Chapter 1 (John H. Mathews).pdf

8/17/2019 Chapter 1 (John H. Mathews).pdf

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#complex" numliers

OverviewGet r ead y f or a tr eat. You'r e a bout to begin stud ying some of the most beautiful

id eas in mathematics. They ar e ideas with surpr ises. They evolved over sever al

centuries, yet they gr eatly simplif y extr emely d if f icult com putations, making

some as easy as sliding a hot k nife thr ough butter . They also have a pplicationsin a var iety of ar eas, r anging f rom fluid flow, to electr ic cir cuits, to the myster ious

quantum wor ld . Gener ally, they belong t o the ar ea of mathematics k nown as

complex analysis, which is the sub ject of this book . This cha pter f ocuses on the

d evelo pment of entities we now call complex numbers.

Com plex analysis can r oughly be thought of as the su bject that a p plies the theor y

of calculus to imaginar y num ber s. But what exactly ar e imaginar y num ber s?

Usually, stud ents lear n a bout them in high school with intr oductor y remar k s

f r om their teacher s along the following lines: "We can't take the squar e r oot of

a negative number . But let's pr et end we can and begin by using the sym bol

i= v = I . " Rules ar e then learned f or doing ar ithmetic with these number s. Atsome level the rules mak e sense. If i= A ,it stand s to r eason that i2 = -l.However , it is not uncommon for stud ents to wond er whether they ar e r eally

doing magic r ather than mathematics.

If you ever f elt that way, congr atulate yourself! You'r e in the company of some of the great mathematicians f rom the sixteenth thr ough the nineteenth

centuries. They, too, wer e per plexed by the notion of r oots of negative num ber s.

Our purpose in this section is to highlight some of the e pisod es in the ver y

colorful histor y of how thinking a bout imaginar y num ber s d evelo ped . We intend

to show you that, contr ar y to po pular belief , ther e is r eally nothing imaginar y

a bout "imaginar y num ber s." They ar e just as r eal as "r eal number s."


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Our story begins in 1545. In that year , the Italian mathematician Girolamo

Cardano pu blished Ars Magna (The Great Art), a 40-cha pter master piece in

which he gave for the fir st time a method for solving the gener al cubic eq uation

Card ano d id not have at his d isposal the power of tod ay's algebraic notation,

and he tend ed to think of cubes or squar es as geometric o b jects rather than al-

ge braic q uantities. Essentially, however, his solution began with the substitution

z = x- a ; . This move tr ansformed Eq uation (1-1) into a cu bic eq uation withouta squared term, which is called a depressed cubic. To illustrate, begin with

z3 + 9 z2 + 24z + 20 = 0 and substitute z = x - a32 = X - * = x - 3. The equationthen becomes (x - 3)3 + 9 (x - 3)2 + 24 ( x - 3) + 20 =0, which sim plif ies to x3 ~ 3x + 2 =O .

You need not wor ry a bout the com putational details here, but in general the

substitution z = x - a32 transf orms Equation (1-1) into

h b- 1 2 d _ 1 2 3wer e - al - 3a2, an c - -3a1a2 + 27a2 + ao.If Card ano could get any value of x that solved a depr essed cu bic, he could

easily get a cor res ponding solution to Eq uation (1-1) from the id entity z = x - a3

2.

Ha p pily, Car d ano knew how to solve a depressed cubic. The techniq ue had been

communicated to him by Niccolo Fontana who, unf ortunately, came to be known

as Tar taglia (the stammerer) d ue to a speaking disord er . The procedure was also

ind e pendently d iscovered some 30 year s earlier by Scipione d el Ferro of Bologna.

Ferro and Tartaglia showed that one of the solutions to Equation (1-2) is

x = 3 _ ~ + J c2 + b 3 + 3 _ ~ _ J c2 + b 3 .2 4 272 4 27

Although Car d ano would not have reasoned in the following way, tod ay we

can tak e this value f or x and use it to f actor the depressed cubic into a linear

and quadr atic term. The remaining r oots can then be f ound with the quad r atic

formula. For exam ple, to solve z3 + 9z2 + 24z + 20 = 0, use the substitution z = x-3 to get x3 -3x+2 = 0, which is a d epressed cubic in the f orm of Equation(1-2). Next, apply the "Fer ro-Tartaglia" f ormula with b = -3 and c = 2 to get

« 2

V 22 ( _ 3)3

« 2

V 22 ( _ 3)3 R' R S· x = - - +- + -- + - - - - + -- = -1+ -1= -2 mce2 4 27 2 4 27 .

X =-2 is a r oot, x + 2 must be a f actor of x3 - 3x + 2. Divid ing x + 2 into x3 - 3x + 2 gives x2 - 2x + 1, which yields the r emaining (duplicate) roots of x = 1. The solutions to z3+9z2+24z+20 = 0 ar e o btained by r ecalling z = x-3 ,

which yield s the thr ee roots Zl = -2 - 3 = -5, and Z2 = Z3 = 1 - 3 = -2.So, by using Tartaglia's work and a clever transfor mation techniq ue, Cardano

was a ble to crack what had seemed to be the impossi ble task of solving the


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gener al cubic equation. Surprisingly, this develo pment played a signif icant r ole

in helping to esta blish the legitimacy of imaginar y number s. Roots of negative

number s, of course, had come up ear lier in the simplest of quadr atic equations

such as x2 + 1 = O. The solutions we know tod ay as x = ± j= I , however ,wer e easy f or mathematicians to ignor e. In Cardano's time, negative number s

wer e still being tr eated with some sus picion, as it was d if ficult to conceive of any

physical r eality corr es pond ing to them. Taking squar e roots of such quantities

was sur ely all the mor e ludicr ous. Never theless, Card ano mad e some genuine

attempts to d eal with j=I. Unf or tunately, his geometr ic thinking mad e ithard to mak e much head way. At one point he commented that the pr ocess of

arithmetic that d eals with q uantities such as j=I"involves mental tor tur es and is truly sophisticated ." At another point he condud ed that the process is "as

r efined as it is useless." Many mathematicians held this view, but f inally there

was a breakthr ough.

In his 1572 tr eatise L ' Algebr a , R af ael Bom belli showed that roots of negative

numbers have gr eat utility ind eed . Consid er the d e pr essed cu bic x3 -15 x-4 =

O.

Using For mula (1-3), we compute x =\/2 + J=TIT + {h- J -121 or , in asomewhat d iff er ent f orm. x = {h+ II j=I+ \/2 - II j=I.

Sim plif ying this ex pression would have been ver y diff icult if Bom belli had

not come up with what he called a "wild thought." He sus pected that if the

or iginal depr essed cubic had r eal solutions. then the two parts of x in the pr e-

ceding eq uation could be written as u + vj=I and u - v j= I f or some realnumber s u and v. That is, Bom belli believed u + vj= I = {h+ ll A and u - v A = \/2 -ll A, which would mean (u + vA)3 = 2 + ll j=I,

and (u - v j=I)3 = 2 - ll j=I.Then, using the well-known algebr aic id entity

(a + b) 3 = a3

+ 3a 2

b + 3ab2

+ b3

, and assuming that r oots of negative number sobey the r ules of alge br a, he obtained

(u + v";=l f = u3 + 3(u2)vv=1 + 3(u)(v H )2 + (vv=1)3

= u3 + 3(u) (vv=1)2 + 3(u2)vv=1 + (v H )3

= (u3 - 3uv2) + (3u2v - v3)v=1

=u(u2 - 3v2) + v(3u2 - v2)H

=2+11H.

( J - j)

(1-;) )

By eq uating lik e par ts of Eq uations (1-4) and 1-.)) Bombelli r easoned that

u(u2 - 3v2) = 2 and v(3u2 - v2) = ll. Perha ps think ing even mor e wild ly,

Bombelli then su pposed that u and v wer e integer s. ThE' only integer f actor s of

2 ar e 2 and 1, so the equation u( u2 - 3v2) = 2 led Bombelli to conclud e thatu = 2 and u2 - 3v2 = 1. Fr om this conclusion it follows that v2 = 1. or v = ±1.Amazingly, u = 2 and v = 1 solve the second equation v(3u2 - v2) = ll, soBombelli d eclar ed the values for u and v to be u = 2 and v = 1, res pectively.


