Complex Numbers and Geometry - Liang-shin Hahn.pdf

OMPLEX NUMBERS & EOMETRY

Liang-shin Hahn

Complex Numbersand

Geometry

Liang-shin Hahn

SPECTRUM SERIES

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All the Moth That's Fit to Print. by Keith DevlinCircles: A Mathematical l1ew. by Dan PedoeComplex Numbers and Geometry. by Liang-shin HahnCryptology. by Albrecht BeUielspacherFive Hundred Mathematical Challenges. Edward J. Barbeau, Murray S. Klamkln. and

William O. J. MoserFrom Zero to Infinity. by Constance ReidI Wont to be a Mathematician. by Paul R. HalmosJourney into Geometries. by Marta SvedThe Last Problem. by E. T. Bell (revised and updated by Underwood Dudley)The Lighter Side of Mathemotics: Proceedings ofthe Eugene Sirens Memorial Conference

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ii

Complex Numbersand

Geometry

Liang-shin Hahn

Published byThe Mathematical Association of America

Iv

IHE" macros by Michael Downes

@J994byThe Mathematical Association ofAmerica (Incorporated)

Library ofCongress Catalog Cord Number 93-79038ISBN 0-88385-510-0

Printed in the United States ofAmericaCurrent Printing (last digit):

109876543 2

To my parents

Shyr-Chyuan Hahn, M.D., Ph.D.Shiu-Luan Tsung Hahn

And to my wife

Hwei-Shien Lee Hahn, M.D.

Preface

The shortest path between two truths in the real domain passes throughthe complex domain. - J. Hadamard

This book is the outcome of lectures that I gave to prospective high-school teachers at the University of New Mexico during the Springsemester of 1991. I believe that while the axiomatic approach is veryimportant, too much emphasis on it in a beginning course in geometryturns off students' interest in this subject, and the chance for them to ap-preciate the beauty and excitement of geometry may be forever lost. Inour high schools the complex numbers are introduced in order to solvequadratic equations, and then no more is said about them. Studentsare left with the impression that complex numbers are artificial and notreally useful and that they were invented for the sole purpose of beingable to claim that we can solve every quadratic equation. In reality, thestudy ofcomplex numbers is an ideal subject for prospective high-schoolteachers or students to pursue in depth. The study of complex numbersgives students a chance to review number systems, vectors, trigonome-try, geometry, and many other topics that are discussed in high school,not to mention an introduction to a unified view ofelementary functionsthat one encounters in calculus.

Unfortunately, complex numbers and geometry are almost totallyneglected in our high-school mathematics curriculum. The purpose

vii

vIII COMPLEX NUMBERS AND GEOMETRY

of the book is to demonstrate that these two subjects can be blendedtogether beautifully, resulting in easy proofs and natural generalizationsof many theorems in plane geometry-such as the Napoleon theorem,the Simson theorem, and the Morley theorem. In fact, one of mystudents told me that she can not imagine that anyone who fails tobecome excited about the material in this book could ever becomeinterested in mathematics.

The book is self-contained-no background in complex numbers isassumed-and can be covered at a leisurely pace in a one-semestercourse. Chapters 2 and 3 can be read independently. There are over100 exercises, ranging from muscle exercises to brain exercises andreaders are strongly urged to try at least half of these exercises. Allthe elementary geometry one needs to read this book can be foundin Appendix A. The most sophisticated tools used in the book are theaddition formulas for the sine and cosine functions and determinantsof order 3. On several occasions matrices are mentioned, but theseare supplementary in nature and those readers who are unfamiliar withmatrices may safely skip these paragraphs. It is my belief that the bookcan be used profitably by high-school students as enrichment reading.

It is my pleasure to express heartfelt appreciation to my colleaguesand friends, Professors Jeff Davis, Bernard Epstein, Reuben Hersh,Frank Kelly, and Ms. Moira Robertson, all of whom helped me with myawkward English on numerous occasions. (English is not my mothertongue.) Also, I want to express gratitude to my three sons, Shin-Yi,Shin-Jen and Shin-Hong, who read the entire manuscript in spite oftheir own very heavy schedules, corrected my English grammar, andmade comments from quite different perspectives, which resulted inconsiderable improvement. Furthermore, I want to thank Ms. LindaCicarella and Ms. Gloria Lopez, who helped me with Jb.Tp

Contents

Preface

1 Complex Numbers1.1 Introduction to Imaginary Numbers1.2 Definition of Complex Numbers . .I.3 Quadratic Equations . . . . . . . .1.4 Significance of the Complex Numbers1.5 Order Relation in the Complex Field1.6 The Triangle Inequality . . . . . . . .1.7 The Complex Plane . . . . . . . . . .1.8 Polar Representation of Complex Numbers .1.9 The nth Roots of 1 .....1.10 The Exponential Function

Exercises .

2 Applications to Geometry2.1 Triangles............2.2 The Ptolemy-Euler Theorem2.3 The Clifford Theorems2.4 The Nine-Point Circle .... ..2.5 The Simson Line . . . . . . . . . . .2.6 Generalizations of the Simson Theorem .

vii

1139

1315171925313842

55556467717683Ix

x COMPLEX NUMBERS AND GEOMETRY

2.7 The Cantor Theorems 902.8 The Feuerbach Theorem 962.9 The Morley Theorem . 103

Exercises llO

3 Mobius lhmsrormations 1213.1 Stereographic Projection 1213.2 Mobius Transformations 1243.3 Cross Ratios. . . . . . . 1293A The Symmetry Principle 1333.5 A Pair of Circles. . . . . . 1373.6 Pencils of Circles . . . . . . 1413.7 Fixed Points and the Classification

of Mobius Transformations . . . . . 1433.8 Inversions.............. 1483.9 The Poincare Model of a Non-Euclidean Geometry 156

Exercises 158

Epilogue 165

A Preliminaries in Geometry 167A.1 Centers of a Triangle 167

ALl The Centroid . 167A1.2 The Circumcenter 169AU The Orthocenter . 171AlA The Incenter and the Three Excenters . 172A1.5 Theorems of Ceva and Menelaus 173

A2 Angles Subtended by an Arc 177A3 The Napoleon Theorem 180A4 The Apollonius Circle 181

B New Year Puzzles 185

Index 191

CHAPTER IComplex Numbers

1.1 Introduction to Imaginary Numbers

One of the most important properties of the real numbers is that theoperations of addition, subtraction, multiplication and division can becarried out freely (with the exception of division by 0). Because of this,an arbitrary linear equation

ax+b=O (a t- 0)

can be solved within the realm of real numbers as x = -bfa. However,the situation is quite different for quadratic equations. For example, aquadratic equation

cannot be solved for x within the realm of real numbers. The square ofany real number cannot be negative, so

x2 + 1 > 1 > 0 for any real number x.

Therefore, x2 +1 =0 is impossible for any real number x. In a situationlike this, we extend the realm of the number system so that the equationbecomes solvable. For instance, for a kindergarten child who knowsonly positive integers, an equation such as

I

COMPlEX NUMBERS AND GEOMETRY

7+ 0 =3is unreasonable, and for persons who know only integers, 5x = 2and x 2 = 17 have no solution. But by extending our number systemto include negative numbers, fractions, and irrational numbers, theseequations have solutions -4, 2/5, Ji7, respectively.

The situation is pretty much the same for x 2 + 1 = O. We extendour number system to include numbers such as J=T; i.e., a numberwhose square is -1. Such numbers are not quite in agreement with ourintuition, and many mathematicians in the past objected to introducingsuch monsters, so they are called imaginary numbers. It was not untilthe 18th century, with skillful manipulations of imaginary numbers,that L. Euler (1707-1783) obtained numerous interesting results. Byrepresenting the imaginary numbers as points in a plane, C. F. Gauss(1777-1855) renamed them as complex numbers, and applied them toobtain many remarkable results in number theory, thus establishing thecitizenship of complex numbers in the number system. About the sametime, in trying to find a uniform way of computing definite integrals,A. L. Cauchy (1789-1857) investigated the differential and integralcalculus of functions with complex numbers as variables. This is thegenesis of the theory of functions that cultivated the environment forN. H. Abel (1802-1829) and C. G. J. Jacobi (1804-1851) to discover theelliptic functions. Furthermore, the development of projective geome-try shows that the complex numbers are indispensable in geometry aswell. As research progresses, it is now clear that to truly understandmathematics, even merely calculus, the realm of the real numbers isunnaturally narrow, and it is imperative that we work with complexnumbers to allain uniformity and harmonicity.

It is customary to use the first leller i of the word 'imaginary' for J=T.Thus complex numbers are numbers of the form a + ib, where a and bare real numbers, and computations with them are carried out just aswith real numbers if we remember to replace e by -1. For example,

(a + ib) (c + id) = (a c) + i(b d),

(a + ib) (c + id) = ac + ibc + iad + i2bd

= (ac - bd) + i(bc + ad).

ComplEx Numbers 3

The division of two complex numbers, (a + ib)/(c+ id), involves findinga complex number x + iy satisfying

a + ib = (c + id) . (x + iy).

Hence, by the above computation, we get

a + ib = (ex - dy) + i(dx + cy);

and so it is sufficient to find x and y satisfying

ex - dy = a, dx + cy = b.

This system of simultaneous equations has a unique solution

ac+bdx= CZ+d2'

unless c = d = O. Hence

be-ady= CZ+d2'

a + ib ac + bd .be - ad-c-+-i---:d = CZ + d2 + ~ CZ + d2 .

Of course, this can be obtained by multiplying both the numerator andthe denominator by c - id.

But why are such operations justified? Isn't the addition of a realnumber a and an imaginary number ib to get a+ibsimilar to the additionof 17m2 and 4kg to get 21C? Also, x2 + 1 = 0 has two solutions, andwhich -one of them is i? Note that x 2 - 1 = 0 also has two solutions,the positive one is 1 and the other -1. But is it meaningful to say i ispositive?

1.2 Definition of ComplEX Numbers

To answer the criticism at the end of the last section, we now givea formal definition of complex numbers. But first let us recall theproperties of the real number system lit

4 COMPLEX NUMBERS AND GEOMETRY

(I) Properties concerning addition.Tho arbitrary real numbers a and b uniquely determine a third num-

ber called their sum, denoted by a + b, with the following properties:

AI- Commutative law: a + b = b + a for all a, b E Ill.

Az- Associative law: (a + b) + c = a + (b + c) for all a, b, c E lItA3. Additive identity: There is a unique real number, denoted 0, such

that

a + 0 =0 +a = a for all a E lIt

A4. Additive inverse: For every a E JR, there is a unique x E JR satisfying

a + x = x + a = O.

This unique solution will be denoted by -a.

(II) Properties concerning multiplication.Tho arbitrary real numbers a and b uniquely determine a third num-

ber called their product, denoted by ab, with the following properties:

MI _ Commutative law: ab = ba for all a, b E JR.

Mz. Associative law: (ab)c = a(bc) for all a, b, c E JR.

M3. Multiplicative identity: There is a unique real number, denoted 1,such that

a} = }a=a forall aEIIt

M4 Multiplicative inverse: For every a E JR, a 'I 0, there is a uniquenumber x E JR satisfying

ax=xa=l.

This unique solution will be denoted by ~ or a-I.

ComplEX NumbErs

(III) Distributive law.

a(b + c) = ab + ac for all a, b, c E IR.

5

Any set that satisfies these properties is called a field. Thus the set IRof all real numbers is a field. Similarly, the set Q of all rational numbersforms a field. However, neither the set Z of all integers, nor the set Nof all natural numbers is a field.

