+z) = !(a+ib+a-ib) = a= Re(z)
fi<z-z) = ii<a+ib-a+ib) = 2 ;: = b = Im(z)
{12)
(13)
(14)
(15)
From eqs.(3), (4) and (5), it follows that if R(z1,z2, Z3, •.. ) is a rational expression of complex numbers z1,z2, z 3, ..• , then
R(z1 ,z2 ,z3, ... ) RCz1,z2,z3, ... > c16)
• PROBLEM 1-19
Prove that if tbe product of two complex numbers z 1 and z 2 is real and different from zero, tben there exists a real number a such tbat
(l)
Z2 = a2 + ib2
The product of z 1 and z 2 is real and different from zero
Z1Z2 r 0
and we can write
Find two numbe
t:__
(7)
(8)
(9)
(10)
, ... ) (16)
(l)
l . t
- secause z1z2 is real and different from zero and a~ + b~ is real and different from zero, we obtain, from eq.(6)
• PROBLEM 1-20
Find two numbers whose sum is 5 and whose product is 9 .
Solution: Let us denote the numbers by z 1 and z 2 • Then
Z¡ + Z2 = 5 and
(7)
(1)
At this point, we don't know if z 1 and z 2 are real or comp'lex.
Thus,
Z2
Zi,1
5 5+1· IIT Z1 ,2 = - Z2,2 = 2 ~
1 f b s .IIT d There is on y one set o num ers z1 = 2 - 1 ~ an
~ + i ~ satisfying eq.(1). It is clear that
Z1 = Zi
ABSOLUTE VALUE
Definition
The modulus or absolute value of a complex number z =a+ ib, written fzl, is given by
21
(1)
- l
The absolute value of a complex number is a nonnegative real number.
l. Evaluate the absolute value of the following numbers
3 + 2i
-6
2. Show that if z 1 and z 2 are complex numbers, then
lzl = lzl lz 12 = z z jRe(z) 1 ~ lzl, IIm(z)I < lzl
lz1llz2I = fz!z2l
l3+2i 1 19+4 = II3
¡si( = 125 = 5
l-6 I = ¡gs = 6
2. lz 1 = /az+b 2 = {a2+(-b) 2 = lzl
z z = ( a+ib )( a-ib) = ª2 + b2 =
Sin ce fRe(z) ¡ 2 + (Im(z) l 2
i t follows that
lzl-~ IIm(z) ¡2
(2)
(3)
(4)
(5)
1136
Now, we shall prove the last identity 1z 1z2 I = ¡z 1 ¡ ¡z 2 ¡. lz1z21 2 = (z1z2Hz1z2)
( zl zl )(z2z2)
Eq.(15) leads
lz1z2. · .z 1Z 1 = lzil lz2 l 0 •• lz 1 l lz 1 n- n n- n (16)
Thus,
(17)
22
t ~
r
Note that the r therefore the l z ~ R. Thus, \\
or
Prove the the0r absol ute val ue exceed the sum
Solution: Sine
Since 2Re(z) =
• PROBLEM 1-22
Show that the only solution of the equation 1 2 - - = z z
is z = l.
2z 1 = zz = lzl 2
or 2z = f zl 2 + 1
(1)
(2)
(3)
Note that the right-hand side of eq.(3) is a real number, therefore the left-hand side also has to be a real number, z GR. Thus, we can drop the absolute value syrnbol and·write
2z = z 2 + 1
The only solution is
• PROBLEM 1-23
Prove the theorem known as the triangle inequality. The absolute value of the sum of two complex numbers cannot exceed the sum of their absolute values, i.e.
(1)
Sin ce 2Re(z) = z + z, eq. ( 2) yields
fz1+
Hence,
and
lzil 2 + 2lz1llz2I +
Clz1l+lz2l) 2
lz1+z2f < lz1I + lz2I
(2)
(3)
(4)
(5)
~" _L__ __
2J
Using mathematical induction, we can extend eq.(5) to any fi­ nite number of complex numbers
1 ~ z 1· < ~ lz 1
K=l K - K=l K (6)
e PROBLEM 1-24
Prove that if z1 and z2 are complex numbers, then
11z11 - 1 z2 l I ~ 1 z1 - z2 I
Solution: Applying
we find
-lzl ~ Re(z) ~ lzl
lzl ~ -Re(z) ~ -lzl Therefore, eq.(4) leads to
lz1-z2l 2 = lz1l 2 - 2Re(z1z2) + lz2l 2
> 1 z l l 2 - 21 Zt Z2 1 + 1 Z2 l 2 = ( 1z1 1 ._ 1 Z2 1 ) 2
Hence,
(1)
(2)
(3)
(4)
(5)
(6)
(7)
. (8)
(9)
(1)
Solution: Note that f al indicates the absolute value of a, while la+ibl is the modulus of a complex number.
From eq.(1), we obtain
a 2 + 2lal (bl + b 2 _.:: 2(a 2 +b 2
) (3)
24
, I
t•
[ ~
Eq. ( 6), an 'ª 1 = lb 1. To show th
we observe tha'.
and
Tbe similari ty lus of a compl.e a is not accide a + i b , then w·h
lzl
Show l that
Solution: By a
(2)
(3)
Hence,
(4)
(5)
(6)
Eq.(6), and therefore eq.(1), becomes an equality wben 'ª 1 = lb 1. To show the other part i.e.
we observe that
la+ibf = {a2+b 2 < lal + lbl
a 2 + b 2 < a 2 + 2labl + b 2
O~ 2labl
{7)
(8)
(9)
(10)
Tbe similarity between the notation la+ibl denoting the modu­ lus of a complex number and f al denoting the absolute value of a is not accidental. If z is a complex nurnber such that z = a + ib, then when b = O, we obtain
lzl = la+ibl = la2+b 2 = la2 = lal (11)
Thus, the modulus reduces to the absólute value when the complex number has no imaginary part. The absolüte value or modulus notation should not cause any dbnfusion.