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Since (2 + -/=1 :)3 = 2 + l1 A ,we clear ly have 2 + A = {h+ l1 A .Similarly, Bom belli showed that 2 - A= {h-11 A. But this means that

~2 +l1R+

, / 2-l1R=(2 +R)+ (2 - R)=4,

which was a pr overbial bombshell. Prior to Bombelli, mathematicians could

easily scoff at imaginar y number s when they ar ose as solutions to quadr atic

equations. With cubic equations, they no longer had this luxur y. That x =4 was

a corr ect solution to the equation x3 - 15x - 4 = 0 was indis putable, as it could be check ed easily. However , to arrive at this ver y real solution, mathematicians

had to tak e a d etour thr ough the uncharted ter r itor y of "imaginar y number s."

Thus, whatever else might have been said a bout these num ber s (which, tod ay,

we call com plex num bers), their utility could no longer be ignor ed .

Admittedly, Bombelli's technique a pplies only to a f ew s pecialized cases, and

lots of work r emained to be d one even if Bom belli's r esults could be extend ed.

Af ter all, today we r e pr esent r eal num ber s geometrically on the number line.

What possi ble r e pr esentation could com plex number s have? In 1673 John Wallis

mad e a sta b at a geometric pictur e of com plex number s that comes close to what

we use tod ay. He was inter ested in r e pr esenting solutions to gener al quadr atic

equations, which we can write as x2+2b x+c2 = a to mak e the f ollowing discussioneasier to follow. When we use the quadratic formula with this equation, we get

Wallis imagined these solutions as dis placements to the lef t and right fr om

the point -b. He saw each d is placement, whose value is J b2 - c2, as the length

of the sid es of the r ight triangles shown in Figur e 1.1.The points PI and P2 r e pr esent the solutions to our eq uation, which is clear ly

corr ect if b 2 - c2 2 ': O. But how should we pictur e PI and P2 when negative r ootsarise (i.e., when f )L . - c L .


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b 2 - C2 < O. He evid ently thought that, because b is shor ter than c, it could nolonger be the hy potenuse of the right triangle as it had been ear lier . The side of

length c would now have to tak e that role.

Wallis's method has the und esir able consequence that - Ais r e pr esented by the same point as is A. Never theless, this interpr etation hel ped set thestage f or thinking of complex num ber s as " points on the plane." By 1732, the

gr eat Swiss mathematician Leonard Euler ( pr onounced "oiler ") adopted this

view concerning the n solutions to the eq uation xn - 1 =O. You will learn

shortly that these solutions can be ex pr essed as cos e + Asin e f or var iousvalues of e ; Euler thought of them as being located at the vertices of a regular

polygon in the plane. Euler was also the fir st to use the symbol if or A.Tod ay, this notation is still the most popular , although some electrical engineers

pref er the symbol j instead so that they can use ito re pr esent cur rent.

Two additional mathematicians d eser ve mention. The Frenchman Augustin-

Louis Cauchy (1789-1857) formulated many of the classic theor ems that ar e now

par t of the corpus of com plex analysis. The Ger man Carl Fr iedr ich Gauss (1777-1855) r einfor ced the utility of com plex number s by using them in his several

proof s of the f und amental theorem of alge br a (see Cha pter 6). In an 1831 pa per ,

he pr oduced a clear geometr ic r e pr esentation of x + i y by id entif ying it with the point (x , y ) in the coor d inate plane. He also d escri bed how to perf orm ar ithmetic

oper ations with these new number s.

It would be a mistak e, however , to conclud e that in 1831 complex num-

ber s were tr ansf or med into legitimacy. In that same year the prolif ic logician

Augustus De Mor gan commented in his book , On the Study and Di fficult ies of

M athematics, "We have shown the sym bol v = a to be void of meaning, or r ather self -contrad ictor y and a bsurd. Nevertheless, by means of such symbols, a par t

of alge br a is esta blished which is of gr eat utility."

Ther e ar e, ind eed , genuine logical pr o blems associated with complex num-

ber s. For exam ple, with r eal numbers V O J j = y'liV b so long as both sides of the eq uation ar e d efined . Ap plying this id entity to com plex numbers leads to

1 = V I = vi (-1) (-1) = AA = -1. Plausible answer s to these pr oblemscan be given, however , and you will lear n how to r esolve this a pparent contr ad ic-

tion in Section 2.2. De Mor gan's r emark illustr ates that many f actor s ar e needed

to per suad e mathematicians to adopt new theories. In this case, as always, a


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f irm logical foundation was crucial, but so, too, was a willingness to modif y some

id eas concerning cer tain well-esta blished pr o perties of number s.

As time passed, mathematicians gr ad ually r efined their thinking, and by the

end of the nineteenth centur y complex number s wer e fir mly entr enched . Thus,

as it is with many new mathematical or scientific innovations, the .theor y of

complex number s evolved by way of a ver y intricate process. But what is the

theor y that Tar taglia, Fer ro, Car d ano, Bombelli, Wallis, Euler , Cauchy, Gauss,

and so many other s helped produce? That is, how do we now think of complex

number s? We ex plor e this q uestion in the r emaind er of this chapter .

2. Ex plain why cubic equations, r ather than q uadr atic equations, played a pivotal

r ole in helping to obtain the acce ptance of complex number s.

(a) 27x3 - 9 x - 2 = O. H int: Get an eq uivalent monic polynomial.

(b) x3-27 x+54=0.

4. Explain why Wallis's view of com plex number s results in - v = T being re pr esented by the same point as is v = T .

5. Use Bombelli's technique to get all solutions to the following d e pr essed cubics.

(a) x3 - 30 x - 36 = O .

( b) x3 - 8 7 x - 130 =O.

(c) x3-6 0x-32=0.

(a) Z 3 - 6 z2 - 3z + 18 = O .

( b) z3 + 3 z2

- 24 z + 28 =O.

7. Is it possible to modif y slightly Wallis's picture of com plex number s so that it is

consistent with the r e pr esentation used tod ay? To help you answer this q uestion,

ref er to the article by Alec Norton and Ben jamin Lotto, "Com plex Roots Mad e

Visi ble," The C ollege Mathemat ics Journal , 15(3), June 1984, p p. 248-249.

8. Investigate librar y or we b r esources and wr ite up a d etailed d escription ex plaining

why the solution to the d e pr essed cubic, Equation (1-3), is valid .


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We have shown that complex number s came to be viewed as ord ered pairs of

r eal number s. That is, a complex number Z is d efined to be

wher e x and yare both r eal number s.

The reason we say or d er ed pair is because we ar e think ing of a point in the

plane. The point (2, 3), for example, is not the same as (3, 2). The ord er in

which we write x and Y in Equation (1-7) mak es a dif f er ence. Clearly, then, two

complex number s ar e equal if and only if their x coor d inates ar e equal and their

Y coordinates ar e equal. In other word s,

(Throughout this text, iff means i f and only if .)

A meaningful number system r eq uires a method for combining order ed pair s.

The definition of alge br aic oper ations must be consistent so that the sum, d if -

ference, product, and quotient of any two order ed pair s will again be an order ed

pair . The key to defining how these number s should be manipulated is to f ol-

low Gauss's lead and equate (x, y ) with x + i y . Then, if Zl = (Xl, y d and Z2 =( X 2 , Y 2 ) ar e arbitr ar y complex number s, we have

Zl + Z2 = (Xl, y d + ( X 2 , Y 2 )

= (X l + i y d + (X 2 + iY 2 )

= (Xl + X 2 ) + i(Y I + Y 2 )

=(X l + X 2 , Y I + Y 2 ) .

Zl + Z2 = (Xl, Y l) + (X 2 , Y 2 )

= (Xl + X 2 , Y I + Y 2 ) . (1-8) I

Zl - Z2 = (Xl, yd - (X 2 , Y 2 )

= (X l - X 2 , Y I - Y 2 ) . (1-9) I


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Zl + Z2 =(3, 7) + (5, -6) =(8, 1) and

Zl - Z 2 = (3. 7) - (5. -6) = (-2. 13).

Zl + Z 2 = (3 + 7i) + (5 - 6i) = 8 + i and

Zl - Z 2 =(3 + 7i) - (5 - 6i) =-2 + 13i.

Given the rationale we devised f or ad dition and subtraction, it is tempting

to define the pr od uct Z lZ 2 as ZlZ2 = ( X 1X2 , Y1Y2). It turns out, however , thatthis is not a good d ef inition, and we ask you in the exercises for this section to

explain why. How, then, should pr oducts be d efined ? Again, if we eq uate (x , y )

with x + iy and assume, f or the moment, that i= Amak es sense (so thati2 = -1), we have

Z l Z 2 = (Xl, Y l)(X 2 , Y 2)

=(X l + i yd(X2 + iY2)

= X 1X2 + iX1Y2 + iX 2Yl + i2Y1Y2

= X1X 2 - Y1Y2 + i(X 1Y2 + X 2Y l)

=(X1X 2 - Y1Y 2, X1Y 2

+ x2yd ··

Z l Z 2 =(Xl , yd( X 2' Y2)

=(X1X2 - Y1Y2 , X1Y 2 + X2Yl). (1-10) I

Z l Z 2 =(3, 7)(5, -6)

= (3·5 - 7· (-6), 3· (-6) + 5 . 7)= (15 + 42. -18 + 35)= (57, 17).