In the previous section, we said complex numbers are numbers of theform a + ib, where a and bare real numbers. Thus complex numbers areessentially a pair of real numbers a and b. Therefore, we give a formaldefinition as follows.

DEFINITION 1.2.1. A complex number is an ordered pair (a, b) ofreal num-bers with the following properties: Two complex numbers (a, b) and (c, d)are equal if and only if a = c and b = d. The sum and product of twocomplex numbers (a, b) and (c, d) are defined by

(a, b) + (c,d) = (a + c, b + d),

(a, b) . (c,d) = (ac-bd, bc+ ad).

Note that our definition of equality for complex numbers has thefollowing properties:

(a) Reflexive: (a, b) = (a, b) for every complex number (a, b);(b)Symmetric: (a,b) = (c,d) (c,d) = (a, b);(c) Transitive: (a, b) = (c, d), (c, d) = (e, I) => (a, b) = (e, I).

THEOREM 1.2.2. With addition and multiplication defined as above, the setIt ofall complex numbers is a field.Proof A muscle exercise. 0

Now, if we consider complex numbers of the form (a, 0), then

(a, 0) (b,O) = (a b,O);(a,O) (b,O) = (ab,O);

6 COMPlEX NUMBERS AND GEOMETRY

(a,O) = (~ 0)(b,O) b' (provided b f- 0),

which is identical to the operations between two real numbers a andb. In other words, there will be no confusion if we regard a complexnumber of the form (a,O) as a real number a. Consequently, we shallconsider real numbers to be particular complex numbers whose secondcomponent is zero.

Next, consider the complex number (0, I). We have

(0,1)2 =(0,1) (0, 1) = (-1,0) = -1.

Namely, the complex number (0,1) corresponds to A in the previoussection. Naturally, the square of (0, -1) is also -1, but if we denote(0,1) = i, then an arbitrary complex number (a, b) can be rewritten as

(a, b) = (a, 0) + (0, b) = (a,O) + (0,1) (b,O),

which justifies the expression a + ib.The complex number i is called the imaginary unit. For a complex

number a = (a, b) = a + ib (a, b E IR), a is called the real part ofthe complex number a and is denoted by ~a; similarly, b is called theimaginary part of a, and is denoted by ~a. Thus real numbers arecomplex numbers whose imaginary parts are O. On the other hand,complex numbers whose real parts are 0 are called purely imaginary.Note carefully, both the real part and the imaginary part of a complexnumber are real numbers!

For a complex number a = (a, b) = a + ib, the number (a, -b) =a - ib is called the complex conjugate or the conjugate complex numberof a, and is denoted by a. The following relations are easy to verify:

a /3 =a /3;

0/3 =a /3;

(;) =; (/3 f- 0);

ComplEX NumlMirs

\0 0+0:nO = .2 '

0=0.

0-0~o = .2i '

7

For any complex number 0 = a + ib (a, bE IR), the product

is always real and nonnegative. Its nonnegative square root is called themodulus or the absolute value of the complex number 0, and is denotedby 101. Thus

THEOREM 1.2.3. 101 = 0ifand only if0 = O.Proof Let 0 = a + ib (a, bE IR). Then 1012 = a2 + Il-. Therefore,

But a2 > 0, b2 > 0 for any real numbers a and b, hence

=a =0, b= 0=0=0.

o

Note that we have used the fact that a and b are real numbers,otherwise a2 + b2 = 0would not imply a = b = O. For example,let a = 1, b = i, then a2 + b2 = 0, but neither a = 0 nor b = O.

It is simple to verify that

1!R01 < 101, ISSol < /01, 101 = 101;10131 = 101 113\ (in particular, 1- 01 = 101);

o = el (provided 13 'f 0).13 1131


THEOREM 1.2.4. For any complex numbers a and f3,

af3 =0 =} a =0 or f3 = O.

Equivalently,af3 f 0 =} a f 0 and f3 f O.

Proof By the previous theorem,

af3 = 0 =} laf31 = 0=} lal . 1f31 = O.

Since lal and 1f31 are real numbers,

lal . 1f31 =0 =} lal =0 or 1f31 =0=} a =0 or f3 =o.

o

Remark. In a set where multiplication is defined, if af3 = 0 eventhough a f 0,f3 f O,thenaandf3arecalledzerodiviso~. The previoustheorem asserts that the complex field iC does not have a zero divisor.

There are algebraic systems that have zero divisors. For example,consider the set of all 2 x 2 matrices:

Its addition and multiplication are defined as

(~ :)+(~(~ :).(~

b')=(a+a' b+b')d' c+c'd+d"

b' ) = ( aa' + be' ab' + bel! )d' ca' + de' cb' + dd

'.

ComplEx NumbErs

The zero element is

O=(~ ~).Then, even though

9

( 00 1)2 fO andwe have

(~ ~). (~ ~) =O.Note that in the proof of the theorem, we have used the fact that the

real field lR has no zero divisor.

13 Quadratic Equations

As we saw in 1.2, the complex number i = (0,1) satisfies the equationXl + 1 = O. But how about other quadratic equations? Do they havesolutions in Co? Or, do we have to keep on adding new members to ournumber system? Consider the following.EXAMPLE. Let us find the roots of

~Xl + (1 + i)x - i = O.Solution. By completing the square, we get

{x + (1 + i)}2 =2i + (1 + i)2 =4i.

:. x + (1 + i) = 20.

x = -(1 + i) 20.

But what is the square root of the imaginary unit i? It must be a numberwhose square is i. So set

u +iv = 0 (u, v E lR).

10

Then, squaring both sides, we get

COMPLEX NUMBERS AND GEOMETRY

1uv = 2'

From the first equation, we get u = v. Suppose u = v, then from thesecond equation, we get

.j2u =v =-'2 '

and so

../i = u + iv = V;(I + i).The case u = -v would not happen since it would imply u2 = -!, butthis is impossible for u real. l It follows that

x = -(I + i) J2(1 + i) = (-I J2)(1 + i).

So, it was not necessary to extend our complex number system inorder to solve this quadratic equation. (The reader should verify thatthe results obtained actually satisfy the quadratic equation.)

Let us now consider a general quadratic equation

ax2 +bx+c=0 (a .,. 0).

I Neglecting the condition that u is real, if we proceed to solve u2 = -!, we get,,12.

u=-,2 ',,12.

v = 'fT"

Hence,

Vi =u + iv =V;i(t - i) =V;(I + i).so we obtain the same result as before.

ComplEx NumbErs

Since we have extended our number system to complex numbers, weshould discuss the case a, b, c E Co Since complex numbers obey thesame rules as real numbers as far as addition, subtraction, multiplicationand division are concerned, by dividing both sides of the above equationby a and completing the square as in the case of real coefficients, we get

Settingb

z=x+ 2a ,

our problem becomes whether z2complex number (. Let

z = u + iv,

( =b2

-4ac4a2 '

= (can be solved for an arbitrary

( = 0: + if3.Our problem can be restated as: Can we always find a pair of realnumbers u and v such that

(u + iV)2 = 0: + if3

for an arbitrary pair of real numbers 0: and f3? Rewriting the lastequality, we get

Hence our problem reduces to solving the system ofsimultaneous equa-tions

2uv = f3.

Since

and u2 + v2 > 0,0:2 + f32 > 0 (": u,v,o:,f3 E JR.), we get


It follows that

Note that

and so

)

1/2

(0+ J 0 2+{32

U = 2 ' )

1/2

(-0 + J 0 2 + {32

V = 2 '

where the signs must be chosen to satisfy 2uv = {3. That is, the squareroot J( =u + iv is given by

.j(=( )

IP ( )IP) (0+';;'+13' + i -o+~o'+ff' , ( _ ( 0+';;'+13') 1/2 + i (-O+~O'+ff')1/2),JO, for {3 =0, 0 > 0;i'; 0, for {3 =0, 0 < O.

for {3 > 0;

for {3 < 0;

We have shown that every (nonzero) complex number has two squareroots.

Remark. For { E JR, the notation ..,ff, was used for the nonnegativesquare root when { > 0, and ..,ff, = i'; { (= iM) when { < o.However, in the sequel, for a nonreal complex number {, the notation..,ff, shall simply mean a square root of { and shall not stand for oneparticular square root.

By our above convention, when { and TJ are negative real numbers,the equality

ComplEX NumbErs B

is no longer valid. It is valid if we interpret both sides as sets of complexnumbers.

We have established the following

THEOREM 1.3.1. A quadratic equation

ax2 +bx+c=0,

has two roots in iC, which are given by

a, b, c E C, a f 0,

-b Jb2 - 4ac2a

1.4 SignificancE of thE ComplEX Numbers

In the previous section, we have seen that every quadratic equation hassolutions in the complex field C. But how about cubic equations, quarticequations, etc.? Do we have to expand our number system each time wedeal with higher degree equations? One of the beauties of the complexnumber system is the validity of the following

THEOREM 1.4.1 (Fuodameotal Theorem of Algebra). A polynomialequa-tion

where ak E C (k =0,1,2, . .. ,n~ ao f 0, n > I, has a solution in C. Inother words,

C is algebraically closed.

The above equation is called a polynomial equation of degree n(when ao f 0). From the fundamental theorem of algebra, we havethe following

COROllARY 1.4.2. A polynomial equation ofdegree n has exactly n rootsin C taking the multiplicities into account.

EXAMPLE. Solve the cubic equation z3 + i = O.Solution. Let z = u + iv (u, v E JR.). Then, since

14

we obtain


From the first of these equations, we get

Henceu=o

When u = 0, from the second equation, we get

v3 - 1 = 0, (v - 1)(v2 + v + 1) = 0.

Since v E JR, v2 +v +1 'f 0, and so v = 1,:. z = i. When u2 - 3v2 =0,u = v'3v. Substituting this into the second equation, we get

23 ( v'3v) v - v3 + 1 = 0, i.e., 8v3 + 1 = 0.

:. (2v + 1)(4v2 - 2v + 1) = 0.

Since v E JR,4v2 - 2v + 1 'f 0, and so v = -!. Hencev'3

u = 'fT'

Therefore, we get 3 solutions

z = z,

v'3 - iz= 2

v'3 - i2

C. E Gauss (1777-1855) gave several proofs of the fundamentaltheorem of algebra in his dissertation. Readers who are interested inits proof may consult standard textbooks in complex analysis such as J.Bak and D. J. Newman's Complex Analysis [Springer-Verlag, New York,1982), or R. P. Boas's Invitation to Complex Analysis [Random House,New York, 1987]. Note that the fundamental theorem of algebra asserts

Complo NumbErs 15

the existence of solutions in C, but does not tell us how to find them. Infact, there is noalgebraic formula that works for every quintic polynomial(or higher).

1.5 OrdEr RElation in thE ComplEX Field

We can always compare the magnitudes of any two real numbers. Thatis, given a, b E JR, either a > b, or a = b, or b > a. Can this resultbe extended to complex numbers? To answer such a question, we firstreexamine the order relation in JR.

PI' (llichotomy) For any a E JR, one and only one of the followingthree relations holds:

a> 0,

For a, b E JR, if we define

a=O, -a>O.

a > b if and only if a - b > 0,

then PI is equivalent to the assertion that for any a, b E JR, one andonly one of the following three relations holds:

a> b, a = b, b> a.

Furthermore, the order relation in JR satisfies the following:

Pz. a> 0, b > 0 => a + b > O.