• PROBLEM 1-26
Show ihat
lz1+z2t 2 + lz1+z2l 2 = 2Clz11 2+lz2l 2 )+4Rez1Rez2 (1)
Solution: By applying
we shall transform the left-hand side of eq.(1) ..
lz1+z2l 2 + lz1+z?l 2 = (z1+z2Hz1+z2)
2lz11 2 + 2lz2l 2 + z1(z2+z2) + z1(z2+z2)
2clz1l 2+lz2l 2> + Cz1+z1)Cz2+z2)
Since z 1 = a 1 + ib1 and z 1 = a 1 - ib1, we obtain
z1 + z1 = 2a1 = 2Rez1
Substituting eqs.(4) and (5) into eq.(3) results in
lz1+z 2l 2 + lz1+z21 2 = 2clz1 l 2 +lz2l 2 ) + 4Rez1Rez2
INEQUALITIES AND IDENTITIES
1 z-1 1 + 1 z+ 1 I _: 212 (1)
Solution: Since both sides·of eq.(1) are positive, we can evaluate the second power of eq.{1).
lz-11 1 + lz+ll 2 + 2lz-ll lz+ll _: 8
Hence,
and
or
(z-l)(z-1) + (z+l)(z+l) + 2f(z-l)(z+l)1 < s
lzl 2 - z - z + 1 + lzl 2 + z + z + 1 + 2lz 2 ~11 < 8
2lzl 2 + 2 + 2lz 2-ll < 8
lzl 2 + lz 2 -ll < 3
lzl 2 + lz 2 +(-l)I _: lzl 2 + fz 2 I + l-11 = lzl 2 + lzl 2 + 1
Note that
Since - lzl _:l., also lzl 2 _: 1, and thus
lzl 2 + lzl 2 + 1 < 1 + 1 + 1 = 3
and
1 ~ z.w.f z =
- E lz.w -z w.1 2
j=l J j=l J l2j<K_:n J K K J
(1)
26
Solution: 'We
The right-hand
1<. -· (z1z1+ ..
j,K=l
That completes
*Joseph Lo in Paris in 181 ian of the eig astronomer.
(5)
,.., "
Solution: We shall utilize the identity
f zl 2 = zz The right-hand side of eq.(1) is equal to
(z1z1+z2z2+ •.• +z z )(w1w1+w2w2+ ••• +w w) - n n n n
r Cz .w -z w. )(z .w -z w. > 12j<K2n J K K J J K K J
¿
K K J J
r 12_j<K2_Il
At this point let us observe that n
E a.b + E a b. = E a.b l<j<K<n J K l<j<K<n K J j,K=l J K
- - - - j1K
Hence, eq.(3) leads to n n n E z.z.w w -
j, K=l J J K K
¿ z.z.w w + ¿ z.z w.w J J K K j,K=l _J K J K
n L
j=l
. E z.zw.w J, K=l J K J K
n
n . E (zJ.w.) (zw) J ,K=l J K K
jf K
(z1w1+z2w2+ ••• +z w )(z1w1+z2w2+ ••• +Z-W) n n n n
n = ( E
j=l
J J j=l J J
That completes tbe proof.
1 ~ z .w · 12
j=l J J
(2)
(3)
(4)
(5)
*Joseph Louis Lagrange was born in Turin in 1736 and died in Paris in 1813. Lagrange was one of the greatest mathematic­ ian of the eighteenth century. He was also a physicist and an astronomer.
L 27.
Prove Cauchy's inequality for the complex numbers.
Solution: This is a well known inequality and appears in dif­ ferent forms and in various branches of mathematics. For the complex numbers, it can be stated as follows:
1 ~ z.w.(
j=l J J j=l J J=l J
For any complex number, lzl 2 is a real number and lzl 2 ~O. Thus, in Lagrange's identity (see Problem 1-28, eq.(1))
We obtain
<
lz -1 2
j=l • PROBLEM 1-30
Prove tbe triangle or Minkowski* inequality for the com­ plex numbers:
/~ ¡z.+w.¡• < G• + ~ Jj=l J J -Jj!l ¡zj¡; Jj~l ¡wjl' (1)
Solution: First, let us consider the left-hand side of eq.(1) and utilize inequality
¡z. +w. I < ¡z · I + ¡wj 1 J J - J (2)
n n E ¡z.+w.¡2 = E ¡z. +w. l Jz. +w. ¡
j=l J J j=l J J J J
n < E ¡z.+w.¡(jz.f + Jw · I >
j=l J J J J
n n = E ¡z.+w.¡¡z.¡ + E ¡z. + w. ¡ ¡w. ¡
j=l J J J j=l J J J (3)
Applying Cauchy's inequality to the last two sums in eq.(3), we f ind
28
For those of y01 reasoning in obi
Cauchy's ir
n E ¡z. +w.
j=l J J
From eq . ( 8) , we
*Hermann Mi laid the foundat ski investigateJ closely connecte crystallography. theory of relati and time. The s gree relative dt:· ª sj.ngle absoJ\.' .. These ideas were 1909.
•BLEM 1-29
Ul!lbers. 1 tnd appears in dif- 1ematics. For the ; :
e r an d 1 z 1 2 ~ O . -28, eq.(1))
(1)
(2)
(3)
-- ~--· ---- ··-··. ···-······-·····---·· .
~ 1 zJ. + w J. 11 zJ. 1 ~ J [ . ~ 1 z . +w . 1 2 ] [ ~ 1 z . 1 2
] j=l "1 J=l J J j=l J
(4)
~ ¡z.+w.¡¡w.¡ < ·1 ~ ¡z.+w.¡~-][ ~-·~:~·,·-:] j=l J J J -"1 j~l J J j=l J
(5)
For those of you who are somewhat confused, let us repeat the reasoning in qbtaining eqs.(4) and (5).
Cauchy's inequality can be written in the form
1 ~ 1 n 1
~ ZJ. VJ. < E 1 Z • V • 1 j=l j=l J J
~ /[.~ ¡z.¡ 2 ][.~ ¡v.1 2 ] '1 J=l J J=l J
Hence,
J J J
J ~ J
< J[ ~ 1 z . +w . 12] [ ~ ·, z . l 2) - j=l J J j=l J
From eqs.(3), (4) and (5) we conclude that
n n n E j Z • + W . 1 2 < E 1 Z • + W • 11 Z • 1 + E 1 ZJ. + WJ. , j WJ. 1
j=l J J - j=l J J J j=l
lz. + w -1 2
(6)
(7)
(8)
(9)
*Hermann Minkow~ki (1864-1909), a German mathematician, laid the foundation for the theory of convex bodies. Minkow­ ski investigated tbe problem of regular lattices, wbich is closely connected with the theory oí numbers and geometric crystallograpby. In 1905, after Einstein developed the special tbeory of relativity, Minkowski showed the mutual link of space and time. The separation of space from time is to a great de­ gree relative depending on tbe system of reference. There is a single absolute forro of existence of matter: space-time. These ideas were developed in bis book "Raum and Zeit", Leipzig 1909.