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Z l Z 2 = (3, 7)(5, -6)

= (3 + 7i)(5 ~ 6i)

= 15 - 18i + 35i - 42i2

= 15 - 42( -1) + (-18 + 35)i

=57 + 17i

=(57, 17).

Of cour se, it mak es sense that the answer came out as we expected because

we used the notation x + iy as motivation f or our d efinition in the f ir st place.

To motivate our d ef inition f or division, we pr oceed along the same lines aswe did f or multiplication, assuming that Z 2 - I - 0:

Z l (X l , yd

Z2 (X 2 , Y 2)

( X l + i yd ( X 2 + iY 2) .

We need to figur e out a way to write the pr eceding quantity in the for m

X + i y. To d o so, we use a stand ard tr ick and multiply the numer ator and d enominator by X 2 - iY 2 , which gives

Z l (X l + iyd( X 2 - iY 2) Z 2 ( X 2 + iY2)( X 2 - iY 2)

X I X 2 + Y IY 2 + i( - X I Y 2 + x 2vd

X~ + v~ X IX 2 + V I V 2 . - X I Y2 + X 2Y l----- + t-----

X~ + y~ X~ + yi

= ( X IX 2 + V I Y2 - X I V2 + X2VI )X~ + y~ , X~ + y~ .

( x I , yd

(X2,Y 2)

( X IX 2 + VIV2 - X IV 2 + X2VI )

X~ + y~ , X~ + V~ .


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• EXAM PLE 1.3 If 21 =(3, 7) and 22 =(5, -6), then

21 =~= (15-4218+35) = (-27 53)22 (5, -6) 25 + 36' 25 + 36 61 '61 .

As with the example for multiplication, we also get this answer if we use thenotation x + i y:

21 (3,7)

22 (5, -6)

3 + 7i5 - 6i

3 + 7i 5 + 6i----5 - 6i 5 + 6i

15 + 18i + 35i + 42i2

25 + 30i - 30i - 36i2

15 - 42 + (18 + 35) i

25 + 3 6-27 53.

=51+ 61 t

= ( - 2 7 5 3 ) .61 '61

To per f or m oper ations on complex numbers, most mathematicians would use

the notation x + iy and engage in alge br aic manipulations, as we d id her e, r ather than a pply the complicated -looking d ef initions we gave for those oper ations on

order ed pair s. This procedure is valid because we used the x + i y notationas a guid e f or d efining the oper ations in the fir st place. R emem ber , though,

that the x + iy notation is nothing mor e than a convenient bookk ee ping d evicefor k ee ping tr ack of how to manipulate ord ered pair s. It is the or d ered pair

alge braic d efinitions that form the r eal found ation on which the com plex number

system is based. In f act, if you wer e to pr ogr am a computer to d o arithmetic on

complex numbers, your program would perf or m calculations on or d er ed pair s,

using exactly the def initions that we gave.

Our alge br aic d ef initions give complex num ber s all the properties we nor -

mally ascr i be to the r eal number system. Tak en together , they d escr i be what

algebr aists call a field . In f or mal terms, a f ield is a set (in this case, the com-

plex num ber s) together with two binar y o perations (in this case, addition and

multiplication) having the f ollowing properties.

(PI) Commutative law f or ad d ition: 21 + 22 =22 + 21.

(P2) Associative law for addition: 21 + (22 + 23) = (21 + 22) + 23·


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(P3) Ad ditive id entity: Ther e is a complex num ber w such that z+w = z f or all complex num ber s z. The number w is obviously the ord er ed pair (0, 0).

(P4) Ad ditive inverses: For any complex number z , ther e is a unique com plex

number 1] (d e pending on z) with the property that z+1] =(0, 0). Obviously,

if z =( x , y) = x + i y , the number 1] will be (-x , - y) =- x - iy =- z.

(P5) Commutative law f or multi plication: Zl Z2 = Z 2Z1'

(P6) Associative law for multiplication: Z l(Z2 Z3) =( ZlZ 2)Z 3.

(P7) Multi plicative identity: There is a complex number (such that z( = zfor all complex number s z. As you might ex pect, (1, 0) is the unique complex

number ( having this pr o perty. We ask you to verify this id entity in the

exercises f or this section.

(P8) Multiplicative inverses: For any complex number z = ( x , y ) other than the number (0, 0), ther e is a complex num ber (d e pending on z ) , whichwe denote Z-l, having the property that ZZ -l = (1, 0) = 1. Based on our def inition f or division, it seems r easonable that the number z-l would be

z-l = (1,0) = 1 = _1 _ = ~ = _ x_ + i-- -= .1L - = (_ x_ -- -= .1L -) We z z x+i y x2+ y2 xY+ y2 x2+ y2 x2+ y2 , x2+ y2 .

ask you to conf irm this r esult in the exercises f or this section.

None of these pr oper ties is dif f icult to prove. Most of the proof s mak e use

of corr es ponding facts in the real num ber system. To illustr ate, we give a pr oof

of pr o per ty (PI).

Proof of the commutative law for addition: Let Zl

Z2 =( X2 , Y2) be arbitr ar y com plex num ber s. Then,

( Xl , yd + (X 2 , Y2)( X l + X 2 , Y1 + Y2)(X 2+ X 1 , Y2+Y 1)

( X2, Y 2) + ( X l , yd Z 2 + zl·

( by definition of add ition of complex numbers)

( by the commutative law for real numbers)

( by definition of add ition of complex num ber s)

Actually, you can think of the real number system as a su bset of the com-

plex number system. To see why, let's agr ee that, as any com plex num ber of the

f or m ( t , 0) is on the x-axis, we can id entify it with the real num ber t. With this

cor res pond ence, we can easily verif y that our def initions f or addition, subtrac-

tion, multi plication, and d ivision of com plex numbers ar e consistent with the

corr es ponding o perations on r eal number s. For exam ple, if Xl and X2 ar e r eal

num ber s, then


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(X l , 0 )(X 2 , 0 )

(XIX 2 - 0, 0 + 0)(XIX2. 0 )

( by our agr eed cor r es pondence)

(by definition of multi plication of com plex num bers)

(confirming the consistence of our correspondence).

It is now time to show s pecif ically how the sym bol ir elates to the q uantity

A. Note that

(0, 1)(0, 1)

(0 - 1, 0 + 0) (by d efinition of multiplication of complex num bers)(-1. 0)

-1 ( by our agr eed corr espond ence).

which means i= (0, 1) = R.SO, the next time you ar e having a d iscussionwith your friends and they scof f when you claim that A is not imaginar y,

calmly put your pencil on the point (0. 1) of the coord inate plane and ask them

if ther e is anything imaginary a bout it. When they agree there isn't, you can tell

them that this point, in fact, r e presents the mysterious Rin the same waythat (1, 0) r e pr esents 1.

We can also see mor e clearly now how the notation X + iy equates to (x . y).Using the preced ing conventions (i.e., X = (x , 0), etc.), we have

(x , 0) + ( 0 . l)(y, 0)(x , 0) + (0, y )(x , y)

( by our previously discussed conventions)

( by d efinition of multiplication of complex numbers)

( by d ef inition of addition of complex numbers).

Thus, we may move f reely between the notations x + iy and (x , y) , d e- pending on which is mor e convenient f or the context in which we ar e working.

Stud ents sometimes wonder whether it matter s wher e the "i" is located in writ-

ing a complex number. It d oes not. Gener ally, most texts place terms containing

an "i" at the end of an ex pr ession, and place the "i" before a var ia ble but after

a constant. Thus. we wr ite x + iy. u + iv , etc .. but 3 + 7i. 5 - 6i. and so f or th.

Because letter s lower in the alpha bet gener ally d enote constants, you will usually( but not always) see the ex pr ession a + bi instead of a + ib. Many authors wr iteq uantities lik e 1 + iv'3 instead of 1+ v'3i to mak e sur e the "i" is not mistak enlythought to be insid e the sq uare r oot symbol. Additionally, if there is concern

that the "i" might be missed , it is sometimes placed bef or e a lengthy expression,

as in 2 cos (-~7 r + 2m ]") + i 2 sin (-~7 r + 2m]").We close this section with three important d ef initions and a theor em involv-

ing them. We ask you for a pr oof of the theorem in the exer cises.