Pl. a > 0, b > 0 => ab > O.

It turns out that all the properties of the order relation in JR, such as

a > b, b > c => a > C;

a>b, c>O=>ac>be;

a > b, c < 0 =9 ac < be


follow from the positivity postulates Ph Pz, and P3. In other words, anorder relation is useful only if all three postulates PI, Pz, P3 are satisfied.

THEOREM 1.5.1. It is possible to extend the order relation in IR to ce suchthat PI and Pz are satisfied, but it is impossible to satisfy P3.Proof For 0 =a + ib (a, bE IR), define

O {a> 0, or

0> = a =0 and b > O.

Pl. For any 0 =a + ib (a, bE IR), one of the following must hold:

a> 0, a =0, -a >0.

(a) Ifa>O=}o>O.(b) If -a > 0 =} -0 > O.

{b>O=}o>O;

(c) If a =0, b =0 =} 0 =0;-b>O=} -0>0.

We have shown that one and only one of

0> 0, 0=0, -0 >0

holds for an arbitrary 0 E ce.Pz Suppose 0 > 0,0' > 0, where

Then

0= a + ib, 0' =a' + ib' (a, b, a', b' E IR).

a> 0 or {a =0 and b> O},

a' > 0 or {a' = 0 and b' > O}.

We must check all combinations of these cases:

(a) a> 0, a' > 0 =} a + a' > 0 =} 0 -I- 0' > O.

ComplEx NumbErs 17

(b) a> 0, {a' =0 and b' > O} ===> a + a' > 0 ===> a + a' > O.(c) a' > 0, {a =Oandb > O} ===> a + a' > 0==* ~ +0' > O.(d) Finally, {a = 0 and b > O} and {a' = 0 and b' > O}, then

a + a' = 0 and b + b' > 0, and so a + a' > o.

Pl. Suppose the order relation in JR. can be extended to iC preservingthe postulate Pl. Then since i f 0, by PI, we must have either i > 0 or-i > O. If i > 0, by Pl , we get i . i > 0, but this means -1 > 0, which isabsurd. Similarly, if -i > 0, again by Pl, we get (-i) . (-i) > 0, whichimplies again the absurd result -1 > O. 0

Remark. Actually, there are infinitely many ways to define order rela-tions in iC satisfying PI and P2. Here we have chosen a lexicographicone. As a consequence, we should not use inequalities such as < or > fornonreal (complex) numbers.

1.6 The TrIangle Inequality

In 1.2, we defined the absolute value of a complex number a = a + ib(a, bE JR.) to be

and mentioned some simple properties of the absolute value. We nowprove the important triangle inequality.THEOREM 1.6.1 (The Triangle Inequality). For any Zl> Z2 E iC,

Proof

= Izd2 + 2!R(Z\Z2) + IZ212< Izd2 + 21zIZ21 + IZ212 (":!Ro < 101 for all a E iC)


= IZI12+ 21zlllz21 + IZ212 ('.'1%21 = IZ21>= (lzll + IZ21>2 .

Since both IZI + z21 and IZII + IZ21 are nonnegative, we obtain

To prove the other inequality, note that ZI = (ZI + Z2) + (-Z2) .. IztI = I(zi + Z2) + (-z2)1 < IZI + z21 + 1- z21

= IZI + z21 + IZ21

It follows that

Interchanging the roles of ZI and Z2, we get

IZ21-lzd $ IZI + z21Iizll-lz211 < IZI + z21

o

We now discuss the situation where the triangle inequality becomesan equality. This is trivially the case if ZI = 0 or Z2 = O. Hence, weconsider the case Zj . Z2 f O. Looking back at the proof, we see that itbecomes an equality if and only if

But a complex number a satisfies ~a = lal if and only if a is a nonneg-ative real number. Hence the above inequality becomes an equality ifand only if

Since we are assuming Z2 f 0, dividing both sides by IZ212 (= Z2%2 > 0),we obtain that :!. > O.

%2

ComplEX NumbEl'5 19

Summing up, Iz\ + Z21 = IZII + IZ21 if and only if

ZI 0->Z2

unless Z\ = 0 or Z2 = 0;

in other words, one of ZI and Z2 has the property that the other is itspositive multiple. We shall give another proof of this fact at the endof 1.8 Polar Representation of Complex Numbers. The reason for thename 'triangle inequality' will become clear in the next section.

1.7 The ComplEX Plane

We have defined a complex number to be an ordered pair of real num-bers, but the set of all ordered pairs of real numbers has a one-to-onecorrespondence with the (x, y)-plane 1R2. So it is natural to let a complexnumber Z = x + iy correspond to the point (x, y) in the plane 1R2

y yz = x + iy

11 2 + i........ ~

o...

x--...,------t-:---+

ox

----jf----+'

-1 - i

FIGURE IJ

In the above correspondence, a real number x = x + iO correspondsto the point (x,O) on the x-axis, and a purely imaginary number iy =o+ iy corresponds to the point (0, y) on the y-axis, so we call the x-axisthe real axis, and the y-axis the imaginary axis. A plane equipped withthe real and imaginary axes is called the complex plane or the Gaussianplane.

COMPl.EX NUMBERS AND GeoMETRY

Let us consider a sum of complex numbers in the complex plane C.Suppose

then

z = x + iy, w =u + iv (x, y, u, v E JR),

z + w = (x + u) + i(y + v).

This suggests that it is best to consider complex numbers as vectors; i.e.,we regard a complex number z = x +iy as a vector whose initial point isthe origin and whose terminal point is the complex number z = x + iy.In other words, we consider a complex number z = x + iy to be avector whose orthogonal projections to the coordinate axes are x and y.Naturally, similar considerations apply to the complex number w. Thenthe sum z + w corresponds to the diagonal vector (from the origin) ofthe parallelogram formed by two vectors z and w. Equivalently, drawthe vector z with the initial point at the origin, and then draw the vectorw with the initial point at the terminal point of z, then the vector withthe initial point at the origin and the terminal point of w as its terminalpoint represents the vector z + w. (See Figure 1.2.)

From now on we shall identify a complex number with a point ora vector in the complex plane, whichever is most convenient for theparticular situation.

We now consider the real mUltiple of a complex number. For z =x + iy and c E JR, we have cz = ex + icy. Thus, in the complex plane,

y

ox

v

FIGURE 1.2

ComplEX NumbErs

if c > 0, simply multiply the length of the vector by e (keeping the samedirection), while if e< 0, multiply the length of the vector by lei andchange to the opposite direction.

y y

z - w

: - \1.'w

___-----o*"'__---1~xo

c: (c > 0)

cz(c


BL-----(CFIGURE 1.4

Solution. Let 6ABC be in the complex plane, and z(, Z2, Z3 are thecomplex numbers representing the vertices A, B, C, respectively. Thenthe midpoints D and E of the sides AB and AC are given by

and

-respectively. Hence the vector DE is given by

But Z3 - Z2 is precisely the vector BC, hence the desired result.EXAMPLE. The point Z that divides the segment joining the points z\and Z2 into the ratio m : n internally is given by

nZI + mZ2z=

n+m '

where m and n are positive real numbers. For, it is easy to see that

Z - ZI

m

Z2 - Z-

n

which gives the desired relation.Equivalently, suppose Z is an arbitrary point on the line segment

joining the points Z\ and Z2, then

Z - ZI = t(Z2 - ZI) for some t E III (0 < t < 1);

ComplEX NumbErs

7~,

,.

i1 t /-'/

II II:, + m:,Il+m

m

FIGURE IS

:. Z = (1 - t)ZI + tZ2

/

//,= == (I - 1)z, + t=,(/

(0 < t < 1).

Conversely, suppose this relation holds. Then since our argument isreversible, we can conclude that Z must be a point in the line segmentjoining ZI and Z2.EXAMPLE. Let z., Z2, Z3 be three arbitrary points in the complex plane.Then the midpoint of the line segment joining Z2 and Z3 is (Z2 + z3)/2,and so the parametric equation of the median through the vertex ZI of

~ZIZ2Z3 is

Z2 + Z3Z = (1 - t)ZI + t . --=--::2~ (0 < t < 1).

Hence the point that divides this median into 2 : 1 internally, is obtainedby substituting t = ~ into the above expression; i.e.,

But this expression is symmetric with respect to ZI, Zz, Z3, and hence thispoint also divides the medians from Z2 and from Z3 into 2 : 1 internally.Therefore, three medians of an arbitrary triangle intersect at a point.We call this point the centroid or the center ofgravity of ~ZJZ2Z3.


...).::.. ,. fG .

o

.. (z, + z)/2...

AGURE 1.6

For z = x + iy (x, Y E IR), we defined the absolute value of z to be

i.e., Izi is the length of the vector z. In other words, Izi is the distancefrom the point z to the origin, which is in agreement with the case whenz is real.

Note that IZtl, jzzl, IZI +zzl are the lengths ofthree sides ofa triangle,and hence the name triangle inequality (Theorem 1.6.1):

It is geometrically obvious that the inequality becomes an equality onlyif the triangle degenerates to a line segment. We shall return to thispoint at the end of the next section.

1.8 Polar Representation of Complex Numbers

So far we have used only the vector aspect of complex numbers, hencewe haven't seen the real power of complex numbers. Multiplication iswell defined for complex numbers while it is not defined for vectors-the dot product (inner product) of two vectors is a scalar, not a vector,while the cross product of two vectors in a plane is a vector that is nolonger in the plane. (The cross product is useful only in 3-dimensional

ComplEX NumbErs

space.) The essence of applications of complex numbers to plane geom-etry lies in the fact that the products of complex numbers are complexnumbers.

For multiplication of complex numbers, it is convenient to use thepolar representation of complex numbers. For a point P = (x, y)

-->(= x + iy) on a coordinate plane, consider the vector OP (where 0is the origin). Let 0 be the angle between OF and the positive x-axis,andT = OP. Then x = TCOSO,y = TsinO.

Naturally, 0 is determined up to mod 27T; i.e., 0 is determined uniquelyifwe neglect the difference of an integer multiple of27T. The angle 0 iscalled the argument of the complex number z.

Throughout this book, unless stated explicitly to the contrary, equali-ties involving arguments will always be interpreted as congruence mod 27T;ie., we shall neglect the difference ofinteger multiples of27T.

We call (T,O) the polarcoordinates of the point P.

y

P:: = I"(cose + isine)

o I" case

I" sine

FIGURE 1.7

The origin is the unique point where T = 0; the argument 0 of thepoint at the origin is not defined.

If z = x + iy (f 0), then we can write the polar representation of z:

z = T(COSO + isinO).

That is, T = Izi and 0 =arg zis the argument of z.


EXAMPLE. For z E C. z f o.

z is real {=} argz = mr (n E Z);

z is purely imaginary {=} arg z = ; + mr

y

I

-1-~---,..k-----i- .... x

-I1 - i

FIGURE IS

EXAMPLE.

(n E Z).

1-11 =1, arg(-I) =(2n + 1)11" (n E Z);

Iii =1, arg(i) = ; + 2mr (n E Z);

11- il = V12 + (-1)2 = ../i, arg(l- i) = - ; + 2n1l" (n E Z).

EXAMPLE. Let w = -I + iV32 .Then

ComplEX NumbErs

Set () = arg w, then

1cos(} = --2' sin () = v'32'

21rand so () = 3" + 2mI' (n E Z). Moreover,

Note that

Iwl = 1, 27rargw =-3 + 2mr.