29
POLYNOMIALS • PROBLEM 1-31
Let a ,a , •.. ,a1,ao be real numbers. If a is a complex n n-1 root of the polynomial equation
a zn + a zn- l + ..• + a 1 z + a 0 = O n n-1 (1)
then show that a is also a root of the equation.
Solution: Since a is a root of eq.(1), we get
n n-1 ªnª +a a + ... + a1a + aó D-1
o (2)
we f ind
we f ind
a an +a a n- 1 + ••• + a1a + ªº = o n n-1
Hence, if a is a complex root of a polynomial equation with real coefficients, then a is also a root of the equation.
(3)
(4)
(5)
(6)
(7)
Verify that 2 + i is a root of the equation
z 4 - 5z 3 + 3z 2 + 19z - 30 = O
and find the other three roots.
(1)
Solution: Since 2 + i is one of the roots of the equation,
2 + i = 2 - i is also a root of the equa tion { see Problern 1-31). Let us denote the roots of eq.(1) by z1, z2, Z3, z~, then
JO
Thus,
Solving ·
a e n
Dividing by a,
a e n
• PROBLEM 1-31
e PROBLEM 1-32
Far z1 = 2+ i, z2 = 2-i, we have
(z-2-i)(z-2+i) = z 2 -4z+5 (3)
To find (z-z3)(z-z .. ), we have to divide z 4-5z 3 +3z 2 +19z-30 by z 2 -4z+5.
Thus,
z 2 -4z+5 ~4-5zl+3z2+19z-30 -(z .. -4z 3 +5z 2 )
-zl-2z2 +19z -(-z 3 +4z 2 -5z)
-6z2+24z-30 -(-6z2 +24z-30)
(z-3)(z+2) O
Therefore the four roots of eq.(1) are
Z1 = 2 + i, Z2 = 2 - i 1 Z3 = 3, Zt+ = -2
(4)
(5)
(6)
(7)
• PROBLEM 1-33
Show that if .the rational number % (where a and 8 have no
common factor except ±1) satisfies the polynomial equa­ tion
o
factors of ao and ªn' respectively.
(1)
a ""n + n-1 13n- 1 6n 0 n.... a 0
_ 1 a B + ... + a 1 a + a 0 = (2)
Dividing by a, we get
n-1 n-2 n-1 a 0Bn a a + a a B + a1B = -~~ n n-1 a
(3)
Since a , •.. ,a1 anda and B are integers, the left-hand side n
of eq.(3) is an integer. Hence, the right-hand side is also an integer. Let us note that a has no common factor with B.
J1
!..! is an integer. a
(4)
T~e left-hand side is, of course, an integer, thus tbe. right­ hand side is an integer. The real numbers a and B bave no common factor. Tberefore, B divides a .
n a ; is an integer.
• PROBlEM 1-34
6z~ - 47z 3 + 148z2 - 167z + 52 = O (1)
Solution: The integer factors of 6 and 52 are, respectively (see Problem 1-33),
f or 6 ±1, ±2, ±3, ±6 (2)
f or 52 ±1, ±2, ±4, ±13
Hence, the possible solutions of eq.(1) are
+! "1 +.!. +.?.. +i +13 +13 +13 ±1, +.=. ±4, ±2, ±13, -21 -3, -6' -31 -3, - 2 ' - 3 , -6
1 4 Given the proper time, one could easily verify that 2 and 3 a~e·the solutions. Thus the polynomial
(3)
(4)
is a factor of 6z"-47z 3+i48z2 -167z+52. Multiplying eq.(4) by 6 ·to obtain integer coefficients, we find
6z 2 - llz + 4 (5)
By dividing,. we find the other factor
~-zi.-: 17z 3 +148~ 2 -167z+ 52 = (6z 2 -llz+4)(z 2 -6z+l3) (6)
Splv~ng t~e equation
we f ind z 2
- 6z + 13 = O
Zi, = Z3 (3+2i) = 3 - 2i
1 4 m~nce, 2' 3 1 3 + 2i, 3 - 2i are the solutions of eq. (1).
J2
(7)
(8)
(9)
(4)
• PROBLEM 1-34
1 4 y that 2 and 3
{4)
• PROBLEM 1-35
Prove that the sum and the product of all the roots of n n-1
ªn z + ªn- 1 z + ... + a i z + a o = O, ªn 'f O ( 1)
is equal to
n -a ¿; z. ~
Solution: The product symbol "Il" is defined as
k JI
K
(2)
(3)
If z 1,z2, ... ,zn are all the roots, then eq.(1) can be written
in the factored form
j=l
Z 1 + Z2 + •.. + Z n
(-1)~ ~ a n
• PRQBLEM 1-3.6
A number is called an algebraic nwnber if it is a solu­ tion of a polynomial equation with integer coefficients.
(1) n n-1 a z +a z + ... + a 1z + a 0 =O n n-1
JJ
1. Show that all rational numbers are algebraic.
2. Show that /2 + 1 and 315 + 2i are algebraic numbers.
Solution: l. Any rational number can be represented in the
f orm ~ where a and b are integers. The solution of the equa-
tion bz - a = o (2)
is z = E· Hence, all rational numbers are algebraic numbers.
2. Letz=/2+1, then
z - 1 = /2 (3)
The second power of eq.(3) is
z 2 - 2z + 1 = 2 (4)
The number 12+ 1 is an algebraic number, since it is a solution of
z 2 - 2z - 1 o
Now, let z V'5 + 2i, then
z - 2i
z 3 - 6iz2
- 8)
Taking the second power of eq.(8), we find
z 6 + 12z 1t - lOz 3 + 48z 2 + 120z + 89 = O
(5)
(6)
(7)
(8)
(9)
Thus, 315 + 2i is an algebraic number. It has been proven that n = 3.1415 ... ande= 2.7182 ... are transcendental numbers. It is still known whether or not numbers such as ne or n+e are transcendental. The algebraic theorv of numbers was de­ veloped by Lagrange, Gauss, Dedekind, and Kummer among others.
34
This chart is 1 interrelationsl ter. Also shm subject matte:r-
REPRESEN'!
2) Sketch the
3 + 4i, 2 -
3) Vlhat is th comp lex number ing the narnber
Solution: 1) terpretation o
Any cqmpl two-dimensionP­ system. The n where the rea: imaginary par complex plane, diagram.
raic.
numbers.