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Definition 1.5: Real part

The real part of z, d enoted R e ( z). is the r eal number x.

I Definiti on 1.6: Imaginary part

The- imaginary part of z. denoted Im ( z) . is the r eal num ber y.

I Definiti on 1.7: ConjugateThe conjugate of z , denoted z , is the complex number (x , - y) = x - iy .

• EXAMPLE 1.4 a) Re (-3+ 7i) = -3 and R e[(9, 4)] = 9. b) Im( -3+7i) = 7 and Im[(9, 4)] = 4. c) -3 + 7i = -3 - 7i and (9, 4) = (9, -4) .

• Theorem 1.1 S uppose t hat z , Zl, and Z2 are ar bitrary com plex numbers.

Then

Re( z)=z+ z.2

z- zIm(z) =~.


14/47

Because of what it erroneously connotes, it is a shame that the ter m imagi-

nary is used in Definition (1.6). It was coined by the br illiant mathematician and

philosopher R ene Descar tes (1596-1650) d ur ing an era when quantities such as

Rwer e thought to be just that. Gauss, who was successful in getting math-ematicians to adopt the phr ase complex number r ather than imaginar y number ,

also suggested that they use lat eral par t of z in place of imaginar y part of z. Un-

f ortunately, that suggestion never caught on, and it a ppear s we ar e stuck with

what histor y has hand ed d own to us.

(a) i275.

(b) ~.

(c) R e(i).

(d ) Im(2).

(e) (i - 1)3.

(f ) (7 - 2i)(3i + 5).

(g) R e (7 + 6i) + 1m (5 - 4i) .

(h) 1m ( ;~~~) .

(i) (4-i)(1-3i)-1+2i

(a) (1 + i)(2 + i)(3 + i).

(b) (3+i)/(2+i).

(c) R e [(i - 1)3] .

(d ) Im[(l + i)-2].(e) ;~~: _ ~ - = - ~ i .(f ) (1 + i)-2

(g) R e [(x - iy)2] .

(h) 1m ( X ~ i Y ) .


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(i) Re [( x + iy ) (x - iy)].

( j) 1m [ (x + iy )3] .

3. Show that zz is always a real number .

4. Ver if y Id entities (1-12)-(1-19).

5. Let P (z) =anzn + an _ Izn-1 + ... + alZ + ao be a polynomial of d egr ee n.

(a) Suppose that an, an-I, ... , aI, ao ar e all r eal. Show that if Zl is a root

of P , then Zl is also a r oot. In other word s, the roots must be complex

conjugates, something you lik ely learned without proof in high school.

( b) Suppose not all of an, an-I, ... , aI, ao ar e r eal. Show that P has at

least one root whose complex con jugate is not a root. Hint: Pr ove the

contr a positive.

(c) Find an example of a polynomial that has some roots occur ring as

complex con jugates, and some not.

6. Let Zl = (Xl, Yl) and Z2 = (X 2 , Y2 ) be arbitr ar y com plex number s. Pr ove or d is pr ove the f ollowing.

(a) R e(Zl + Z 2 ) =Re(zl) + R e(z2).

( b) Re(zlz2) =R e (Zl) R e (Z 2).

(c) 1m (Zl + Z 2 ) =1m (Zl) + 1m ( Z 2 ) '

(d ) Im(zlz2) =1m (Zl) 1m ( Z 2).

7. Prove that the complex num ber (1, 0) (which we id entif y with the real number 1)

is the multi plicative id entity f or com plex num ber s.

8. Use mathematical induction to show that the binomial theor em is valid for com plex

num ber s. In other wor d s, show that if z and war e ar bitrar y complex number s and n

. 't" t th (+)n '\' (n) k n-k h (n) n!nISa POSI Ive In eger , en z w =6 k Z W , wer e k =k!(n-k)!'k= O

9. Let's use the sym bol * f or a new ty pe of multiplication of com plex num ber s d ef ined by Zl*Z 2 =(X I X 2 , Yl Y2 ). This exercise shows why this is an unfor tunate d ef inition.

(a) Use the d efinition given in pr o per ty (P7) and state what the multi-

plicative id entity ( would have to be for this new multiplication.

( b) Show that if you use this new multiplication, nonzer o com plex num ber s

of the for m (0, a) have no inver se. That is, show that if Z = (0, a),ther e is no complex num ber w with the pr o per ty that z * w =(, wher e( is the multi plicative id entity you f ound in par t (a).

10. Ex plain why the com plex num ber (0, 0) (which, you recall, we id entify with the

r eal number 0) has no multi plicative inver se.

11. Prove pr oper ty (P9), the distr i butive law for complex number s.


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12. Verif y that if z = (x, y ) , with x and y not both 0, then Z -l = (l , z 0) (i.e., Z -l = ~).

H int: Let z =(x, y ) and use the (ord er ed pair) d efinition f or division to compute

Z-l = ((1,0». Then, with the r esult you o btained , use the (or dered pair ) d efinition x , y

for multi plication to confir m that zz-l =(1, 0) =1.

13. Fr om Exer cise 12 and basic cancellation laws, it f ollows that z-l =~~~. Thenumerator here, Z, is trivial to calculate and , as the d enominator zz is a realnum ber (Exer cise 3), com puting the quotient ~ should be r ather straightfor war d.

Use this fact to com pute Z-l if z = 2 + 3i and again if z = 7 - 5i.

14. Show, by equating the r eal number s Xl and X2 with (Xl , 0) and (X2 , 0), r espec-

tively, that the complex definition f or d ivision is consistent with the real d efinition

f or d ivision. H int: Mimic the ar gument given in the text f or multi plication.

Complex numbers ar e ord er ed pair s of real num ber s, so they can be re pr esented

by points in the plane. In this section we show the ef f ect that alge br aic o per ations

on com plex number s have on their geometric repr esentations.

\lVecan re present the num ber z =x + iy =(x , y) by a position vector in the x y plane whose tail is at the or igin and whose head is at the point ( x , y). When

the xy plane is used for dis playing complex num ber s, it is called the complex

plane, or mor e sim ply, the z plane. Recall that Re(z) = x and Im(z) = y.Geometr ically, Re(z) is the projection of z =(x , y) onto the x-axis, and Im(z)

is the pr o jection of z onto the y-axis. It mak es sense, then, to call the x-axis the

real axis and the y-axis the imaginar y axis, as Figur e 1.3 illustrates.

Add ition of complex numbers is analogous to add ition of vector s in the plane. As we saw in Section 1.2, the sum of Zl = Xl + iY l = (Xl, Y l) and Z2 = X 2 + iY 2 = ( X 2 , Y 2) is (Xl + X 2 , Y l + Y2) . Hence Zl + Z2 can be obtained vectorially by using the " par allelogr am law," wher e the vector sum is the vector

r e pr esented by the d iagonal of the par allelogram f or med by the two original

vector s. Figure 1.4 illustr ates this notion.

The diff er ence Zl - Z2 can be r e pr esented by the d is placement vector f r om

the point Z2 = (X 2 , Y 2 ) to the point Zl = (Xl, Y l), as Figur e 1.5 shows.

C Co py of vector 21

Ci y(positioned at the tail of vector 22)

Z ••• .,..~.ZI+Z22. . • •

" Copy of vector 22

. 'z~ (positioned at the tail of vector 2t)

Imaginary axisY

n ; , ? , f '• I I I I I I

X


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y ~ Co py of vector ZI- Z2

~

~..l..~( positioned at the tail of vector z : J ,. ZI

X----~ r ~ Co py of vector - Z2

Z • - z7. 1(positioned at the tail of vector Zl)1 - 2

4 + 3i

/LFigur e 1.6 The r eal and imaginar y par ts of a complex number .

Definition 1.8: Modulus

The modulus, or absolute value, of the com plex num ber z = x + iy is anonnegative r eal number denoted by Izl and d efined by the r elation

The number Izi is the d istance between the origin and the point z = (x , y) .The only complex number with modulus zero is the number o . The number z =4 + 3i has mod ulus 14+ 3il =J42 + 33 =v '25 =5 and is de picted in Figur e1.6. The number s IR e(z)l, IIm(z)l, and Izi ar e the lengths of the sid es of the right

triangle OPQ shown in Figur e 1.7. The inequality IZ11< IZ21means that the point Z1 is closer to the origin than the point Z2. Although o bvious from Figur e

1.7, it is still prof ita ble to work out alge braically the stand ard r esults that

which we leave as an exercise.