The points wand w2 together with 1form three vertices of an equilateraltriangle inscribed in the unit circle.

y

............

..

...... 1 x

......

,

or

FIGURE 1.9

Polar representation is convenient for multiplication ofcomplex num-bers due to the following.

COMPlex NUMBeRS AND GeOMeTRY

THEOREM 1.8.1. Suppose

ZI = TI(Cos81 + isin8d.

then

Zz = TZ(cos 8z + i sin 8z).

ZIZZ =TITZ {cos(81 + 8z) + isin(81 + 8z)};

i.e.,arg(zIZZ) = argz) + argzz.

In other words, the absolute value ofthe product is the product ofthe abso-lute values, and the argument ofthe product is the sum ofthe arguments.

Proof By the addition formulas, we have

ZIZZ =TI(cos81 + isin81) Tz(cos8z + isin8z)=TITZ{ (cos 8) . cos8z - sin81 . sin8z)

+i(sin 81 . cos 8z + cos81 . sin 8z)}= TITZ{cos(81 + 8z) + isin(81 + 8z)}.

o

Remark. By restricting argz to the interval [O,27r) or (-71".71"), argzwill be uniquely determined for all Z E C (except Z = 0), but then therelation

arg(zlzz) = argzi + argzz

will not be valid, and arg Zwill not be a continuous function of z.

COROLLARY 1.8.2.

arg(zlzz ... zn) =arg Zl + arg Zz + ... + arg Zn.

ComplEx NumbErs

COROUARY 1.8.3 (DeMoivre). For z = r(cos8 + isin 8) and n E Z,

zn = rn(cosn8 + isin n8),

viL,arg(zn) =n arg(z).

Proof We prove only the case n = -1.

(cos 8 + isin8)(cos8 - isin8) = cos28 + sin28 = 1,

dividing both sides by r(cos 8 + i sin 8), we get

( 8 1 .. 8) = !(cos8 - isin8)r cos + tsm r

= ~ {cos(-8) + i sin(-8)},since cos(-8) = cos 8, sin(-8) = - sin 8.

arg (z-I) = - arg(z).o

COROUARY 1.8.4.

Zlarg- =argzl-argz2'

Z2

provided Z2 f 0.EXAMPLE. From the DeMoivre formula, we can derive formulas for thesine and cosine functions. Choose r = 1, then it becomes

(cos 8 + i sin 8r = cosn8 + isinn8.

In particular, for n =3,

cos 38 + isin38 = (cos 8 + isin8)3


cos 3/1 = COS3 /I - 3cos /I . sinz /I = 4 cos3 /I - 3 cos /I,

sin 3/1 = 3 cosz /I . sin /I - sin3 /I = 3 sin /I - 4 sin3 /I.

Our theorem says that multiplying by Z in the complex plane meansmagnifying (or contracting) a figure by the factor 14 and rotating(counterclockwise) by the angle argz. In particular, multiplying by imeans rotating by 1 (counterclockwise).

With this preparation, given points Zl and Zz on the complex plane,we can construct the product Z3 = ZI Zz geometrically. All we needis an observation that lIOlz1 and lIOZZZ3 are similar (with the sameorientation).

v- - - -

....' - -,-~

r

y

.r------1';;,--1------..

FIGURE 1.10

Similarly, to construct the quotient Z4 = ~ geometrically, we need to'2

observe that lIOzzzl and lIOlz4 are similar (with the same orientation).EXAMPLE. Construct 1/(2 + i) geometrically. Let ZI = 2 + i, Z2 =1/(2+i), then lIOlz1 and lIOz21are similar (with the same orientation),and so we construct Zz as in Figure 1.11.

We now return to the case when the triangle inequality (Theorem1.6.1 ),

ComplEx NumbErs

y

12 + i

AGURE IJI

becomes an equality. Examining the proof, we notice that equality holdsif and only if anyone of the following equivalent conditions holds:

1. !R(ZIZ2) = IZlz21.2. Zl Z2 is a nonnegative real number.3. z\ is a positive real number or ZI Z2 = o.

Z2

4. Zl and Z2 have the same argument (mod 21T).5. ZI and Z2 are on the same ray from the origin.

--+ --+ . .6. Vectors OZI and OZ2 have the same dlrecllon.

1.9 The nth Roots of 1

The unit circle-the circle with center at the origin and radius l-canbe expressed as

1(1 = 1 ( E C).


In terms of polar representation, this can be wrinen as

(=cosO+isinO

For this (, if z E C, then we have

(0 E IR).

I(zl = 1(!lzl = Izl, arg(z) = arg( + argz.

It follows that multiplying by ( simply means the rotation of z with respectto the origin by the angle 0 (= arg().

Suppose z = X + iy, (z = x' + iy'. Then

x' + iy' = (cosO + i sinO)(x + iy)= (x cosO - ysinO) + i(xsinO + ycos 0).

x' = xcosO - ysinO,

y' = xsinO + ycosO.

This may remind us of a linear transformation. If we regard x + iy andx' + iy' as vectors (:) and (::), respectively, then the above relations canbe wrinen as

(X') = (C?SO - SinO) (x).y' smO cosO yConversely, if this relation holds, then

x' + iy' = (cosO + i sinO)(x + iy),

and so to multiply ( = cos O+i sin 0 and z = x+iy is the same as multiplyingthe rotation matrix

( cosO -SinO)sinO cosO

ComplEX NumbErs

y y

33

FIGURE 1.12

x

EXAMPLE. For ( = cos8 + isin8, (2 is obtained by rotating (by theangle 8. Another rotation by 8 gives (3, etc.EXAMPLE. For 0 = r(cos 8 + i sin 8), 0 2 is obtained by rotating 0 bythe angle 8 and magnifying (or contracting) its length by the factor r.EXAMPLE. We now want to find alI the cube roots of 1; i.e., we want tosolve z3 = 1.

Set z = r(cos 9 + i sin 8). Then, by the DeMoivre fonnula,

r 3( cos 38 + i sin 39) = 1.

cos 38 + isin38 = 1.

Since r is a positive real number, we get

r = 1, cos39 = 1, sin 38 = O.

. 38 = 2k1f'. . , i.e., 8 = 2k1f3(k =0, 1, 2, 3, ...).

According to the fundamental theorem of algebra (Theorem 1.4.1),there must be exactly 3 roots, yet it appears that we have found infinitely


many roots. However, by the periodicity of the sine and cosine func-tions, we have

wo = 1 + iO =W3 =W6 = W9 = ... = W-3 = W-6 = ... ,

211" . 211"WI =cos - + ism - =3 3

-1 + iV32

=W4 = W7 = WlO = ... = W-2 = w-s = ... ,

411" . 411"W2 =cos- + ism- =3 3

-1 - iV32

=Ws = Wg = WII = ... = W_l = W-4 = ....

These three points are the venices of the equilateral triangle inscribedin the unit circle with one venex at 1. Note that if we set W = WI, thenW2 = w2 = W, w2 + W + 1 = O. (See Figure 1.13.)

We now embark on the task of finding all the solutions of the equation

w....

...

1 .r..

...

....

FIGURE I.B

ComplEX Numbers

Let a solution be given in the polar form

Z = r(cosll + isinll).

By the DeMoivre formula,

3S

:. r = I,

Therefore,

cosnll + i sin nil = 1.

nil = 2k7r, II = 2k1rn

(k =0, t, 2, ...).

Conversely, it is simple to verify that

2k1r . 2k1rzk=cos-+ism-

n n

satisfy the given equation. We claim that

(k =0, t, 2, ...)

if k' =k (mod n), then

Without loss of generality, we may assume that

Then

k' = k + jn (j E Z, 0 < k < n).

2k'1r .. 2k'1rZk' =cos - + ~ SID --

n n

( 2k1r 2 ) .. (2k1r 2)= cos ---;- + j1r + ~ SID ---;- + J1r2k1r .. 2k1r

=cos - + ~ SID - = Zk,n n


so there are at most n distinct roots Zk corresponding to k = 0, 1, ... ,I

n - 1. Conversely, if Zk' = Zk; that is, if

2k'1f 2k'1f 2k7r . 2k7rcos-- + isin- = cos- + ism-,

n n n n

then we have

2k'1f = 2k1f + 2m1fn n

for some m E Z,

which implies that k' =k (mod n).Summing up, there are exactly n roots

2k7f .' 2k7rcos- +ism-,

n n

corresponding to k = 0, 1, 2, ... ,n - 1. They form the vertices of theregular polygon with n sides inscribed in the unit circle with one of thevertices at 1.

EXAMPLE. Find the roots of zS = 12 + 5i.

y

12 + 5;

. "'-.

+--tn--:-:r------------------+xo

FIGURE 1.14

ComplEX NumbEl's

Solution. Let rp = arg(12 + 5i) = arctan 152. Since 112 + 5il = 13, ifwe set z = r(cos0 + i sin 0), then

r5(cos50 + i sin 50) = 13(cosrp + isinrp).

:. r =m, 0=rp+2k1r5 5 (k =0,1,2,3,4).

Thus, the five roots form vertices of the regular pentagon inscribed in acircle with the center at the origin and one vertex at m(cos! +i sin!)(see Figure 1.14). A reader should compare the polar representationapproach with the method of 1.3 and 1.4.EXAMPLE. Let z = cos 2; + i sin 2;. Then since z5 = 1, but z f 1, wehave

Dividing both sides by z2 (z2 f 0), we get

i.e.,

But z + ~ = 2 cos 2; ,so we get

2 21T 21T4cos - + 2 cos - - 1 = O.5 5

But cos 25" > 0, so we obtain

21T -1 v's:.cos S = 4

21T v's - 1cos -5 = -'--4=--

This result implies that a regular pentagon can be constructed with acompass and straightedge.


. V5+I.QUIZ: Where does the other value 4 come from? Is It a mean-ingless number simply to be discarded?

1.10 The exponential Fundion

In calculus2 we learned that

XX2 X 3 xneZ=I+_+_+_++_+I! 2! 3! n!

is valid for all x E lIt What if x is replaced by i9? The left-handside becomes ei9 But what is this exponential function with a complexvariable? We don't know. So let us look at the right-hand side first.

(iO)"Every term -- makes perfectly good sense, so we collect the termsn!

according to whether the term has i or not (we are changing the orderof summation in an infinite series which is justified since our series isabsolutely convergent), and we get

{92 9" (J6 92n }

1- 2! + 4! - 6! + ... +(-I)n(2n)! + ...

{93 es 97 92n+ l }+i 9 - - + - + - + ... + (- I)n + ...3! 5! 7! (2n + I)!

= cos 9 + isin9,

again by what we learned in calculus. Since ei9 has no meaning, wemight as well use this result to define it:

ei9 := cos9 + isin9.

2 A reader who has no background in calculus may skip this section, treating e'9 in whatfollows as a shonhand for cos 0 + i sin 0, with the understanding that e'9 . e i ", = e i (9+",),which is just a rewriting of the identity

(cosO + isinO) (COS'l' + isin'l') = COS(O + '1') + isin(O + '1').

ComplEx NumbErs

Then the DeMoivre formula(cos 0 + i sin o)n = cos nO + i sin nO

reduces to(ei8)n = ein8,

a triviality. Substituting 0 = 71" in the equality defining eiB , we get

39

ein

=cos 71" + i sin 71" = -1 , :. e in + 1 =0,

which ties together the five numbers O,J, 71", e, i; arguably, these are thefive most important numbers in mathematics.