89 = o
(5)
(6)
(7)
(8)
(9)
; been proven that 1dental numbers. !h as ne or lT+e f numbers was de- 1mmer among others.
- CHAPTER2
Complex Numbers as Vectors 2-2 , 2-3
Applications in Geometry 2-4 - 2-25
Complex Numbers as Points 2-1
Center of Mass 2-26, 2-27
This chart is provided to facilitate rapid understanding of the interrelationships of the topics and subject matter in this chap­ ter. Also shown are the problem numbers associated with the subject matter.
REPRESENTATIONS OF COMPLEX NUMBERS
_• PROBlEM 2-1
1) Discuss the geometrical interpretation of the complex number and describe the complex plane.
2) Sketch the following numbers in the complex plane
3 + 4i, 2 - i, 6i, 5, -1 - i, 2i
3) What is the relationship between the modulus of a complex number z and the distance of the point represent­ ing the narnber from the origin of the coordinate system?
Solution: 1) We will now briefly outline the geometrical in­ terpretation of the complex nurnbers.
Any complex number can be represented by a point in a two-dimensional plane endowed with a rectangular coordinate system. The number z = x + iy corresponds to the point P(x, y), where the real part x of tbe number z is the abscissa and the imaginary part y is the ordinate. This plane is called the complex plane, also referred to as the z plane, or the Argand* diagram.
35
1 X Re Z
In this plane, the x axis is the real axis and the y axis is the imaginary axis. This procedure establishes a one-to­ one and onto correspondence be~ween the field of complex num­ bers and the two-dimensional plane.
2) Since x is the real part and y is the imaginary part, the six numbers
3 + 4i, 2 - i, 6i, 5' -1 - i, 2i .
are located as shown in Fig. 2. We use the terms "complex number" and "point" interchangeably.
-2 -1
• 2-i
*Jean Robert Argand was born in Geneva in 1768,and died in Paris iQ 1822. Argand was the first to give the geometrical rep­ resentation of a complex number and applied it to show that every algebraic equation has a root. In 1806 Argand published bis major book "Essai".
• PROBLEM 2-2
A complex number z = x + iy can be represented by a point P(x,y). The number x + iy can also be considered as a vector OP whose initial point is the origin O and whose terminal point is P(x,y). The vector OP is called the positiop vector of point P.
Im
P(x,y)
M
Fig.
J6
a) Z1 + Z2 (
Fj
Through P1, draw ates of P3 are ( z 1+z2. In vecto
~--··- -·-· ·-~---------___;.----r-----
:; and the y axis ;hes a one-to­ Jf complex num-
inary part, the
Re
1768, and died in le geometrical rep­ to show that every published his
• PROBLEM 2-2
by a point red as a and whose lled the
- l.
2.
Two vectors OM and NR having the same magnitude (length) and direction are considered equal
OM = NR (1) even though they may have different initial points. We write OP = x + iy Based on vector representation of complex nurnbers, find graphically
a) z 1 + z2 (z1 ,z2 are any two arbitrary number~)
b) (2-2i) + (1+5i)
Find graphically.the difference
(2)
Solution: l. Let P1 and P2 be the points z1 and z2.
Im
Fig. 2
Through P 1 , draw P1P1 equ~l and parallel to OP2. The coordin­ ates of P3 are (x1+x2, y1+y2). I!ence, P3 represents the point z 1+z 2 • In vector notation,
OP1 + OP2 = OP3 (3)
37
b) Graphically the surn of the numbers (2-2i) and (1+5i) is found to be (3+3i), as shown in Fig. 3. This result can be easily verified analytically.
Im
e
Fig. 4
To find the sum (1+2i) + (2-3i) + (-5+4i), we first dra.w vec·tor OA = (1+2i). Then, from the point A we draw the vector AB = 2 - 3i. From point B, we then draw BC = -5 + 4i. Vector OC is the sum
OC = OA + AB + BC
Im
Re
Fig. 5
we draw a vector from the point z 1 to the point z2, z Therefore,
OP1 + P1P2 = OP2 = z2
Give the graphical interpretation of the following in­ equali ties:
l. lz1+z2l ~ (z1I + lz2l
2. lz1+z2+z3 I ~ lz11 + lz2 I + lz3 I 3. lz11 - lz2 I ~ lz1-z2 I
J8
(1)
(2)
(3)
-
·.-
--
solution: 1. 1engths of the the lengths of to the length e
Thus,
3. We bave
APPLICATIC lN GEOMET:
Find the equa
- 4i. Vector OC is
(5)
1 .....
Solution: l. N~te that lz1f, lz2I and lz 1+z 2 1 represent the lengths of the s1des of a triangle (see Fig. 1). The sum of the lengths of two sides of a triangle is greater than or equal to the length of the tbird side.
Tbus,
Im
Re Fig. l
Re Fig. 2
The geometric interpretation of inequality (2) is tbe statement that in a plane a straigbt line is the sbortest distance be­ tween two points O and A.
3. We bave ( 4)
Thus, . (5)
Tbe geometrical interpretation of inequality (5) is that a side of a triangle has a length greater than or equal to the difference in lengths of the other two sides.
APPLICA TIONS OF COMPLEX NUMBERS IN GEOMETRY
• PROBlEM 2-4
Find the equation of
•
39
2. an ellipse wi th major axis of length 2a and foci at (c,O) and (-c,O).
Solution: l. A circle of radius r and center at zo is the collection of all points whose distance froM zo is r, as shown in Fig. l.
Im
Re Fig. l
Therefore, the equation of a circle in the complex plane is
lz-zol = r (1)
The unit circle is the circle with radius r ~ 1 and cEnter at the origin. A disk of radius r and center z 0 is the set of all points z satisfying the inequality,
1 z - zo 1 < r (2)
2. An ellips.e is the set of all points z such that .. the sum of the distances from z to the f oci must be equal to the majar axis of the ellipse (see Fig. 2).
Im
-a
Fig. 2
Thus, the equation of an ellipse in the complex plane is
(z+cl +~z-cl = 2a (3)
• PROBLEM 2-5
l. Re(z) > 3
3. lz I < 4
) = p 5.
s. lz 2-zl 2
3. Tbe region
Fig. 3
. foci at
1 · at z 0 is tbe ;o is r, as s bown
ple x plane i s
(1)
(2)
tha t ·t h e sum o f to t h e majar
x plane is
) p
7 _ Jz 2 -1J = s
8 . Jz 2 - zJ
Solution: l. Fig . l.