The differ ence Z1 - Z2 r e presents the dis placement vector from Z2 to Z1, so

the distance between Z1 and Z2 is given by 1Z1 - z21. We can obtain this distance

by using Definitions (1.2) and (1.3) to obtain the familiar f or mula

If z =(x, y) = x + iy , then - z =(- x , - y) =- x - iy is the r ef lection of z

through the or igin, and z = (x, - y) = x - iy is the r eflection of z thr ough thex-axis, as illustr ated in Figur e 1.8.

We can use an im por tant alge br aic r elationship to establish pr o per ties of the

absolute value that have geometric a pplications. Its proof is rather str aightfor -

war d , and we ask you to give it in the exer cises for this section.


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z = (x, y)=x+ i y

~--+-----+-----


19/47

• EXAMPL E 1.5 To pr oduce an example of which Figur e 1.9 is a r eason-

a ble illustr ation, we let Z l = 7 + i and Z 2 = 3 + 5i. Then I Z l l = )49 + 1 =v'5Oand I Z 2 1 = )9 + 25 = V34. Clear ly. Zl + Z 2 = 10+ 6i: hence I Z l + z 2 1 =)100 + 36 =J136. In this case. we can verify the triangle ineq uality withoutr ecour se to computation of sq uar e roots because I Z l + z 2 1 )136 = 2V34 =V34 + V34


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(a) 1(1+i)(2+i)l.

(b) I ~ - = - ~ i l .(c) 1(1+i)

50I.

(d) Izzl, wher e z = x + iy.

(e) Iz-112,wher ez=x+iy.

(a) Zl = 2 + 3i and Z2 = 4 + i.

( b) Zl = -1 + 2i and Z2 = -2 + 3i.

(c) Zl =1 + iV 3 and Z2 =-1 + iV 3 .

(a) ~+i.

( b) V2+i(V2+1).

(c) 2+3i.

(d) -./ + iV 3.

(a) (1-21)-

( b) (1-22).

(c) (1-26).


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(a) Iz + 1 - 2il =2.

(b) Re(z+ I )=O.

(c) Iz+2il::':;1.

(d ) 1m (z - 2i) > 6.

7. Prove that v'2lzl 2 : : IRe(z)1 + IIm(z)l·

8. Show that the point z! ~Z2 is the midpoint of the line segment joining Zl to Z2.

9. Show that IZI - z21 ::.:;IZll + IZ21.

10. Prove that Izi =0 if f z =O.

11. Show that if z # 0, the four points z , 2, - z , and -2 are the ver tices of a r ectanglewith its center at the origin.

12. Show that if z # 0, the f our points z, i z , - z , and -iz ar e the vertices of a squarewith its center at the origin.

13. Show that the equation of the line thr ough the points Zl and Z2 can be ex pressed

in the form z = Zl + t (Z 2 - Zl), where t is a r eal number.

14. Show that the nonzer o vector s Zl and Z2 are par allel iff Im(zlz2) = O.

15. Show that IZIZ2Z31=IZlllz21lz31.

16. Show that Iznl =Izln

, wher e n is an integer .

17. Suppose that either Izi =1 or Iwl =1. Prove that Iz - wi =11- 2wl·

18. Pr ove the Cauchy-Schwar z inequality: I k ~l ZkU!"k I : : . : ;

19. Show Ilzll-lz211 ::.:;IZl - z21.

20. Show that ZlZ2 + ZlZ2 is a r eal number.

21. If you stud y carefully the pr oof of the triangle inequality, you will note that the

r easons for the inequality hinge onRe(zlz2) ::.:;IZIZ21. Und er what conditions will

these two quantities be equal, thus turning the triangle inequality into an equality?

22. Prove that IZl - z212=IZll2 - 2 Re(zlz2) + IZ212.

23. Use induction to pr ove that I t Zk I : : . : ; t IZk I for all natur al number s n.k =l k =l

24. Let Zl and Z2 be two distinct points in the com plex plane, and let K be a positive

r eal constant that is less than the d istance between Zl and Z2.

(a) Show that the set of points { z: Iz - zll- Iz - z21 = K} is a hy perbolawith f oci Zl and Z2·

( b) Find the equation of the hy perbola with foci ±2 that goes thr ough the

point 2 + 3i.

(c) Find the equation of the hy per bola with f oci ±25 that goes thr ough

the point 7 + 24i.


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25. Let Zl and Z2 be two distinct points in the complex plane, and let K be a positive

r eal constant that is gr eater than the d istance between Zl and Z 2.

(a) Show that the set of points { z : [z - z l l + I z - z 2 1 =K} is an ellipsewith foci Zl and Z2.

( b) Find the equation of the ellipse with foci ±3i that goes through .the point 8 - 3i.

(c) Find the eq uation of the elli pse with f oci ±2i that goes through the

point 3 + 2i.

1.4 THE GEOMETRY OF COMPLEX NUMBERS,

CONTINUED

In Section 1.3 we saw that a complex num ber z = x + i y could be viewed as avector in the xy plane with its tail at the origin and its head at the point (x, y).

A vector can be uniquely s pecif ied by giving its magnitud e (i.e., its length) and

dir ection (i.e., the angle it mak es with the positive x-axis). In this section, we

f ocus on these two geometr ic aspects of com plex numbers.

Let r be the mod ulus of z (i.e., r =Iz l) , and let e be the angle that the linefrom the or igin to the complex number z mak es with the positive x-axis. (Note:

The number e is und efined if z =0.) Then, as Figur e 1.11(a) shows,

Definition 1. 9: Polar representation

Id entity (1-TI) is known as a polar representation of z , and the values T and

e ar e called polar coordinates of z.

z = ( x ,y) = x+ i y

- -'1=(r cos e , r sin e ) = r (cos e + isin e )1

1

e I(x, 0)

z = (x, y) = x + iy

(0, y) .... : =(r cos e, r sin e) = r (cos e+ isin e)

r :

e

(x, 0)


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• EXAMPLE 1.7 If z = 1 + i, then r = .;2 and z = (.;2cos % ' .;2 sin % ) =.;2 (cos % + i sin % ) is a polar re pr esentation of z. The polar coordinates in thiscase ar e r =.;2, and B =% .

As Figur e 1.11 ( b) shows, B can be any value f or which the id entities cos B =~

and sin B = ~ hold. For z = 1 = 0, the collection of all values of B f or whichz =r(cosB+isinB) is d enoted ar g z. For mally, we have the following d efinitions.

Definition 1.10: arg z

If z = 1 = 0,

Note that we wr ite B E ar g z as o pposed to 8 =ar g z. We do so because

ar g z is a set, and the d esignation e E ar g z indicates that 8 belongs to that set. Note also that, if 81 E ar g z and 82 E arg z, then ther e exists some integer n

such that

{7r } { 77r 7r 97r 177r }

ar g (1 + i) = ~ + 2n7 r : n is an inteO"er = ... -- - - -- ... .4 b' 4'4'4'4'

Mathematicians have agr eed to single out a s pecial choice of 8 E ar g z. It is

that value of e f or which -7r


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yar g z C ar ctan -,

x

wher e ar ctan; = { B : tan B =;} . Note that, as with ar g, ar ctan is a set (as

o pposed to Ar c tan. which is a num ber ). We s pecif ically id entif y ar g z as a

pr o pe r su bse t of ar ctan ~ because tan B has per iod n, wher eas cos B and sin B

have per iod 2n. In selecting the proper values f or arg z, we must be car ef ul in

s pecif ying the choices of ar ctan ~ so that the point z associated with rand B lies

in the ap propr iate q uadr ant. _

• EX AM PLE 1.10 If z = -J3 - i = T (c os B + isinB), then r = [ z l =1-J3-il = 2 and BE ar ctan; = ar ctan _ -~ = {~+nn:nisaninteger }.

It would be a mistak e to use ~ as an acce pta ble value for B , as the point z asso-

ciated with r = 2 and B = ~ is in the f ir st quadrant, wher eas -J3 - i is in thethird quadr ant. A cor rect choice f or B is B =~- n =-g " . Thus.

-5n -5n- J 3 - i = 2cos -- + i2sin--

6 6

(-5n ) (-5n )= 2 cas -6- + 2mr + i2 sin -6- + 2nn .

Ar g (- J 3 - i)

ar g (-v3 - i)

6

{ -5n }

= -6- + 2nn : n is an integer .