Now the exponential function f(x) = eX (x E JR) is characterized bythe property

f(x + y) = f(x) . f(y); i.e.,

(and the initial condition f' (0) = 1). Is this still valid when extended toa complex variable? The answer turns out to be affirmative, and thereis a very slick proof that appeared in T. Takagi [Mathematical Analysis,Iwanami, Tokyo, 1986, p. 190), if we know that a complex power seriescan be differentiated termwise within the disc of convergence. Let

Z Z2 znf(z)=l+-+-++-+.I! 2! n!By tennwise differentiation, it is immediate that

for all zEC andall nEJ\l.

Therefore, by the Taylor series expansion of f(z + w) around z, we get

f'(z) J"(z) f


Adding and subtracting the equalities

eiB = cosO + isinO

ande- i8 = cosO - i sin O.

we getei8 + e-iB

cosO = Z

Multiplying, we get

eiB _ e-iBsinO = Zi

EXAMPLE. Show that

d" (e'" cos x) = zn/2e", cos (x +~)~n 4 (n=O.I.Z... ).Solution. This can be done by mathematical induction, but the follow-ing computation gives insight, and allows generalization.

= !R { :n e(l+i)'" }= !R { (I + it .e(l+i)'" }

= !R { ( ,fi . ei (1r/4)r.e(l+i)'" }= zn/2 . !R { e'" . ei ("'+(n1r/4))}

=2n / 2 e'" . cos (x + n41l") .

Complo NumbErs

EXAMPLE. Show that for 0 < r < I,

41

2 1- r 21 + 2 (rcosO + r cos20 + ... + rncosnO + ...) = .1 - 2rcosO + r 2

Solution.

00 00

1 + 2L rncosnO = 1 + 2L~ (rneinS )n=1 n=1

{reiS }

= 1 + 2~ 1 _ reiS_ ~ {reiS (1 - re-iS )}- 1 + 2( S ( S1 - re' ) 1 - re-' )

=1+ 2(rcosO-r2)1 - 2rcosO + r 2

1- r 2- -------,---=

1-2rcosO + r2 '

which is the important Poisson kernel. Taking the imaginary part insteadof the real part in the above computation, we obtain the conjugatePoisson kernel:

00 0L n. _ rsmr smnO - 1 2 0 2

- rcos + rn=1

EXAMPLE. It is simple to verify that

(0 < r < 1).

(n =0);(n = I, 2, 3, .).


12" 12" (ei6 + e-i6 )4. cos4 (J d(J = d(J.. 2o 0= _I [2" (eM + 4e2i6 + 6 + 4e-2i6 + e-4i6 ) d(J

24 Jo=~ [2" d(J = 311".

16 Jo 4More generally, for n E N, we have

121' 12" (2n) 211"sin2n (J d(J = cos2n (J d(J = . -o 0 n 22n(2n)! 211"

= (n!)2 . 22n

= 1.3.S ... (2n-3).(2n-l).27I" (Wallis).246 .. (2n - 2) . (2n)

Exercises

1. Perform the indicated operations, and reduce each of the followingnumbers to the form x + iy (x, Y E IR):

(a) (1 - i)(2 - i)(3 - i);

(b) (J3 +it4 + 3i

(c) 3-4i'S-z(d) S ' where z = 4 + 3i.

+z

2. Find the real numbers x, y, u, v satisfying

z = x + i, W =3 + iy,

z + W = U - i, zw = 14 + iv.

ComplEx NumbErs 43

3. Letz = a + ib(a,b E lR).

(a) Express IlR(z2)12 + 1~(z2)12 = Izl4 in tenns of a and b.(b) If z = 2 + i, then this gives us 32 + 42 = 52. Can you find other

Pythagorean triples?

4. Show that any complex number z with Izl = I, but z '" -1 can beexpressed as

1 + itz = -:---

1- it

with an appropriate choice of the real parameter t.

5. Let z = a + ib (a, bE lR). Find conditions on a and b such that

(a) z4 is real;

(b) z4 is purely imaginary.

6. Find the absolute values of

(a) 3 + 2i;(b) -1 + iV3;(c) -i(1 + i)(2 - 3i)(4 + 3i);(d) (3 - i)(-I.+ 2i).

2 - 31

7. Let z = a + ib, W = c + id (a, b, c, d E lR).

(a) Express Izwl = Izllwl = Izwl in tenns of a, b, c, d.(b) Choosingz = 2+i,w = 2+3iin(a),wegetS2 + 1 = 42 +72

Find other positive integers p, q, r (p '" I, q '" 1) satisfying theequation

(c) Show that the set

S = {p EN; P =m2 + n2 for some m, n E N}

44 COMPLEX NU~8ERS AND GEOMETRY

is closed under multiplication; i.e., p, q E S ==> pq E S.

8. (a) Prove the parallelogram law:

for two arbitrary complex numbers 0 and f3.(b) Suppose that 101 = 1f31. Show that for any 'Y E C,

(c) Interpret the equalities in (a) and (b) geometrically.

9. If 101 < I and Izi < I, show that Itz::z I ~z;

(b) Iz - 1+3il < 4;(c) Iz-II+lz+il =2;

(d) Iz - 1 + il-Iz + 1 - il > 2.

II. Let 101 = 1f31 = I'YI = I.

(a) Suppose 0 + f3 + 'Y '10. Show that I{3""'( + ""'(a + 0{3I= I.a+{3+""'(

12. Solve the following quadratic equations :

(a) ~z2+(I_ i)z+i = o. (Compare with the example in I.3. Whatconclusion can you draw?)

(b) (I - i)z2 - 3z - (I + i) = O. (What is the discriminant? Are theroots real?)

ComplEX Numbers 45

13. (a) Show that if Ct is a root of a polynomial equation with realcoefficients (i.e., all the coefficients are real), then Ct is also aroot.

(b) Show that a polynomial equation with real coefficients and odddegree must have at least one real root.

14. What is wrong with the following 'proof'?

:. 2 =O.

15. Solve

(a) z3 - i =0;

(b) Z4 + 1 =0;

(c) ii + 32 =0;(d) z6 -1 = O.

16. Give another example of an order relation in iC satisfying the pos-tulates PI and ~ in 1.5.

17. Let a, b, c E JR. Prove

(a) a> b, b> c =} a > c;(b) a> b, c> 0 =} ac> be;(c) a> b, c < 0 =} ac < be,

from the postulates Pt, ~,and P3 in 1.5.18. If (5 = I, show that

() ( + (2 + (3 + ~ = 2;a 1 + (2 1 + (4 1 +( 1 + (3

(b) ( + (2 + (3 + (4 =0 ( ., I).1 - (2 1 _ (4 1 - ( 1 _ (3

46

19. Let wZ + w + 1 = O.


(a) Show that every complex number Z E C can be expresseduniquely in the form

z=a+bw (a, bE 1R).

(b) Find a and b (a, b E 1R) such that 7 + 5w + 3w2

= a + bw.1-2w

20. Show that for arbitrary Zl, Zz, ... ,Zn E C, we have

IZI + Zz + ... + znl < Izd + Izzi + ... + 1z,.1

When does equality hold?

21. Given three vertices 3 + i, 1 - 2i, -2 + 4i of a parallelogram, findthe fourth vertex. How many solutions are there?

22. Show that the diagonals of a parallelogram bisect each other.

23. Show that in an arbitrary quadrangle, the midpoints of the foursides are the vertices of a parallelogram.

24. We have seen in 1.7, that a point Z is on the line segment joiningthe points Zl and Zz if and only if

Z =(1 - t)ZI + tzz for some t E (0, 1).

What if t E IR is not in this range?

25. (a) Show that ZI. zz, Z3 E C are collinear if and only if there arethree real numbers a,l3, 'Y, not all zero, such that

aZI + I3zz + 'YZ3 = 0, a + 13 + 'Y = O.

(b) Can this be extended to four points or more?

26. Given a hexagon, if we choose the midpoints of alternate sides, weobtain the vertices of two triangles. Show that the centroids of thesetwo triangles coincide.

Complu NumbErs 47

27. Let A', B', C' be points on (the extensions of) the respective sidesBC, CA, AB of 6ABC such that

BA' CB' AC'BC = CA = AB'

where the orientations of the line segments are taken into account;

i.e., ~-:; > 0 if""iii and BC are of the same direction, and ~-:; < 0if they are of the opposite direction; and similarly for the otherratios. Show that the centroids of 6A'B'C' and 6ABC coincide.

28. (a) Given four arbitrary points ZI. Z2, Z3, Z4 in the complex plane,let WI. W2, Wl, W4 be the centroids of

respectively. (If you so wish, you may assume that no threeof these four points are collinear.) Show that the four linesegments joining the points Zt and WI. Z2 and W2, Z3 and W3,Z4 and W4 intersect at one point.

(b) Generalize.29. (a) Given 3 points, construct a triangle for which these points are

the midpoints of the sides.

(b) Given 5 points, construct a pentagon (may be self-intersecting)for which these points are the midpoints of the sides.

(c) Generalize.(d) What if the number of the given points is even?

30. Let Zt. Z2, Z3 be three arbitrary points in the complex plane.

(a) Show that a point Z is in the interior or on the boundary of6ZtZ2Z3 if and only if there are nonnegative real numbers (t, {3,'Y such that

(t + {3 + 'Y = 1.

48 COMPlEX NUMBeRS AND GeOMeTRY

(b) Find the locus of the points for which 0 = ~.(c) Show also that if the triangle L>zlz2z3 does not degenerate,

then the correspondence between the set of points z in the(closed) triangle (i.e., the boundary included) and the set ofordered triples

is one-ta-one. [(0, {3, 1') are called the barycentric coordinatesof the point z.J

31. (a) Let z" Z2,' .. ,Zn be arbitrary points in the complex plane (n >2). Show that a point Z is in the smallest closed convex polygoncontaining these n points if and only if there are nonnegativereal numbers 0" 02, ... ,On such that

n

Z = '"ozLJ J l'j=l

n

(b) Assuming z" Z2, . . ,Zn are vertices of a convex n-gon, do wehave uniqueness of representation as in the previous problem?

32. (a) Suppose 0 < arg w - arg Z < 7r, show that the area of L>Ozw isgiven by

12~(zw).

(b) Suppose 0 < arg Zl < arg Z2 < ... < arg Zn < 27r, show thatthe area of the polygon whose vertices are at Zl, Z2, ,Zn isgiven by

(c) Show that the result in (b) can also be written as

1 n4i L(Zk - Zk-l)(Zk +Zk-l).

k=l

ComplEX NumbErs 49

33. (a) Given z E C, show that there exist 0,.8 E C with 101 = 1.81 = 1such that z = a + .8 if and only if Izi < 2.

(b) Given z E C, show that there exist a,.8,oy E Cwith 101 = 1.81 =loyl = 1 such that z =a + .8 + ")' if and only if Izi < 3.

(c) Generalize.

34. (a) Find the condition on a, b E JR for which the system of simulta-neous equations

cos x + cosy = a,

has solutions x, y E JR.

sinx + siny = b,

.,fia = 2'(b) Solve the above system of equations for the case

b = v!62 .

()SI . v!6. v!6cove cos x + smy =2' smx + cosy =-2'

(d) Solve 5cosx + 3siny = - 7V;. 5sinx - 3cosy = 7V;.35. (a) If Zl + Zz + Z3 = 0 and IZII = Izzi = IZ31 = 1, show that z.,

zz, Z3 are the vertices of an equilateral triangle inscribed in theunit circle.