2
> o
The r egion Re(z) > 3 is the hal f - plane shown in
Im
Fig. 1
2 . The commo n part of t he hal f-plane Re(z) > 1 and the half­ plane Im( z) < 1 is s hown in Fig . 2 .
Im
1
Im(z) _;l
Fiq . 2
3 . Tbe r egion Jz l < 4 i s a disk of radius 4 with center at the
I m
origin as shown in Fig. 3.
4. The region lzl > 4 is also shown in Fig. 3. Tbe sets lzl < 4 and lzl ~ 4-have no common elements.
5. Let z = x + iy, then
Re ( z 2 ) = x 2 - y 2 = p (1)
This equation represents a hyperbola (see Fig. 4). it becomes a pair of straight lines.
For p = O,
6. We have Im(z 2 ) = 2xy = p.
The equation xy = ~ represents a rectangular {equilateral) hyperbola, as shown in Fig. 5.
Fig. 5
/ /
42
(2)
(3)
(4)
1 2 . z -
8. Let lz2 -zl lz2-
is the product <
• ~e
represents the lemniscate of Bernoulli, as shown in Fig. 6.
Im
Re
Fig. 6
This conclusion can be reached without long calculations. It follows that
lz2 -ll = f (z+l)(z-l)f = fz+lf fz-11 = s
lz2-ll is the product of the distances of z f rom 1 and -1.
8. Let lz2-zl = 2. Tben
1 z2-zl = 1 z(z-1) J = lzl lz-11 = 2
(5)
(6)
is the product of the distances of z from O and l. Since it is constant (equal to 2), eq.(6) represents the lemniscate.
• PROBLEM 2-6
lz+ll = 2 z-1
lz+1¡ 2
or (z+l)(z+l) = 4(z-l)(z-1) (3)
so that 5 5 - zz - 3 z - 3 z + 1 = o. (4)
Eq. (4) can be written in the f orm
5 - 5 16 - o (5) e z - 3H z - 3 ) - 9- or
lz §.12 = 16 3 9
(6)
Eq.(7) represents a circle of radius ~ with center at (~,~). as
4J
shown in Fig. l. Thus, all the points z lying on the circle satisfy eq.(1).
Im
Fig. l
The solution to this problem can also be found if we rep­ resent z by
Z = X + iy
lx+l + iyf = 2lx-1 + iyf (8) or
(x+l) 2 + y 2 = 4[(x-1) 2 + i 2 J (9)
( X - i) 2 + y 2 = 196 ( 10 )
As expected, eq.(10) represents a circle of radius ~ with 5 center at (3,0).
• PROBLEM 2-7
Find the equation of the straight line passing through two given points A(x 1 ,y 1 ) and B(x 2 ,y 2 ).
Solution: Points A and B can be represented by the complex nurnbers
and
Re Fig. l
(1)
Let z be the position vector of any point C on the line passing through A and B.
44 tw~-- --~·. ~ ---·.
The so;.1~!'
z = or
or
Solution: Let
(8)
(9)
PROBLEM 2-7
through two
the complex
OB + BC = OC or z2 + BC = z.
Hence,
and
Hence,
Vectors BC and BA are collinear, therefore, we can write
BC = t BA or
The sought equation is
X -.X2 = Y - y2 X1 ..;. X2 Yl - Y2
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Eq~(lO) is the standard equation of the s~~aight lin~. Eq.(9) is the parametric equati?n of the line.
Another way of describi.ng a straight line is the syrmnet­ ric equation. Since BC and CA are collinear, we get
a. BC = B CA ( 11 ) or
Solving eq.(12) for z, we find
or
X y
• PROBLEM 2-8
What requirement must be fulfilled for the points z 1 , z 2 and z 3 to be collinear?
(12)
(13)
( 14)
Solution: Let us assume that three complex numbers z1, z2 and
z 3 are collinear, that is, that all three numbers lie on one line, as shown in Fig. l.
Im
Re
Pig. 1 p
In such case, the vectors z 1 - Z3 and z2 - z 3 are parallel and we can write
where t is a real parameter. From eq.(1), we conclude that z1, z2, Z3 are collinear if and only if
is a real number.
• PROBLEM 2-9
Show that if the points z1, z 2 , Z3 are collinear, then there exist real number$ a, S, y ~ R, such that
lal + IBI + IYI >O
a + S + y o
o
(1)
(2)
(3)
(1)
(2)
Solution: As established in the preceeding problem (Problem 2-8), three complex numbers z 1 , z 2 , Z3 are coilinear if and only if the ratio
Z¡ - Z3 Z2 - Z3
(5)
and
(6)
46
2
Thus
Solution: Fig.
(4)
(5)
(6)
(7)
and Z¡ - tz2 + (t-l)Z3
• PROBLEM 2-10
Find the point z which divides the line segment (z 1 ,z2 )
ª1 in the ratio -, (a 1 +a 2 ) 1 O. ª2
(8)
(9)
(10)
Solution: The points z1, z2 and z divides the segment (z1,z2) in the
tm
are collinear. The point z ratio ~' as shown in Fig .l.
<l2
and
e
• PROBLEM 2-11
Find the equation of the straigh't line perpendicular to a complex vector zo.
Solution: Fig. 1 illustrates the situation. Im
Re
Fig.
47
(1)
(2)
(3)
Let z be an arbitrary point on the line p. We bave
1 OB 1 2 = f OA f 2 + 1 AB 1 2 (1)
. Substituting
into eq.(1), we find
(4)
Hence,
zz zozo + (z-zo )(z-zo)
From eq.(4), we find
zzo + zoz 2zozo (5)
Thus, the equation of a straight line perpendicular to Zo is
zoz + zoz = 2fzol 2 (6)
• PROBLEM 2-12
Próve that two vectors ·z 1 and z 2 are perpendicular if, and only if
(1)
Solution: The statement a if and only if B, means that a im­ plies 8 and tbat B implies a. We shall accomodate the follow­ ing notation:
a => B reads a implies B
-a<=> B reads a if, and only if B
We bave to prove that
[z i and z 2 are) ( - + - = o) perpendicular <=> Ziz 2 ziz 2
First we sball prove ~>
We assume that z1 and z 2 are perpendicular as shown in Fig. l.
Im
lz2-Z1l2 :
Therefore,
If
From Fil
We bave
( 5)
(6)
(1 )
Then
lz2 - z1l 2
lz1f 2 + lz2l 2
Z¡Zz + Z¡Z2
If
Then
Z¡?.¡ + Z2Z2 - Z¡Zz - Z¡Zz
which imp l i es tbat Z ¡ and Z2 are pe rpe n dicu l ar .