Note that ar g (-J3 - i) is indeed a proper subset of ar ctan ~~~ .

• EXA M PLE 1.11 If z = x + iy = 0 + 4i , it would be a mistak e to attem pt tofind Ar g z by look ing at ar ctan ~, as x = 0, so ~ is und ef ined . If z i = 0 is on they-axis, then

Ii

Ar g z = " 2 if Imz > O. and n

Ar g z =-"2 if Imz < O .


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As you will see in Chapter 2, Ar g z is a discontinuous f unction of z because

it " jumps" by an amount of 27 T as z crosses the negative r eal axis.

In Chapter 5 we d efine e Z f or any com plex num ber z. You will see that this

com plex ex ponential has all the pr o per ties of r eal ex ponentials that you studied

in ear lier mathematics cour ses. That is. e Z l e Z 2 =e Zl +Z 2. and so on. You will

also see, amazingly, that if z = x + iy , then

We will esta blish this r esult r igorously in Cha pter 5, but ther e is a plausible

ex planation we can give now. If eZ has the normal pr operties of an exponential, it

must be that e x

+iy

= eXeiy

. Now, r ecall f r om calculus the values of thr ee infinite00 00 ( )n 00 ( )n

series: e X = L trXk C OS X = L (~~)! x2n. and sinx = L (2~~l)lx2n+l. Su b-k =O n= O n=O

00 00

stituting i y f or x in the infinite series f or eX gives eiy = L t r (iy l= 2 : = t r ik yk k =O k =O

At this point. our ar gument loses r igor because we haye not talk ed a bout infinite

ser ies of complex number s. let alone whether such series conver ge. Never theless.

if we mer ely tak e the last series as a f ormal expr ession and s plit it into two ser ies

accor ding to whether the ind ex k is even (k =2n) or od d (k =2 n + 1), we get

co 1 co 1=L __ i2n y2n +L i2n+1 y2n+ l

n=O (2n)! n=O (2 n + I)!

= f _ 1 _ (i2t y2n + f 1 (i2ti y2n +ln= O (2n)' n=O (2n + 1)1

=f _l _(-lty2n+if 1 (_lt y2n+1n=O (2 n)! n=O (2 n + I)!

=cos y + isin y.

Thus, it seems the only possible value for eZ is that given by Eq uation (1-:11).

We will use this r esult f r eely fr om now on and. as stated . supply a r igor ous pr oof

in Cha pter 5.

If we set x =0 and let e tak e the r ole of y in Eq uation (1~31 . we get af amous r esult known as Euler's formula:


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y e'~= (O,I)=i

ei"=(-1, 0) =-1 t/lei~= G , '1 ) =1+ '1i

\+;eiOr l

=ea

"=(l,O)=1~ - x

I r Li~=e-if =(.f i _ .f i)=.f i- 'ff iThe unit cir cle t 2' 2 2 2

If (j is a real number , eie will be located somewhere on the cir cle with r adius

1 centered at the origin. This assertion is easy to verif y because

Figure 1.12 illustrates the location of the points eie for various values of (j.

Note that, when ( j =1T , we get ei7r =(COS 1T, sinn) =(-1, 0) =-1, so

Euler was the f ir st to discover this r elationshi p; it is r ef err ed to as Euler 's

identity. It has been la beled by many mathematicians as the most amazing

relation in analysis-and with good r eason. Sym bols with a r ich histor y ar e

mir aculously woven together -the constant n used by Hippocr ates as ear ly as

400 B.C.; e, the base of the natur al logarithms; the basic conce pts of ad dition (+)

and equality (=); the f ound ational whole numbers 0 and 1; and i, the number

that is the central focus of this book .

Euler 's f or mula (1-32) is of tr emend ous use in esta blishing im por tant alge-

braic and geometric properties of complex num ber s. You will see shor tly that it

enables you to multiply complex number s with gr eat ease. It also allows you to

expr ess a polar form of the com plex number z in a mor e compact way. R ecall

that if r = Iz i and (j E arg z , then z = r( cos (j + isin (j). Using Euler 's f ormulawe can now wr ite z in its exponential for m:

• EXAMPLE 1.12 With ref er ence to Exam ple 1.10, with zhave z = 2ei( -57r /6) .


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Together with the r ules f or ex ponentiation that we will verif y in Cha pter 5,

Equation (1-35) has inter esting a p plications. If Zl = r 1 ei&, and Z2 = r2ei&2, then

ZlZ2 = r1 ei&l r2ei&2 = r1 r2ei(&, +&2)

= r 1r2 [cos(81 + 82) +isin (81 + 82)] .

Figur e 1.13 illustr ates the geometr ic signif icance of this equation.

We have alread y shown that the modulus of the prod uct is the product of the

moduli; that is, I Z 1 z 2 1 =I Z 1 1 1 z 2 1 . Id entity (1-36) esta blishes that an argument of ZlZ2 is an argument of Zl plus a n ar gument of Z2. It also answer s the question

posed at the end of Section 1.3 r egar ding why the pr oduct ZlZ2 was in a d if f erent

q uad r ant than either Zl or Z2. It further of fers an inter esting explanation as

to why the pr oduct of two negative r eal numbers is a positive r eal number .

The negative number s, each of which has an angular displacement of Jr r adians,

combine to produce a pr od uct that is r otated to a point with an ar gument of

Jr + Jr = 2Jr r adians, coincid ing with the positive real axis.Using ex ponential for m, if Z # 0, we can write ar g z a bit mor e compactly as

Doing so ena bles us to see a nice r elationshi p between the sets ar g (ZlZ2) , ar g Zl,

and ar g Z2.

Befor e pr oceeding with the proof , we r ecall two important facts a bout sets.

Fir st, to esta blish the equality of two sets, we must show that each is a su bset of

the other . Second , the sum of two sets is the sum of all combinations of elements


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fr om the fir st and second sets, r es pectively. In this case, ar g Zl + arg Z 2

{81 + 82 : 81 E arg Zl and 82 E ar g Z 2} .

Proof Let 8 E ar g ( Z l Z 2). Because Z l Z 2 =r1 r2ei(O, +02), it f ollows fr omFor mula (1-37) that 81+ 82 E ar g (Z l Z 2). By Equation (1-29) ther e is someinteger n such that 8 =81 + 82 + 2mr. Fur ther , as Z l =r1 e

iO" 81 E arg Z l·

Lik ewise, Z2 =r2ei02 gives 82 E arg Z2. But if 82 E arg Z 2 , then 82 + 2mr E

ar g Z 2. This r esult shows that 8 = 81+ (82 + 2mr ) E ar g Z l + arg Z 2. Thus,arg(zlz2) ~ arg Z l +ar g Z 2. The proof that ar g Z l +ar g Z 2 ~ arg(zlz2) is lef t

as an exer cise.

1

1 10

Z- = - [cos(-8) +isin(-8)] = _e-t .r r

z = r (cos8 - isin8) = r [cos (-8) + isin (-8)] = r e-iO , and

Z l = r 1 [cos (81

_ 82) + i sin (81 _ 82)] = r1 ei(O,-02). Z 2 r 2 r 2

If z is in the fir st quadrant. the positions of the number s z, z , and Z-l ar eas shown in Figure 1.14 when Izi < 1. Figur e 1.15 d e picts the situation when

I z l > 1.

y

t = (0,1) y

t = (0,1) z

f The unit circle f The unit circle

Figur e 1.14 R elative positions of z , Z,

and z-l when I z [ < 1.Figur e 1.15 R elative positions of z , Z,

and Z-l when I z l > 1.


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• EX AM PL E 1.13 If z = 1 + i, then r = Izi = V 2 and e = Ar gz = % .Ther ef or e, z-l = ~ [cos (~4"') + isin (7)] = ~ [...;;- i...;;]and has modulus

1 _ v'2v'2 - 2'

• EX AM PLE 1.14 If Zl = 8i and Z2 = 1 + iV3, then r e presentative polar f orms f or these num bers ar e Zl = 8 (cas ~ + isin ~) and Z2 = 2 (cas ~ + isin ~) .Hence

Zl 8 [ ( 7 r 7 r ) . . ( 7 r 7 r ) ] ( 7 r . . 7 r )Z2 = 2 cas 2 " - 3 " + ~sm 2 " - 3 " = 4 cas " 6 + Z S111 " 6

=2)3 + 2i.

(a) 1- i.

(b) -/3+ i.

(c) (-1-i/3)2

(d) (1 - i)3 .

(e) 1+~v'3'

(f ) i2 1.

(g)

(a) (v'3- i) (1+ i/3) = 2V3+ 2i.