(b) If Zl + Zz + Z3 + Z4 =0 and Izd = Izzi = IZ31 = 1Z41 = 1, whatcan be said about the quadrangle with vertices at z., Zz. Z3. Z4?

t t36. For any complex number a f 0, show that a. -a, if' - if' 0 arecollinear.

37. Find polar representations for

(a) 1 + i;(b) 4-3i;(c) 1 + w;

50

(d) .!.,w

where w2 +w + 1 =O.

38. Give a counterexample to


if we make the restriction 0 < arg z < 21r. What if we make therestriction -1r < arg z < 1r?

1 _ zn+139. (a) Show that 1 + z + z2 + ... + zn = (z 1'1).l-z

(b) Suppose (17 = 1 ( l' 1). Show that

where k is an arbitrary integer that is not a multiple of 17.

40. Let 9 be the exterior angle of a regular n-gon. Show that

(a) 1 + cos9 + cos 29 + ... + cos(n - 1)9 = 0;(b) sin9 + sin 29 + '" + sin(n - 1)9 = O.

41. (a) Derive the following identities:

n cos n6 . sin (n+I)II~cosk9= 2 2L-J "6'k=O SIn 2

n sin n6 . sin (n+l)8~sink9= 2 2~ . 6k=J SIn 2

(0 < 9 < 21r).

(b) Use the identities in (a) to evaluate the sums:

nL k = 1 + 2 + 3 + ... + n,k=l

ComplEX NumbErs

n

:~:::::>2 = 12+ 22+ 32 + ... + n2.k=1

[This selVes to check our results in (a).]Hint: lim sink/1 =? lim 1 - cosk/1 =?

8_0 /1 8-0 82

42. Show that

(n) (n) (n) n/2 . mr1 - 3 + 5 - + ... = 2 sm 4'Hint: (1 + z)n = (~) +G)z+ ... +(:)zn.

43. (a) Express 1 + ivIJ in polar form.(b) Simplify (1 + ivIJ) 1991 + (1 _ ivIJ) 1991.(c) Simplify (1 + i)I991 + (1 - i)I991, (1 + i)I991 - (1 - i)I99I.

44. Supposexn + iYn = (1 + ivIJr (xn,Yn E 1R). Show that

XnYn+l - Xn+IYn =22nvIJ,

45. Find the smaIlest positive integers m and n satisfying

51

46. Let x = a + b, Y = aw + bw2, Z = aw2+bw, where w2+w + 1 = O.Express x3 + 11 + z3 in terms of a and b.


47. Let W2 +w+ 1= O.

(a) Given any two polynomials p(z) and q(z), show that the polyno-mials

f(z) = p(z)p(wz)p(w2z),

g(z) = p(wz)q(w2z) +p(w2z)q(z) + p(z)q(wz)

have nonzero coefficients ak, bk only if k is a multiple of 3, whereak, bk are the coefficients of zk in f(z) and g(z), respectively.

(b) Prove that every function '1'(z) (defined for all z E q can beuniquely expressed as

rp(z) = f(z) + g(z) + h(z),

where f(wz) = f(z), g(wz) = wg(z), h(wz) = wlh(z) for allz E Co

(c) Generalize.

48. Show that

(a) sin 20 = 2sinO cosO,

Sin30=4sinO.SinG -0) .SinG +0),Sin40=8sinO.cosO.Sin(; -0) .sin(; +0),sin SO = 16sinO sin G-0) . sin Cs11" - 9)

. sin (; +0). sin (2; +0).(b) cos29=2sinG -0) .sinG +0),

cos 30 = 4 cos0.sin (~ - 0) .sin (~ +0),

ComplEx NumbErs

COS40=8sinG -0) 'Sine; -0) ,sinG +0). (37r )

sm g+9,

cos 50 = 16cosO sin (~ - 0) .sin C~ -0). sin (!!:. + 9) .sin (37r + 9)10 10'

(c) Generalize.

49. (a) Let ( = e"i/n, show that

n

TI(z-(k)=zn-l,k=1

n-ITI (z - (k) = I + z + ... +zn-I .k=1

(b) Show that

53

. 11" 211" (n - 1)11" nsm-sm-sm =--

n n n 2n - 1 (n > 2).

50. Construct a regular pentagon using only a compass and straightedge.

J51. Suppose z + - = I. Show that the sequence {Wk}~, where Wk =z

zk + ~, is periodic, and fmd its period.z

52. (a) Show that the polynomial z2n + zn + I (n E N) is divisible byz2 + Z+ I if and only if n is not a multiple of 3.

(b) Find a necessary and sufficient condition on natural numbers pand q such that the polynomial zP+ Z9+ I is divisible by z2+ z+ I.

54

53. Show thatl


(a) For every natural number n, there exist polynomials Pn(x) andqn(x) (with real coefficients) satisfYing

cosnB = Pn(tanO) cosn 0,

sin nO = qn(tanO) . cosn O.

(b) Pn(x) = Ho +ix)n+(I-ix)n},qn(x) = ii {(I + ix)n - (1 - ix)n}.

(c) P'n(x) = -nqn-I(X). q'n(x) = npn_l(x) (n> I).

54. Guess and prove similar relations to the first Example in 1.10 :

rr'(a) dxn (e- X sin x);

(b) ::;n (e,jjx . cos x);

dn(c) -d(e- x . sin V3x).xn

55. Evaluate

(a) H = Jeax . cosbxdx,(b) K = Jeax . sinbxdx.Hint: H + iK = ?

) Problem 53, excluding part (b), is taken from the entrance examination oflbe UniversityofTokyo, Japan, February 25, 1991.

CHAPTER 2Applications to Geometry

2.1 Triangles

We now discuss applications of complex numbers to plane geometry. Itis important to keep in mind that complex numbers are not just vectors;they can be multiplied by each other. In applications to geometry, weshall make full use of this property. Complex numbers are particularlyeffective for certain types of problems, but may be cumbersome forsome problems that can be solved by elementary methods.

In elementary geometry, triangles are the building blocks and thecongruence and similarity of two triangles are the most fundamentalconcepts. We start from the conditions on the similarity of two trian-gles in terms of complex numbers. Let us first present the followingnotational conventions and some review. Throughout this chapter, wesay 6Zt Z2Z3 and 6WtW2W3 are similar, and write

if and only if the angle at Zk is equal to that at Wk (hence Zk correspondto Wk, k = 1,2,3), and they are of the same orientation (i.e., they areboth counterclockwise or both clockwise).

If they are of the opposite orientation (one clockwise, the othercounterclockwise), then we write

55aJ


W,

AGURE 2J

(reversed).

Note that Zk still must correspond to Wk (k = 1,2,3).As usual, we use the notations II, ..L to denote that two lines (line

segments or vectors) are parallel or orthogonal, respectively.Since for distinct points 0, (3, ')' E C,

(3-0arg =arg({3 - 0) - arg(')' - 0)

')'-0

= the oriented angle from the vector cry to "iJ,

0, (3, ')' are collinear {3 - 0 E IR')'-0

{3-0 {3-0 =~-;

')'-0 ')'-0

and

-+ {3-00{3 ..L cry is purely imaginary

')'-0

Applications to GEomEtry

[3-0. [3-Ci~ + =0.

1'-0. 1'-0.

More generally, for four distinct points a, [3, 1', 0 E iC,

O13I1::;8~~-aEIR-1'

[3-0. [3-0.~ = .

0-1' 0-1"

57

-+ -+furthermore, 0.[3 and 1'0 have the same (opposite) direction if and onlyif ~:::~ is a positive (negative) real number; and

-+ -+ [3-0.0.[3 .11'0~ 0 is purely imaginary

-1'

[3-0. [3-0.~ 0 + - = O.-1' 0 -1'

Consequently, if 0.[3 t: 0, then

a a10. + [31 = 10.1 + 1[31~ [3 E IR and [3 > o.

1rHIoc>1tE~ 2.1.1. 6ZIZ2Z3 '" 6w\W2W3

Z2 - Z\ W2 -WI~ -

Z3 - ZI W3 - WI

Z\ WI 1~ Z2 W2 1 = O.

Z3 W3 1

Proof Tho triangles are similar if and only if the ratios of the lengthsof the two corresponding sides are the same and the (corresponding)angles between them are the same (including the orientation). Hence


Z2 - ZI W2 -WI

AppIIcatlons to GEOmEtry S9

___-.('A~(a)

B(P)IL-----4....;..,....---'1......

'.

FIGURE 2.2

(b)z z 1Z\ Z2 1 = O.Z2 z\ 1

EXAMPLE. The perpendicular bisectors of the three sides ofan arbitrarytriangle meet at a point. This point is called the circumcenter of thetriangle.Solution. Let the three vertices A, B, C of the triangle be representedby complex numbers G, 13, 'Y, respectively. Then the equation of theperpendicular bisector of the side BC is

Z Z 113 'Y 1'Y 13 1

=0',

I.e.,

Similarly, those of the sides CA and AB are

60 COMPlEX NUMBeRS AND GEOMETRY

respectively. Adding any two of these three equations gives the thirdone, which implies that the solution of any two of these equationsautomatically satisfies the third. In other words, the intersection of anytwo perpendicular bisectors is on the remaining perpendicular bisector.

Solving a system of simultaneous equations consisting of any two ofthese three equations, we obtain the circumcenter :

Z=1012(.8 - 1') + 1.8/2(')' - 0) + 11'12(0' -.B)

0(.8 - 1') + .8(')' - 0) + 1'(0' - .8)

Note that, by symmetry, we see again that this solution also satisfies theremaining equation.

EXAMPLE. 6z t Z2Z3 is an equilateral triangle

6Zt Z2Z3 ~ 6z3z t Z2

Zt Z3 1

Z2 Zt 1 =0Z3 Z2 1

222 Zt + Z2 + Z3 - Z2Z3 - Z3Zt - ZtZ2 =0

(W2 +W + 1 =0) Zt + WZ2 + w2z3 = 0 or Zt + W2Z2 + WZ3 = 0

Zt 1 1 Zt 1 1 Z2 W 1 =0 or Z2 w2 1 = O.

Z3 if 1 Z3 W 1 6Zt Z2Z3 ~ 61ww2 or 6 ZtZ2Z3 ~ 61w2w.

EXAMPLE. (Napoleon) On each side of an arbitrary triangle, draw anexterior equilateral triangle. Then the centroids of these three equilat-eral triangles are the vertices of a fourth equilateral triangle.

Proof Let 6Zt Z2Z3 be the given triangle, and


,2

w_------'=k

..... :

t,

w,

FIGURE 23

or

v

x

61

be aU equilateral with the same orientation as ~lww2, say, (where w 2 +W + 1 = 0), and with (I. (2, (3 as the centroids of these equilateraltriangles. Then

WI + WZ3 + w 2Z2 = 0,

Z2 + WZI + w 2W) = O.

To prove that ~(1(2(3 is equilateral, we compute

(I +W(2 + W2(31 W if

=3(wl + Z3 + Z2) + 3(Z3 + W2 + ZI) + T(Z2 + ZI + W)

= ~ {(WI + WZ3 + w2Z2) + (Z3 + WW2 + W2ZI) + (Z2 + WZI + w 2W)}=O.