(2)
( 3)
( 4)
(5)
(6)
, mean s t hat a im- d * nod a t e the fo llow- q. e . ·
;; shown i n Fig . 1.
*q.e . d. [Latin quod erat de monstrandum ) which was to be preved .
• PROBLEM 2-13
Using compl e x numbers , fi nd t h e l ength of the median of a t riangl e whose vertices are A(x 1 ,y 1 ) , B( x 2 ,y 2 ) and C(x3,y3) .
Im
B
Re f'ig . 1
Sol u t i on: Le t z1, z2 , Z 3 b e complex number s representin g ver­ tices A, B and C, respect ive l y .
From Fig . 1, we find
49
1 CD! = IBDI
Remembering that
lzl = lx+iyl = lx 2 +y 2
and substituting z1, z 2 , and Z3 into eq.(6), we find
~/(x2+~3-2X1) 2 + (y2+y3-2Y1) 2
(1) :
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Let z1, z2, Z3 be three complex numbers such that
(1)
(2)
Show that these numbers are the vertices of an equilateral triangle inscribed in the unit circle.
Im
Re
as shown in Fig
Let z i , z2, Z3 Prove that
Solution: Sin
an equilateral
Solution: From eq.(2), we see that z.1, z 2 , z 3 are positioned ·0 n the unit circle. Let
as shown in Fig. l.
Then
(3)
To show that the triangle is equilateral we have to prove that
First, let us show that
Since Z1 + Z2 + Z3 = 0,
then
which is equivalent to
4 + 1 4 - 1 o
In a similar manner, we verify that
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Hence, the triangle A1A2A3 is equilateral and inscribed in the unit circle.
• PROBLEM 2-15
Let z 1 , z 2 , Z3 be the vertices of an equilateral triangle. Prove that
lzzl 2 + z1z3+ Z3Z1 = lz31 2 + z1z2 + z2z1 (1)
Solution: Since the triangle is equilateral
or
51
(2)
(3)
Jm
Eq . ( 4) l e ads to (5)
1Z¡ 12 + 1Zzl2 - Z 1Z2 Z1Z2 1z2l2 + 1z31 2 Z3Z2 Z2Z3
1z 1 l 2 + 1z 31 2 Z ¡ Z 3 Z3Z ¡
Eq . ( 5) i s e quiv a l e nt Lo
1z 1 l 2 Z2Z 1 Z1Z2 1z3 l 2 Z3Z2 Z2Z1 ( 6)
1Z2l2 Z3Z2 Z2Z1 1Z ¡l 2 Z¡Z ~ Z3Z¡ (7 )
Subtracting eq . (7) from e q . ( 6) we fi nd
1Z2 l 2 + Z¡ Z3 + Z3Z ¡ = 1z3 l 2 + ;¿ l Z2 + Z2Z1 (8)
q. e .d. • PROBLEM 2 -16
l. Le t z 1 and z 2 b e two non- paralle l vector s . Prove tb a t if a a nd B are r ea l numb e r s and
OZ ¡ + f3z 2 = 0, ( 1)
the n a = e = o 2. Prove t h at the diagonals of a para l lelogram bisect each oth e r.
Solution: l . Le t z 1 = a 1 + ib 1 and z 2
az 1 + f3z2 = ( a a 1+ Ba2 ) + i(ab1+ Bb2 ) o
Eq. ( 3 ) is e quival e n t t o
aa 1 + Ba2 o
Th e solutio n is 8 = o a
l"f __ ~-lb2. T - , l . e .
a1 a2 i f z 1 and z 2 are non-paral l e l.
( 2)
( 3)
(4)
(5)
( 6)
2 . Th e quadril a t e r a l OABC s hown i n F i g . 1 i s a par a llelogram .
52
From the figu1
He nce ,
Hint : Rep r< numbers .
~
(6)
(7)
(8)
A real parameter a exists such that
O < a < 1 (8)
(9)
z1(a-S) + z2(l-a-S) = O (11)
Since z1 and z2 are non-parallel 1 by part I of this problem, we obtain
Hence,
• PROBLEM 2-17
Find the point of intersection of the medians of a tri­ angle.
·Hin~: Represent the vertices of thé triangle by complex numbers.
Solution: The triangle is shown in Fig. l.
53
(12)
(13)
Im
B
D
o Re
Fig . 1
The medians of a triangle intersect at one point. Therefore. it is e no ugh to take . into account two med i ans, say AC and OD. From the fi~ure, we obtain
We can write
(1)
(2)
( 3 )
(4)
( 5 )
(6)
(7)
where a a nd B are pos itive real p a rameters. Again, f rom the ..,.,_.__ figure
or
( 8 )
(9)
(10)
Applyin g t he r esult s of Problem 2-16 ( e q s .(1 ) and (2)), we get
B CL o - 2
Solving for a.
ABCD is a quac the complex .,,, Prove tha t ABC
Solution: We
( AB1 pa
First we shal gram, then i t: length . Us i n ,
From Fig . 1 , ,
Now we wil 1 ABCD i s a p a ·
... ~e
(1)
(2)
(3)
( 4)
(5)
(6)
(7)
( 11)
(12)
So l ving far a and B, we f ind
2 CL = -3,
B = 1 3
Therefore , the med ians of a tri ang l e trisect each other .
(13)
• PROBLEM 2- 18
ABCD is a q uadrilater a l whose vertices are represented by the complex vectors z1 , z2, z 3 , Z4 as shown in Fig . l. Prove t h at ABCD is a parallel ogram if , a nd on l y if,
Z ¡ - Z2 + Z3 - Z4 = Ü (1)
Im
A
Solutio n: We have to prove that
<= > z - z +z -z ( ABCD is a ) ( parallel ogram 1 2 3 4
First we shall show that = > is true . If ABCD is a parallelo­ gram, t h e n its sides AB and DC must be parallel and o f equal length . Using vector nota tion, we can wri te
(2)
l\B Z 2 Z1 ( 3)
DC Z3 Z4 ( 4)
Substituting eqs . (3) a nd ( 4) in to eq . ( 2) results in
Z2 Z1 Z3 Z4 (5)
or Z1 Z2 + Z3 Z4 o (6)
Now we wi ll s how t ha t <= i s true . If z 1 -z
2 +z
.55
q.e.d.