( b) (1+i)3=-2+2i.

(c) 2i(v'3+i) (l+i/3) = -8.(d ) 1~i =4-4i.

(a) -4.

( b) 6-6i.

(c) -7i.


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(d) -2V3-2i.

(e) (1~t)2'

(f ) i+6 .. /3 .

(g) 3 + 4i.

(h) (5 + 5i)3 .

4. Show that ar g Zl + ar g Z2


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1.5 THE ALGEBRA OF COMPLEX NUMBERS,

REVISITED

The real number s ar e d ef icient in the sense that not all algebr aic oper ations on

them pr oduce real num bers. Thus, f or V = T to mak e sense, we must consid er thedomain of com plex num ber s. Do complex number s have this same d eficiency?

That is, if we ar e to mak e sense of ex pr essions such as ~. must we a ppeal

to yet another new number system? The answer to this question is no. In other

word s, any reasonable algebraic operation perfor med on complex numbers gives

com plex num ber s. Later we show how to evaluate intriguing expressions such as

ii. For now we only look at integr al powers and r oots of com plex numbers.

The impor tant player s in this r egard ar e the ex ponential and polar for ms of

a nonzero complex number z = r eie = r( cos B + i sin B ). By the laws of ex ponents(which, you r ecall, we have pr omised to pr ove in Cha pter 5) we have

zn = (reiet = r neine = r n [cos(nB) +isin(nB)], and

z-n = (reie) -n = r -ne-ine = r -n [cos (-nB) + i sin (-nB)] .

• EXA MPLE 1.15 Show that (-V3 - i)3 =-8i in two ways.

Solution (Method 1): The binomial f or mula (Exer cise 14 of Section 1.2) gives

( ; ; : ; ) 3 ( .( _ 5 r .) )3 ( 3 '(-15r.)) ( - 1 5 1 1 - 1 5 1 1 )-v3 - i = 2et -6- = 2 et -6- = 8 cos-6-+ isin -6-= -8i.


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• EXAMPLE 1.16 Evaluate (-V3 - i)30.

Solution (-V3 - i)30 = (2ei(-r ))30 = 2 30e-i251 r = _ 230.

An interesting a p plication of the laws of ex ponents comes fr om putting the

eq uation (eie)n = eine in its polar form. Doing so gives

which is k nown as De Moivre's f or mula, in honor of the French mathematician

A br aham De Moivre (1667-1754).

• EXAMPLE 1.17 Use De Moivr e's f ormula (Equation (1-40)) to show that

cos 5B = cos58 - 10 cos3 8 sin2 8 + 5 cos 8 sin4 8.

Solution If we let n= 5 and use the binomial f or mula to ex pand the left sid eof Eq uation (1--10), we obtain

The r eal part of this ex pr ession is cos5 8 - 10 cos3 e sin 2 e +5 cos e sin 4 e . Equatingthis to the r eal par t of cos 5B + isin 58 on the right sid e of Eq uation (1--10)esta blishes the d esir ed r esult.

A k ey aid in d etermining r oots of complex num ber s is a corollar y to the

f und amental theor em of alge bra. We prove this theor em in Cha pter 6. Our

pr oof s must be ind e pend ent of the conclusions we d er ive here because we are

going to mak e use of the cor ollar y now .

• Theorem 1.4 (Corollary to the fundamental theorem of algebra)

I f P ( z) is a pol ynomial o f d egr ee n (n> 0) with comple x coe ff icient s , then

the equat ion P (z) = 0 has pr ecisel y n (not necessaril y d ist inct ) solut ions.

Proof R ef er to Chapter 6.

• EXAMPLE 1.18 Let P ( z) = z3 + (2 - 2i) z2 + (-1- 4i) z - 2. This polyno-mial of d egr ee 3 can be written as P ( z) =( z - i)2 ( z + 2). Hence the equationP ( z) = 0 has solutions Z l = i, Z 2 = i, and Z 3 = -2. Thus, in accord ance withTheor em 1.4, we have thr ee solutions, with Z l and Z 2 being repeated roots.


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Theor em 1.4 implies that if we can find n d ist inct solutions to the eq uation

zn = c (or zn - C = 0), we will have f ound all the solutions. We begin our

sear ch for these solutions by look ing at the simpler eq uation zn = 1. Solving

this eq uation will ena ble us to hand le the mor e gener al one q uite easily.

To solve zn = 1 we f ir st note that, fr om Id entities (1-29) and (1-37), wecan deduce an im por tant condition that d eter mines when two nonzer o com plex

number s ar e eq ual. If we let Zl = Tl eiB, and Z2 = T2eiB2, then

wher e k is an integer . That is, two nonzer o complex number s ar e equal iff their

mod uli agr ee and an ar gument of one equals an ar gument of the other to within

an integr al multi ple of 21 f .

We now find all solutions to zn = 1 in two stages, with each stage corr e-

s ponding to one d ir ection in the iff par t of R elation (1-41)' Fir st, we show that

if we have a solution to zn = 1, then the solution must have a cer tain f or m.Second , we show that any q uantity with that form is ind eed a solution.

For the f ir st stage, suppose that z = TeiB is a solution to zn = 1. Putting

the latter equation in exponential form gives TneinB = 1· ei'O, so R elation (1-,n)

implies that Tn = 1 and n e = ° + 21f k. In other word s,e =27 r k

,n

wher e k is an integer .

So, i f z = TeiO is a solution to zn = 1, then R elation (1-42) must be true.

This o bser vation completes the f ir st stage of our solution str ategy. For the second 2 k '0' 2~k

stage, we note that i f T = 1, and e = ~, then z = Te' = e'--n is ind eed asolution to zn = 1 because zn = (ei 2';,k) n = ei2r r k = 1. For exam ple, if n = 7

and k = 3, then z = e i¥ is a solution to Z7 = 1 because ( e i¥ ) 7 = ei6r r = 1.Fur thermor e, it is easy to verif y that we get n distinct solutions to zn = 1

(and , ther ef or e, all solutions, by Theor em 1.4) by setting k = 0, 1, 2, , n - 1.

The solutions for k = n, n + 1, ... merely r e peat those for k = 0, 1, , because

the arguments so generated agr ee to within an integr al multiple of 21f. As we

stated in Section 1.1, the n solutions can be ex pr essed as

i2k~ 21f k. . 21f kZk=e n =cos--+zsm--, for k=0,1,2, ... ,n-1.

n n

They ar e called the nth r oots of unity.·2rr " O 0

When k = ° in Equation (1-43), we get Zo = et -n = e = 1, which isa r ather trivial r esult. The f ir st inter esting r oot of unity occur s when k = 1,

giving Zl = ei~. This par ticular value shows up so of ten that mathematicians

have given it a special symbol.


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Definition 1.12: Primitive nth root

For any natur al number n , the value W n given by

.h 2Jr 2Jr W n =e' n =cos - + i sin -

n n

By De Moivr e's f ormula (Eq uation (1--10)), the nth roots of unity can be

ex pr essed as

Geometrically, the nth roots of unity ar e equally s paced points that lie on

the unit circle C1(0) ={ z : Izi =I} and for m the vertices of a r egular polygon

with n sid es .

• E X A M PL E 1.19 The solutions to the equation z8 = 1 ar e given by the eight1 - i2" k - 2 1T k .. 2 1T k r k - 0 1 2 7 Ie· . r vaueszk -e 8 -coss+zsms,lor - , , , ... , . n alteslanlorm,

these solutions ar e ±1, ±i. ± V2tV2) and ± V22 iV2. The primitive 8th r oot of

unity is W8 - ei¥ - - ei~ - cos2': +isin2': - V2 +iV2- - - 4 4- 2 2'

From Ex pressions (1-11) it is clear that W 8 = Z l of Eq uation (I-J3). Figur e

1.18 illustr ates this r esult.

The pr ocedur e f or solving zn = 1 is easy to gener alize in solving zn = c f or any nonzer o com plex number c. If c = pei¢ = p (cas ¢ + isin ¢) and z = r eie ,then zn =C iff r neine = pei¢. But this last eq uation is satisf ied if f

r n = p , and

nO =¢+ 2k J r . wher e k is an integer.

--n + i-n -OJ32 8

2 - i-n _ ,.,72 -~8

~-i=OJ~


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As bef ore, we get n distinct solutions given by

l i4>+2nk l ( ¢+27r k .. ¢+ 27rk ) Z k = pne n = pn cos---+2sm --- ,

n n

f or k = 0, 1, 2, ... ,n - 1.