Therefore, ~(1(2(3 is an equilateral triangle. o

COMPlEX NUMBERS AND GEOMEmY

Alternate Proof. Since 6(tZ3ZZ ~ 601w, we have

(I 0 1Z3 1 1 = 0;ZZ W 1

i.e.,(1 - W)(I - Zz + WZ3 = o.

r _ Zz - WZ3'>1 - l-w

Similarly,r _ Z3 - WZI,>z- 1 '

-w

r _ Z] -WZz'>3 - l-w

(I +w(z +WZ(3 = 1 ~ W {(zz - WZ3) +W(Z3 - WZI) +WZ(ZI - wzz)}=o.

o

The above result is usually attributed to Napoleon. It is well knownthat Napoleon established I'Ecole polytechnique (1794), which pro-duced most of the French mathematicians in the early 19th century,and was fond of mathematics, especially geometry. Yet many peopleare skeptical that Napoleon knew enough geometry to discover thistheorem. Incidentally, Napoleon Bonaparte is one of very few peoplein modem history who are known by their first names. Galileo Galilei(1564-1642) is another example.EXAMPLE. Lines f .. fz, f3 are parallel to each other with fz between f]and f3. The distance between fl and fz is a, and that between fz andf3 is b. Express the area of an equilateral triangle having one vertex oneach of the three parallel lines in terms of a and b.Solution. Choose the coordinates as in Figure 2.4. We fix one vertexat ai, move another vertex t along the real axis, and try to find the locus

Applkatlons to GEOmEtry

v

(a +b=-)'--i-t- -=-;-- ....,....._1.!.-3

63

al b

a

I,

I,----;O


2.2 ThE PtolEmy-Euler Theorem

For any four complex numbers 0,/3,1,6, the following identity is easy toverilY:

(0 - /3), ("( - 6)+(0 - 6) (/3 - 1) = (0 - 1)' (/3 - 6).

By the triangle inequality, we obtain

10 - /31 h - 61 + 10 - 611/3 - 11 > 10 - 111/3 - 61

Let us investigate when the inequality becomes an equality. In the case ofthe triangle inequality,

equality holds if and only if ~ is a positive real number (provided22

ZI Z2 to). Thus we are looking for a condition to ensure that (0 - POky - 6)(0 - 6)(f3 -1')

is a positive real number. But

(0 - /3)("( - 6)(0 - 6)(/3 - 1)

0 - /3/1 - /30-6 1-6

is a positive real number

is a negative real number

arg {O -/3/1 - /3}0-6 1-6

{ 0-/3} {1-/3}=arg -arg =1r0-6 1-6 (mod 21r).

It follows that 0, /3, 1. 6 are cocyclic. i.e., 0, /3, 1, 6 are on the same circleor line (see Corollary A.23 in Appendix A) and 0 and 1 are on the oppositesides of the chord joining /3 and 6, which results in the alphabetical order(clockwise or counterclockwise).

We have proven the following.

Applications to GEOmetry 65

a ~ y 0, , I0 Y @I I

0 a ~ Y,

AGURE 2.5

THEOREM 2.2.1. For any four points A, E, C. D in the plane,

ABCD + BC DA > AC BD.

Equality holds if and only if these four points are cacyclic (or collinear)and are in alphabetical order (clockwise or counterclockwise).

The equality was discovered by C. Ptolemy (ca. 85-165), while thegeneral case was found over a thousand years later by L. Euler (1707-1783). However, using complex numbers, their results can be obtainedin a single stroke.

The expression

( a - 'Y) / (/3 - 'Y)(a,/3;'Y,6):= a-6 /3-6

iscalIed the cross ratio of the four points a, /3, 'Y, 6. It plays an importantrole in various parts of mathematics, especialIy in projective geometry,which is certainly one of the most beautiful branches of mathematics.

COROlLARY 2.2.2. Four points 0', /3, 'Y. 6, are cacyclic (or collinear) ifandonly if

(a,/3; 'Y,6) E JR.


In the sequel, 'collinear' is regarded as a particular (degenerate) caseof 'cocyclic '.

When the inscribing quadrangle is a rectangle, the Ptolemy theoremreduces to

COROLLARY 2.2.3 (Pythagoras). In a right triangle ABC, with the angleat C being the right angle,

EXAMPLE. Let ABCDEbe a regular pentagon of sides t inscribed in a~

circle of radius r, and P be the midpoint of CD, and d the length of adiagonal. Applying the Ptolemy theorem 2.2.1 to the quadrangles ACD Eand ACPD, we get

and 2xd = 2rt,

where x is the length of a side of a regular decagon inscribed in the circleof radius r. It follows that

satisfies

A.-~

....... '

FIGURE 2.6


Therefore, the ratio of the radius T to the side x of the inscribed regulardecagon is the famous golden ratio :

1+ J5cp=

2(... cp > 0).

In particular, a regular pentagon and a regular decagon can be constructedwith a compass and straightedge, as we have already seen at the end ofI.9.

2.3 The Clifford Theorems

In this section, I we prove an infinite sequence of theorems discovered byW. K. Clifford (1845-1879). The crucial step is the following lemma,which we shall use in other sections too.

LEMMA 2.3.1. Suppose there are four circles C.. C2 C3, C4 in a plane.Let C. and C2 intersect at ZI and WI, C2 and C3 intersect at Z2 and W2,C3and C4 intersect at Z3 and W3, C4 and C1 intersect at Z4 and W4. Thenthe points ZI, Z2, Z3, Z4 are cocyclic if and only if WI, W2, W), W4 arecocyclic.

Proof By assumption, the following four cross ratios are real :

) _ Z. - Z2 / Z. - WI~,WI - ,W2 - Z2 W2 - WI

_ Z2 - Z3 / Z2 - W2Z3, W2) - ,

W3-Z3 W3-W2

I Other than the lemma, the material in this section will not be needed for the rest of thebook, and so may be skipped in the fust reading.

68

Therefore,

,


(ZI,W2; Z2,WI) (Z3,W4; Z4,W3)(Z2, W3; Z3, W2) . (Z4, WI; z), W4)

= {(ZI - Z2) / (ZI - Z4)} . {(WI - W2) / (WI =W4)}Z3-Z2 Z3-Z4 W3-W2 WJ W4

is real. Hence (ZI' Z3; Z2, Z4) is real if and only jf (WI. W3; W2, W4) is real.o

c,

.,.'

c,c,

FIGURE 2.7

We say n lines in a plane are in general position if no two of them areparallel and no three of them meet at a point.

Let us call the intersection of two lines in general position, theirClifford point. From three lines in general position, we obtain threeClifford points, by choosing a pair of them each time, and the circlethrough these three points (the circumcircle of the triangle formed bythese three lines) is called the Clifford circle of the three lines.

Applications to~"Y 69

Now. given four lines Ch C2, C3. C4 in general position, let Zjk bethe intersection ofthe lines Cj and Ck (other than 00), and C'mn be thecircumcircle of f>zmnznIZlm. (We disregard the permutation of indices;for example, Zjk = Zkj, C

'mn = Cn1m .) Applying Lemma 2.3.1 to C234,C2, C h C134, and noting that

C234. C 2 intersect at Z23. Z24;C2. C 1 intersect at 00. Z12;C 1, C I34 intersect at %13, %14;

C I34, C234 intersect at Z34, ZI234,

where ZI234 is the 'new' intersection of C234 and CI34 (i.e., other thanZ34), then, since Z23, 00, ZI3, Z34 are collinear (all on C3). we concludethat %24, %12, Z14, %1234 are cocyclic. But the circumcircle of f>z24z12Z14 isthe circle C 124, hence the circles C234, C I34, C124 meet at ZI234

On the other hand, with the same C234, Cz, C" C l 34. if we note thatZ24, 00, Z14, Z34 are collinear (all on C4 ), we conclude that Z23. Z12, Z13,ZI234 are cocyclic. But the circumcircle of f>z23ZI2ZI3 is the circle C123,hence the circles C234, C I34, C123 meet at %1234

------

c.

\\

\\\\IJII

fI

I

/em/

I/

//

-

"\\\\IIIII

fI

II

I/

//

/

I \ II \ I

f Z' ff ,f

II\\\\

\C,,~,,,,

,

FIGURE 2.S


We have shown that the circles C234, C134, C124, Cm meet at one pointZI234, which we call the Clifford point of the four lines C I , C2, C3, C 4.

Before we proceed to the case of five lines, we remark that the pointZjk is at the intersection of the lines Cj and Ck ; the circle Clmn passesthrough the points Zmn, Zln> Zlm ; the point Zklmn is the intersection ofthe circles C 1mn, C kmm Ckln, C k1m . In particular, the circles C 1mn andC kmn intersect at Zklmn and Zmn. Now we are ready for the next step.

Suppose we are given five lines C\, C2, C3, C4, C5 in general position.Preserving our notations above, and taking four lines at a time, weobtain five Clifford points Z2345, Z1345, .ZI245, ZI235, ZI234. We claim thatthese five Clifford points are cocyclic. To prove this, it is sufficientto prove that any four of these five Clifford points are cocyclic. Forexample, take ZI345, Z1245, ZI235, ZI234. These can be considered as theintersections of C I34 and Cm, C m and C 145 , C m and C I25, C I24 andCI34, respectively. The second point of the intersection of these pairs ofcircles are the points ZI3, Z15, Z12, Z14, which are collinear (all on C 1), andso, by Lemma 2.3.1, we get the desired result. The circle so obtained iscalled the Clifford circle of the lines C\, C2, C3, C 4, C 5, and is denotedby C I2345

Now given six lines C\, C2, C 3, C 4, C 5, C 6, in general position. Takingfive lines at a time, we obtain six Clifford circles. We claim these sixcircles meet at a point, the Clifford point of the six lines. To prove this,it is sufficient to show that any three of these six circles meet at a point.Note carefully: it is simple to give an example of four circles, any threeofwhich meet at a point without all four of them meeting at a point, butthis cannot be done if we have five circles or more.

Suppose we want to show that C23456, C I3456, C I2456 meet at a point.Consider a sequence of four circles C23456, C245, C 145 , C 13456. Theseintersect, in pairs, at Z2345 and Z2456, Z45 and Z1245, ZI345 and ZI456, Z3456and ZI23456, where ZI23456 is the intersection of C23456 and C I3456 otherthan Z3456 But the points Z2345, Z45, ZI345, Z3456 are all on the circleC 345 . Hence the points Z2456, ZI245, Z1456, and ZI23456 must be cocyclic.However; the first three of these four points are on the circle CI2456,hence the circle C I2456 passes through the intersection of C23456 andC13456

Now it is simple to carry on the induction argument.

Applications to GEomEtry 11

2.4 The Nine-Point Circle

Given a triangle ABC, choose its circumcenter a to be the origin of thecomplex plane, and let 0, fJ, "( be the complex numbers representingthe vertices A, B, C, respectively. Without loss of generality, we mayassume that the circumcircle has radius 1; i.e., 101 = IfJl = bl = 1.Then it is natural to ask, Where is the point u = 0 + fJ + "(?

Since u- 0 = fJ +"(, and (3 ; -y is the midpoint D of the side BC, u ison the perpendicular from the vertex A to the side BC, and the lengthlu- 01 is twice that of aD. By symmetry, u also is on the perpendicularfrom B to C A, and on that of C to AB; i.e., u is the orthocenter H of!'J.ABC. Note that we have shown that three perpendiculars from thevertices to the opposite sides meet at a point, called the orthocenter of!'J.ABC.

0.i 2

0B(~)\L----'D~*A----"IC(y)

u'

FIGURE 2.9

Now I = !(0 + fJ + "() is the midpoint of the yne segmentjoiningthe circumcenter a and the orthocenter H. The distance from 2 to themidpoint D of the side Be is

~ = I~I = ~.