From eq.(1) we conclude that the rema1n1ng two sides of ABCD, namely AD and BC are also parallel and of equal length. Therefore,
(10)
then (12}
• PROBLEM 2-19
Prove that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
Solution: Let z1~ z 2 , z 3 , z 4 represent vertices A, B, C, D as shown in Fig. l.
Im
e
AE EC and (1)
Thus, we con e
Sol ution : Ql
( 10)
Ne have
where
Substituting eqs .(2), (3) and (5) into eq.(4), we obtain
Z3-Z¡ + Z2 -Z4 Z2 - Z ¡ =
2 2
o r Z¡ - Z2 + Z3 - Zt+ = Ü
It was s hown in Problem 2-18 that if z 1 , z 2 , z 3 and Z4
represent the vertices of a quadrilate ral, then the quadri­ lateral is a parallelogra.1l if and only i f
Z¡ - Zz + Z3 - Zt+ = Ü
Thus, we conclude that ABCD is a parallelogram.
• PROBLEM 2-20
ABCD is a quadrilateral and E, F, G, H are midpoi nts of its sides . Prove that EFGII is a parallelogram.
Solution: Quadrilateral ABCD is s hown in Fig. l.
lm
Re
(3)
( 4)
(5)
(6)
(7)
Let z 1 , z 2 , z 3 , Z4 be the position vectors of A, B, C , D, r e ­ s pec ti vely .
To find the positjon v e ctor of F we compute
(1)
AF AB Z2 - Z1 ( 3) 2 2
Thus,
The posi tion vector of F is Z1 + Z2 2
In the same wa~ we compute the position vector of G as
z2+ z3. 2
Respectively, the position vector of H is z3+ z~, and
the position vector of E is z1+ z~. 2
2
A quactrilateral ABCD is a parallelogram, if and only if,
Substituting the position vectors of E, F, G, H we find
o
• PROBLEM 2-21
Prove the theorem:
If OABC is a parallelogram and AE bisects side OC, then AE trisects the diagonal OB (see Fig. 1).
Im
B
Fig. l
Solution: Let z1 be the position vector of A and z 2 be the position vector of C.
58
(6)
(7)
Again, from th
Substi tuting E
The sum of t equal to the
(2)
( 3)
(4)
(5)
z + z f H i s 3 4 , and
2
(6)
\ and z 2 be t h e
\Ve h ::i.ve OB = z 1 + Zz
A positive r ea l number a e x ists s u ch that
OD = a OB = a(z 1+z2)
Fr om Fi g . 1, we obta in
OE 1 oc Z2 2 2
EA OA OE = Z¡ - Zz 2
A positive r ea l number B ex i sts such t hat
Again, from the figure
OE + ED
Subs tituting eqs .(2), (3) and (5) i nto eq . (6) result s in
~2 + s[z1 - 2;) = a (z1+z2)
o r o
Applying t he r esul ts of Proble m 2-16, pa.r t I, we obtain
B a = O
a nd
ex = f3
2 3
• PROBLEM 2-22
Prove that
What i s the relations hip between t hi s ide ntity and the fol lowing t h eor e m?
(1)
The s um of the squares of the s ides of a para lle l ogram is eq ua l to t he s um of the s quares of the diagonals.
59
Solution: The left-hand side of eq.(1) can be transformed to
(2)
q.e.d.
Let OABC be a parallelogram, as shown in Fig. l. Im
B
loe¡ = lz2I IABI (4)
CA Z2-Z1, !CAi 1 Z2-zil (6)
The left-hand side of eq.(1) is
lz1+z2 l 2 + lz1-z2 l 2 = IOBI 2 + !CAi 2 (7)
which is the sum of the squares of the diagonals. The right­ hand side of eq.(1) is
which is the sum of the squares of the sides of a parallelo­ gram.
By proving identity (1), we have proven the theorem.
• PROBLEM 2-23
Prove that the equation of a circle in the z plane can be written in the form
zz + az + az + b = O (1)
where a is a complex constant and b is a real constant.
60
be transformed to
L
Solution: The equation of a circle of radius r and center at zo is
1 z - Zo 1 = r
(see Problem 2-4) .
lz - zol 2
Let -zo = a
- then -zo a
zz + az + az + zozo - r 2 o
Denoti ng
zz + az + az + b = o
r z
• PROBLEM 2-24
The complex numbers z 1, z 2 , z 3 , z 4 are s uch that
Z ¡ +Zz+Z3 + z 4 = 0 and
(1)
(2)
Show that z 1 , z 2 , z 3, z 4 a r e the ver tices of a rectangle .
Solution: All vertices are located on the unit circle, as s hown in Fig . l.
Fig. l
61
To show that z 1z 2z 3z1t is a rectangle, we have to prove that
and Jz4-Z¡j lz3-Zz I Jz¡-Zzl = ¡z1¡-Z3l
From eq.(1), we obtain
Thus e z 3 +z z > (z 3 +z z >
and
1
3 -z
and
which is eq.(16).
(3)
(4)
(5)
(6)
(7)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
,
Zz, Z3, Z1t are located on a circle, and since they satisfy eqs.(3) and (4), we conclude that z
1 z
2 z
3 z
Show tha t for · termines the le
f or zz - 2b I~
Furthermore, s and only if, t
Solution: The
e have to prove
• PROBLEM 2-25
Let K be a circle passing through the point z = ±a with tbe center at z 0 = ib.
Show that for any point z, the value of zz - 2b Im(z) de­ termines the location of z with respect to K. That is,
for zz - 2b Im(z)=>I: :: z is exterior to K
z is situated on K
z is interior to K
Furthermore, show.that z is interior to the circle K if, and only if, the point a 2 is exterior to K.