Each solution in Equation - - can be consider ed an nth r oot of c. Ge-

ometrically, the nth r oots of c are eq ually s paced points that lie on the circle

Cp'k (0) ={ z : Iz i = p - ! . } and form the ver tices of a r egular polygon with n sid es.Figur e 1.19 illustrates the case for n = 5.

It is inter esting to note that if ( is any par ticular solution to the equation

zn = c, then all solutions can be gener ated by multiplying ( by the var ious nthroots of unity. That is. the solution set is

The r eason f or this is that if (n =c, then f or any j =0, 1, 2, ... , n - 1,

((w~r = (n (w~r = (n (w~)j = (n(l) = c, and that multiplying a number

by W n = ei2; incr eases an ar gument of that num ber by 2 : , so that Ex pressionsl--Hi) contain n d ist inct values .

• E X A M PL E 1.20 Find all cube r oots of 8i =8 (cos ~ + i sin ~).

Solution Formula (1 ±. j I gives

(~ + 2 7 r k ~ + 2 7 r k )

Z k =2 cas ~ 3 +isin 2 3 ' f or k =O. 1. 2.

The Cartesian f orms of the solutions ar e Zo = V3+i. Zl = -v'3+i, and Z2 = -2i.as shown in Figur e 1.20.


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Is the q uadratic formula valid in the complex domain? The answer is yes ,

pr ovid ed we ar e caref ul with our ter ms.

t Theorem 1.5 (Quadratic formula) I f a z2 + b z + c = 0, t hen t he solut ion

set for z is { -b+(b::4ac)! } , wher e by (b2 - 4ac)! we m ean all d ist inct

squar e r oot s o f the number insid e the par ent heses.

Proof The proof is left as an exer cise .

• E X A M P L E 1.21 Find all solutions to the eq uation z2 + (1 + i) z + 5i =O.

Solution The quadr atic f ormula gives

1

-(l+i)+ [(1+i)2-4(1)(5i)r -(1+i)+(-18i)!z - ------------- - --------- 2(1) - 2

. '( _ 1 L ). . . 1 . 1 (-~ + 2 k 1 r )As -18~ = 18 et 2, Eq uatlOns (1-45)gIve (-18~)2 = 182et 2 , f or k = 0and 1. In Cartesian form, this ex pr ession r educes to 3 - 3i and -3 + 3i. Thus,our solution set is {-(Hi)+(3-3i) -(Hi)+(-3+3i)} or {I - 2i -2 + i}

2 ' 2 ' , .


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In Exer cise 5 b of Section 1.2 we ask ed you to show that a polynomial with

nonr eal coefficients must have some roots that do not occur in complex conjugate

pair s. This last example gives an illustr ation of such a phenomenon.

(a) (1 - iv l3 )3 (v13 + i)2 .

(b) (l+i):(l-i) .

(a) by sq uar ing twice.

( b) by using De Moivr e's f ormula, given in Eq uation (1-40).

3. Use the method of Example 1.17 to esta blish tr igonometric identities for cos 3B

and sin3t1.

4. Let z be any nonzero complex number and let n be an integer. Show that zn + ( z)nis a r eal number .

1

(a) (-2+2i)3.

1(b) (-1)5.

1

(c) (-64)4.

1

(d) (8)6.

6. Prove Theorem 1.5, the q uadr atic formula.

7. Find all the r oots of the equation Z4 - 4z3 + 6z2 - 4z + 5 = a if Zl = iis a root.

8. Solve the eq uation (z + 1)3 =z3.

9. Find the thr ee solutions to A =4 Y 2 + i4 Y 2.10. Let m and n be positive integer s that have no common f actor. Show that ther e

ar e n distinct solutions to wn =zm and that they ar e given by

" " ( m(8 +2." .k ) +. . m(8+27I'k )) f k 0 1W k =r n cos n ~Sln n or = , ,.. ,n- 1.


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( b) Use par t (a) and De Moivr e's f or mula to d erive Lagrange's identity:

1 5 in[(n+1)0 ]1 + cas e + cas 2 e + ... + cas n e = 2 " + 2'~ , wher e 0 < e < 2 1 1 .

SIn 2 "

12. If 1 =zo, Z1, ... , Zn-1 ar e the nth roots of unity, prove that

(z - Zl) (z - Z2)'" (z - Zn-1)

= 1

+ Z

+ Z2

+ ... + zn-l

13. Let Zk i = 1 be an nth r oot of unity. Pr ove that 1 + Zk + z~ + ... + Z~-1 = O.

14. Eq uation " De Moivr e's formula. can be esta blished without r ecourse to

pr o per ties of the ex ponential f unction. Note that this id entity is trivially true

f or n =1.

(a) Use basic tr igonometr ic id entities to showthe id entity is valid f or n =2.

( b) Use induction to verif y the id entity for all positive integer s.

(c) How would you verif y this id entity for all negative integer s?

15. Find all f our r oots of Z4 + 4 = 0, and use them to d emonstr ate that Z4 + 4 can bef actor ed into two quadr atics with r eal coef f icients.

16. Ver if y that R elation 1- 1 is valid.

In this section we investigate some basic ideas concer ning sets of points in the

plane. The fir st conce pt is that of a cur ve. Intuitively, we think of a cur ve as

a piece of str ing placed on a f lat sur f ace in some ty pe of meand ering pattern.

Mor e f or mally, we def ine a curve to be the range of a continuous com plex-valued

f unction z ( t ) d ef ined on the interval [a. b ]. That is, a curve C is the r ange of a f unction given by z ( t ) =(x ( t ) . y ( t ) ) = x ( t ) + i y ( t ) . for a ::::::t ::::::b, wher e

both x (t ) and y ( t ) ar e continuous r eal-valued f unctions. If both x ( t ) and y ( t )

ar e diff er entia ble, we say that the cur ve is smooth. A cur ve for which x ( t ) and

y (t) ar e dif f er entia ble exce pt f or a f inite number of points is called piecewise

smooth. We s pecif y a cur ve C as

and say that z (t ) is a parametrization for the cur ve C. Note that, with this

parametrization, we ar e s pecif ying a d ir ection f or the cur ve C, saying that C is

a cur ve that goes f r om the initial point z (a ) =(x (a ), y (a)) =x (a)

+ i y (a)

to the terminal point z (b) = ( x (b). y (b)) = x (b) + i y (b). If we had another function whose r ange was the same set of points as z (t ) but whose initial and

final points wer e r ever sed , we would ind icate the cur ve that this function d ef ines

by -C . .

• E X A M PL E 1.22 Find par ametrizations f or C and -C . where C is the

str aight-line segment beginning at Z o =(xo , Y o ) and end ing at Z l =(X l , yd .


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Solution R ef er to Figur e 1.21. The vector f or m of a line shows that the

d irection of C is Z l - Z o 0 As Z o is a point on C, its vector eq uation is

C : z (t ) =Zo + (Z l - zo ) t , for 0 : s ; t : s ; 1, or C : z (t ) = [ xo + (X l - x o ) t ] +i[Y O + (Y l - Y O ) t ], f or 0 : s ; t : s ; 1.

Note that ")'(t) = z(l- t ). which illustr ates a gener al pr inciple: If C is acur ve par ametr ized by z ( t ) f or 0 : s ; t : s ; 1. then one par ametrization for -Cwill be ") '(t) = z(l- t ), f or O :S ; t:s ; 1.

A cur ve C having the pr o per ty that z (a) = z (b ) is said to be a closed curve.The line segment 1--1 is not a closed cur ve. The r ange of z ( t ) = X (t ) + iy ( t ) ,wher e x (t ) = sin 2 t cas t , and y (t ) = sin 2 t sin t for 0 : s ; t : s ; 21 r is a closed cur ve because z(O ) = (0,0) = Z (21r ) . The r ange of z( t) is the f our -leaved rose shownin Figur e 1.22. Note that, as t goes f rom 0 to ~. the point is on leaf 1: fr om ~

f igurf' 1.22 The cur ve x (t ) =sin 2 t cas t . y (t ) =sin 2 t sin t f or 0 : s ; t : s ; 21 r . whichf orms a four -leaved r ose.


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to 7 l" , it is on leaf 2 ; between 7 l" and 3 ;, it is on leaf 3; and f inally, for t between3 ; and 2 7 l" , it is on leaf 4.

Note fur ther that, at (0,0) , the cur ve has crossed over itself (at points other

than those corr esponding with t =0 and t =2 7l " ) ; we want to be able to d istin-

guish when a curve d oes not cr oss over itself in this way. The curve C is called simple if it does not cr oss over itself

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