Similarly, the distance from I to the midpoint E of the side CA, and tothe midpoint F of the side AB are all equal to ~.

Furthermore, the distance from I to the midpoint of the line segmentjoining the orthocenter H to the vertex A is

10 + (f _ (f = 101 = ~

2 2 2 2'

Similarly, the distance from I to the midpoint of BH, and to that ofCH are also equal to ~.

To find the foot>. of the perpendicular from the vertex A to the sideBC, we first compute the point 0' where this perpendicular meets thecircumcircle again. Thus 0' must satisfy the conditions

- -+00' 1- {Jr,

From the first condition, we get

10'1 = 1, 0' 'f o.

i.e.,

0-0'

{3-'"'( is purely imaginary;

0-0' 0-0/--:-- + =O.{3 - '"'( {3 - '"'(

Substituting the relations 0 = ~,etc., this becomes

Q - 0.'{ I + {3'"'(} = O.00'

Hence, {3'"'(

0=--.o

To check whether our computation is correct, note that 10'1 = 1, andf!.. 'I, = -1, and so arg (f!.) + arg (2.) = 11", which means that;;;; 1-Q 0 Q d-+

{3'"'(; viz., 0' is the point where the perpendicular from A to the side BCmeets the circumcircle again.

Applications to GEomEtry 73

Now the distances from the vertex B to 0.' and to u are

{J-y {J{J+ - = - 10. +-YI = 10. +-yl,0. 0.

lu - {JI = 1(0. + (J + -y) - {J! = 10. + -yl,

respectively. Hence 6{J0.'u is an isosceles triangle, and

" = ~(u + 0.') = ~ (u -~) .It follows that the distance from I to the foot " (of the perpendicular fromthe vertex A to the side Be) is

Similarly, the distances from I to the other two feet of perpendiculars arealso 4.

/ . /. ." ./ , , .,

fiGURE 2.10

Summing up, we have obtained the following.


THEOREM :z.4.1 (The Nine-Point Circle). In any triangle,

(a) the feet of the three perpendiculars from the vertices to the oppositesides;

(b) the midpoints ofthe three sides; and

(c) the midpoints of the segments joining the orthocenter to the threevertices,

are all on the same circle, whose center is at the midpoint of the segmentjoining the orthocenter and the circumcenter, and the radius is one halfofthat ofthe circumcircle.

The line passing through the onhocenter, circumcenter, centroid,and the center of the nine-point circle is known as the Euler line of thetriangle.

Let z., Z2, Z3 be three arbitrary points on the unit circle Izi = 1. Thenthe circumcenter, centroid, the center of the nine-point circle, and theonhocenter of 6Z)Z2Z3 are given by

0,

respectively, and the radius of the nine-point circle is !.Suppose we are given four points Z), Z2, Z3, Z4 on the unit circle.

Choosing three points out of these four points at a time, we obtain fourtriangles (all of which are inscribed in the unit circle).

The center of the nine-point circle of 6Z2Z3Z4 is T) = HZ2 + Z3 + Z4),the center of the nine-point circle of 6ZIZ3Z4 is 7'2 = !(Z\ + Z3 + Z4),the center of the nine-point circle of 6Z\Z2Z4 is T3 = !(Z) + Z2 + Z4),the center of the nine-point circle of 6Z\Z2Z3 is T4 = HZ\ + Z2 + Z3),and their radii are all equal to !.

Consider the point


Then it is immediate that

Hence the nine-point circles of

7S

all pass through the point

in particular, the centers of the four nine-point circles are on the circlewith center at T and radius!. Let us call this circle the nine-point circleofthe quadrangle ZIZ2Z3Z4.

Now, suppose we are given five points Zl> Z2, Z3, Z4, Zs on the unit cir-cle. Then the center of the nine-point circle of the quadrangle Z2Z3Z4ZSIS

1/1-1 = 2(Z2 + Z3 + Z4 + zs), etc.,

and the distances from these centers to the point

are1

1/1--/1-11=2' etc.

Hence the centers of the nine-point circles of the quadrangles

are on the circle with the center at/1- = !(ZI +Z2+Z3 +Z4 +zs) and radius!. Let us call this circle the nine-point circle ofthe pentagon ZIZ2Z3Z4ZS

Next, suppose we are given six points Zt, Z2, Z3, Z4, Zs, Z6 on the unitcircle, ....

Thus we have an infinite sequence of theorems discovered by J. L.Coolidge.

76

2.S The Simson Une

COMPlEX NUMBERS AND GEOMTRY

We start this section with some preparatory comments about the equa-tion of a line.

Given a line e, let a be the unit vector perpendicular to e, and p thedistance from the origin to the line e. Then for any point z on e, z - pais a vector on e, and since a is a vector perpendicular to e, we have

z-paa

is purely imaginary; i.e.,

_z----'pa=__ + z- pa = O.a a

Hence the equation of the line eis given by

z z-+-=2p'

- ,a aI.e., z + kz = 2pa,

where p E JR, k = ~,:. Ikl = 1. And to obtain the equation of a lineperpendicular to e, simply replace a by ia :

z Z-. -==2q

~a ~a

~y

,)/ao

for some q E JR; i.e.,

xI

RGURE all


z Z .- - - = 2qz,a a ' VIZ., Z - kz =2qia, (k = ~) ,

77

where the constant on the right can be adjusted to pass through a specificpoint.

Note that a linear equation in z and z jointly is the equation of a lineif and only if it is self-conjugate; viz., the relation obtained by taking thecomplex conjugate of both sides of the equation must be equivalent to theoriginal equation. For example, taking the complex conjugate of

we obtain

z +kz = 2pa (where pE JR, k= ~),

z+kz =2po.

Substituting k = ~, this becomes

kz+ z = 2pok = 2po,

which is the original equation. Similarly, for the equation

z - kz=2qia

In particular, it is necessary (but not sufficient) that the coefficients a and.8ofz and z in

have the same absolute value; i.e., lal = 1.81. It follows that equations suchas z +z = i or 2z - z= I are not equations of lines.

We are now ready to prove the Simson theorem.

THEOREM 2,5.1. Given b.ABG and a point D. let P. Q. R be the feet ofthe perpendiculars from the point D to (the extensions of) the sides BG.GA, AB. respectively. Then the points P. Q. R are collinear ifand onlyifD is on the circumcircle ofb.ABG.

Proof Without loss of generality, we may assume that b.ABG is in-scribed in the unit circle, and the points A, B, G, D are represented by


P(A) Cry)Q(/l)

A(a)

/7

D(o)FIGURE i!.1i!

the complex numbers cr, fJ, 'Y, 6, respectively. Then the equation of theline BC is

z z 1fJ fJ 1'Y 'Y 1

= O,

I.e.,(fJ - 'Y)z - (fJ - 'Y)z + (fJ'Y - fJ'Y) = O.

Using the relations fJ = Ii, 'Y = ~, this can be rewritten asz + fJ'Yz = fJ + 'Y.

Hence the equation of the perpendicular from D(6) to the side BC is

Therefore, the intersection P(A) of these two lines is obtained bysolving these two equations:

Applications to GEOmEtry

Similarly, Q(tL), R(II) are given by1 _

tL = 2b + 0 + 6 -1'06),

1 _II = 2(0 + (3 + 6 - 0(36).

Now,A-IIP(A), Q(tL), R(II) are collinear {=} E littL- 1I

H "hh' - ,.2owever, WIt t e notatIOn r = 161 (hence 6= "6)' we have

A - II = b - 0)(1 - (36)tL - II b - (3)(1 - 06)

(0-1') / (0 -6r-2 )= {3 - l' {3 - 6r-2= (0, (3; 1',6r-2).

Therefore,

P, Q, R are collinear {=} (0, (3; 1', 6r-2 ) E IR

{=} 0, {3, 1', 6r-2 are cocyclic

{=} 16r-21= 1{=} r = 161 = 1.

79

o

This line is usually called the Simson line of the point D with respectto 6ABC. However, historians have searched in vain for it through theworks of Robert Simson (1687-1768). It seems to have been publishedfirst by William Wallace (1768-1843) in 1797.

We now try to find the equation of the Simson line. We keep the samenotations as before; in particular, we assume 6ABC is inscribed in the

eo COMPLEX NUMBERS AND GEOMETRY

unit circle, and the point D(6) is on the unit circle. Then the foot P ofthe perpendicular from D(6) to the side BG is given by

Let us now introduce the notations

then

(11 =a + {J +1', (12 = {J1' +1'a + a{J, (13 = a{J1';

- 111(120\ =a + {J + "y = - + - + - = -,

a {J l' (13- 1 1(13 = afJ"Y = - = -.

a{J1' (13

Therefore, the above expression for z becomes

1 ( (13 )z = - (11 - a + 6 - -2 6a '

and

_ 1 (_ _ - (13)z = 2 (11 - a + 6 - 60

Eliminating a from these two relations, we get

_ 1(2 (13)6z - (13 Z = 2 6 + (1\6 - (12 - 6' .

This is a relation that must be satisfied by the foot P(>.) of the per-pendicular from D(6) to the side BG. However, since this relationcontains (11. (12, (13 only, and so is symmetric with respect to a, (J, 1'.It follows that this relation is also satisfied by the feet Q(/l) and R(1/)of the perpendiculars from D(6) to the sides GA and AB, respectively.

Applications to GEometry 81

However, this is an equation of a straight line, hence the feet P, Q, Rarecollinear, and the equation we obtained is the equation of the Simsonline. We have given an alternate proof to the if-part of Theorem 2.5.1.

THEOREM 2.5.2. Let L, M, N be three points on the circumcircle off:::.ABC. The necessary and sufficient condition that the Simson lines ofthe points L, M, N with respect to f:::.ABC meet at one point is

AL + BM + CN= 0 (mod 211").

Proof Let the circumcircle of f:::.ABC be the unit circle, and u(, U2, U3the complex numbers representing the points L, M, N, respectively.Then the equations of the three Simson lines under consideration are

_ 1(2 (73)u2z - (73Z = - u2 + (71 U2 - (72 - - ,2 U2

_ 1 ( 2 (73)U3Z - (73Z = - U3 + (71 U3 - (72 - - .2 U3

Hence the intersection of the first two Simson lines is given by

1 ( (73 )Z = -2 U\ + Uz + (7\ + -- ,UIUZ

and that of the last two Simson lines is given by

1 ( (73 )Z = - Uz + U3 + (7\ +-- .2 UZU3

Therefore, the necessary and sufficient condition for these two pointsto coincide is that (73 = U\UZU3; i.e., o.f3'Y = UIUZU3 Since a., f3, 'Y, u(,U2, U3 are all complex numbers with absolute value I, by setting theirarguments as 81. 8z, 83, '1'1. '1'2, '1'3, respectively, we obtain

81 + 8z + 83 = '1'1 + '1'2 + '1'3 (mod 211").


which is the desired condition. o

Note that if this condition is satisfied, then the intersection is givenby

By symmetry, we obtain the following

COROLLARY 2.5.3. Let A, B, C, 4 M, N be sixpoints on a circle. Then theSimson lines ofthe points 4 M, N with respect to 6ABC meet at a pointifand only if the Simson lines of A, B, C with respect to 6LMN meet atapoint. Moreover, in this case, all six Simson lines meet at the midpoint ofthe line segment joining the orthocenters of6ABC and 6LMN.

A

/'/'

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