z
Fig. 1
Consider a point z exterior to K. Then
¡z-z 0 1 > r Eq.(2) can be transformed to
1Z-Zo12 ( z-z o )( z-z o )
•Z
Re
(1)
(2)
zz + ib(z-z) + bz = zz - 2blm(z).;i:.. b2 > a2+b2
Thus, for a point z exterior to K,
zz - 2blm(z) > a2 (4)
¡z-z 0 1 = r (5)
63
Thus, if z is on the circle K, then
zz - 2b1m(z) = a2
(6)
(7)
(8)
In the same manner, we can show that if z is interior to K, then
zz - 2b1m(z) < a2 Now we will prove
( z is interior] (ªz2 J to K <=> is exterior to K
. We shall utilize eqs.(4) and (9). =>
Since z is interior to K and z = x + iy,
zz - 2b1m( z) = {x+iy )(x-iy) - 2by x 2 + y 2 - 2by < a 2
2
is exterior to K, observe that
[ ªz 2
] ( ªz 2
(9)
(10)
<=
The last inequality can be derived from eq.(10), indeed
x 2 + y 2 < a 2 + 2by
a 2(x2+y 2 ) < a 4 + 2ba 2y
a 2 < a 1t + 2ba 2 y x2 + y2
is exterior to K, therefore,
r~i[n - 2blm(ªz 2
) ª2 a 2 + 2by x2+y2
ª2 + 2by > x 2 + y2
x2 + y2 - 2by < ª2
zz - 2b Im(z) < ª2
64
solution: To 1 1, consider thE
Im
Now, the syste placed by one
as shown in Fi
(9)
(12 ) 11. -
APPLICATIONS IN PHYSICS • PRO BLEM 2-26
find t h e mass cen ter of the s ystem of three masses· m1 , m2 ,
m 3
, whose pos i tion vectors are z1 , Z2 , z 3 r esp ectively.
So lution : To find t h e mass center of the system s hown in Fig . 1, consider t h e par t of t he system s hown in F i g . 2.
Im
Fi9. l Fi.q. 2
I f zA is the IDass cente r o f tbi s system, then
ID ¡ (ZA- Z ¡) = ID3( Z3-ZA)
Solvi ng fo r z A we find
z = m1z1 + m3Z3 A m1 + m 3
Re
(1)
(2)
Now , the systeID consist ing of m1 at z 1 and m3 a t z 3 can be r e ­ placed by on e mass e qual ID 1 + m3 locat ed at
ZA m1z1 + ffi3Z 3
as s h own in F i g . 3. Im
m 3
z l
mz In
65
The whole problem reduces to finding the mass center of the system consisting of mass m2 at z 2 and m1 + m3 at zA ·
Modifying eq.(2), we find
(m1+m3 )zA + m2 z 2 zB = ( m 1 +m 3 ) + m 2
+ m1z1+m3Z3 mi m3 m1 +m 3 + m2z 2 _ m1 z1 + m2z2 + m3Z3
m1+m2+m3 - m1+m2+m3
Eq.(3) gives the position of the mass center of the system shown in Fig. l.
• PROBLEM 2-27
Find the mass center of the system of masses m1, m2, ... , mn situated at z1, z2, ... ,zn.
(3)
Solution: First consider mass m1 and mass m2 situated at z1 and z2. The mass center is
z' = m1z1 +m2z2 1 m1 + m2
(1)
Thus m1 and m2 can be replac~d by one mass m1 + m2 located at z~ gi ven by eq. ( 1") •
Now, consider mass m1 + m2 located at z~ and m3 located at Z3. The mass center is
z , = z ; ( m t +m 2 ) + m 3 z 3 2 (m1+m2) + m3
m1z1 + m2z2 + m3Z3 m1+m2+m3
(2)
, m 3
, z 2
, z 3
can be replaced by one mass m1 + m2 +m 3 located at z; given by eq.(2).
For k < n the system m1,m2 ... ,mk with position vectors
z1, z2, ... , zk can be replaced by one mass m1 + m2 + ... + mk lo- cated at "
z' k-1 m1z1 + ... + mkzk
m1 + ••. + mk (3)
For the next r.iass k+l, we obtain the total mass m1 + ... + mk + mk+t located at
z' = k
m1z1+ .•. +mkzk+mk+Izk+1
rn1+ ... +mk+mk+i
Thus, for a system of n rnasses m1, .. ,m 0
situated at z 1 ,z2 1 ••• ,
zn, the mass center is
m1z1+m2z2+ ... +m0 z~ (5) m1+m2+ ... +m
0
66
Applications Geometry 3·21 3-23. 3-24
(;llii 1::::.
Polynomü 3- 46. 3- ~
~his chart is pre interre la t ionshi] ter· Also shown subject matter.
center of the at ZP. .
+m2z2+ m3Z3 1 +m2+m3
of the system
(2)
q . ( 2).
osition vectors
DE MOIVRE'S THEOREM
/ Polar Form of Complex 1 Oot an d Cross Nuwbers 3-1 to 3-3 Product 3-19, 3- 20. 3-22 ~
Produ c t and Quotien t 3- 4. 3-5, 3-53
Euler's Formula 3-6 Geomet ric Applications in to 3- 9, 3- 15 , 3 - 18 , - Problems 3 - 10 to Geometry 3-21. 3-33, 3-34 3-14, 3-16, 3-17 3-23, 3-24
Trigonometric DeMoiv re' s Theo r em Identities 3-34 , 3-25, 3-35.3-37to ,..__. 3-35, 3-38 to 3-39 3-40,3-43 to
3-45
l Geometric Con-
Powers and Roots of structions 3-41, Complex Numbers 3- 25 3-42 to 3-30 . 3-37
/ ·~ INth Roots of Unity Straight-Edge- 3-31, 3-32 and- Compass Con-
s t r uctions 3-32 . 3-36
~Polynomials 1--- Trigonometric Chebyshev 3-46. 3-47,3-51 Series 3- 48 to ¡._., Polynomials
3-50 3- 52
This c h art is provided to facilitate r ap i d u nderstandi ng of t he inter relat i o nships of the topics and sub ject matter in this chap­ ter . Al so shown are the problem numbers associated wit h the subject matter.
67
• PROBLEM 3-1
l. Describe the s ystem of polar coordinates.
2 . Le t z b e a complex number z = x + iy. Re present z in po l a r coordinates .
3. Express t he f ollowing c omple x numbers in polar forrn
Z ¡ 4+4/3i
Z z - 2 +2i
z 3 - 5 i
Solu t i o n: l. We shall int r o duce in the z pl a n e the polar co­ ordi nates . Th e location o f e ach point P of the plane i s unique l y de t e rmined by two polar coordinates r and e, as s hown in Fi g . l.
y
o Fig. 1
o X X
He r e , r i s the l e ng th_2.f th e segme nt OP, a nd 8 i s t he a n g l e be ­ twee n t he x-axis and OP . The r e lat ions hip between the Ca rte ­ sian coord i n ates (x, y ) a nd the p o l a r c oordinates (r,O) i s
x = r cose , y = r s ine (1)
2 . A comple x numbe r z = x + iy c a n b e r epresen ted by a p o int in t h e z p l a