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Unformatted text preview: .Jmﬁn’bw 1H; 4 ‘3‘? CHAP 2] FUNCTIONS, LIMITS AND CONTINUITY
. :21 e2: 0 91: _= £11303 yl + ism yl) . 93, (cos 212+ {sinryz} ' = '3‘" " .9” (605111 + i sin 119(ch ya + {sis yz) r
 . . = #31”! {cos (y; + 119 .+ i shawl + y2)} = gzl+z, (‘0 fezl '= leﬂcosy + isins)! '= lexnco'ay +‘isinyi : 3:. 1 = e:
"i (9} BY Part (a): e=+2k1ﬁ = 32 92hr! = 6413032161: + isin’ 2kg) = e1
3; f _
',f I' This shows that the function 32 has period 2%, In partiéular,_ it has period 2“:
g'; r r r s , ' 9. Prove;
(a) sinzz + (503% =, 1 eCz '__e—iz eiz + e—iz  ' By deﬁnition, rsinz = 721: ; cosrz = .2 . ’ II'ZI‘henr. _ 2 r u _ r
(a) sinzs + c0522 4—: M 2'—— m. 2 r  7 r a 5
‘ ' ‘ . '21? ' 2. ‘ ' ‘
= __ (em—wuss) + (e2i=‘+_2+e,2fz) 2'1
. 4_ ._ 4 
(b) (1) eiz eiz = 2i'sin z,_ (2). eiz+ e4; ="2‘egsz
Adding (1) and (2): _ 29%: i 230;: _+ 2i sinz  '9‘.” =HEbaz + {sins 
Subtifscﬁng (I) from (.2); ' '28"h = 72 cosz —. 2i sin: sud ‘ a“? : cos'z — isinz
_. o ' {[21+z,)%_3fikzi+sg} ashes; ...V 'e~iz,;_3~ié,,j _
+ Ex ' : _._.....____ I ._._.___' I .
(c) sm(z1 22) _ 2i.“ _ . I  21 7 r _ _ I
_ 7 (cos 2] + isin zl).(oos 2'2 isin 32):..7— ’(cros z; {j} sin ZINE.03 22 54‘ isingz)
_ ‘ —. . r _2z' 7 '
' = sin 2; cos 22 {insz, sin 32' 7 " 
' 7‘ _ ei(z1'Fzr,)r+3—_i(zlfz,) _ 621.6% ‘+ (knew, (1 _ _r _  V;
@ c°s(z‘r+‘22,) '_27 r _ _ (Cos 21. 4 {sin 21)(coszz + isin 22) _ (cos 2:, jisinzlxms 22 1+ {sin22')
. . _ :. _ 2
= cos 2, c0322 .—': sin21 sinzz'.
10. Prove that thezeros or. (')_ sing and '{bycbsz'zareaufreai anth them, 
I — —‘z _. ‘I ‘1‘ 3 V 7 7 ‘ _ I ‘. ' ‘_‘ ’I'
‘ (a) If sinz z ﬂagii...— = 0, theme“ '=e’,‘= or 333:1: smut:7; '; 0,:ﬂ,i2, .
3: Hence 21's _= ‘ and. z = ks, i.e. 2 =5 .0, i1, 1—2:, tam5., are the zeros, (b) If was 2 3—2;: 0,‘ then 6“ =e.‘?z or ,e?izr=I——1 =_re‘r_2".+1)’lﬁ, ..k = 0, i1, :2! 2.7;. Home 2% :‘(2k+_1)wi and"g: (in g)”, is. z '='_s[2.‘=35.r_12',;5§/2,‘ . .. as the gems; Il..Prov_e that 7(a) sini(.—z) =Sin«z,‘ (b) cos (,.—'z) :.cosz, (c) :tsn'('¥z)z tan z_. 7 _ ‘ _ __ _7_ git—2);};F'it—‘zy' 3—}: _'e£zi_. sees 84}: '7 ‘ ' 7 (b) sou—z) = we?‘“”+'°"‘*"..=',__e*“+4°‘zg 'me“_+'°“‘".= c555 
:7 ' 2' 7 2 ' ‘: '2 _ .'
ta _ = Sin(_z) :_—Sin'2' : __ , I 77 I ‘ d T' (o) _ z} cos(—,z)'l cos; , _ talfzv r_ _“’j‘y“—‘g=_(‘v') 311,157) . ._ _ Functions of 2 having the propei'ty' that 1142):: ﬂz) larelésJIedbddifuractions, while those1'on  which f('—z) = f(z_) are Quadeven fumt'iom. Thus sin z; ahdtan zfare odd functions, while cosz,
' ' isaneven'function! .7 '  ' '  .  (b) e“ : cosz +_13sin z; 6”“ _= cosz — isinz (d) cos (21 +22) = (:03 z: cos 22 f si 45' (6) sin (2; +522) = sin 21 cog 22 + cos slain 32 n él sin 22 rr————L_ 50 _FUNCTIONS, LIMITS ANDCONTINUITY _ ' [Cases ‘21. Discuss the Riemann surface forthe'function Of Problem 19. . r We can. have diﬁ‘erent Riemann surfaces corresponding to Fig. 223 or 224 of Problem 20.
Referring to Fig. 223, for example, We imagine that the 2 plane consists of two sheets superimposed
on each other and cut along the branch line. Opposite edges of the cut are then joined, forming the
Riemann surface. On making one complete circuit around z = i, we start on one branch and wind up
on the other. Howaver, if we make one circuit about both 2 = i and 2 = i, we do not change branches at all. This agrees with the results of Problem 19.‘ 22.. Discuss the Riemann surface for the function ﬁe) r lnz [see Problem 14]. In this case we imagine the 2 plane to consist of inﬁnitely many sheets superimposed on each
other and out along a branch. line emanating from the origin 2 = 0. We then connect each cut edge to
the opposite cut edge of an adjacent sheet. .Then every time we make a circuit about 2: 0 we are on
another sheet corresponding to a different branch of the function. The collection of sheets is the
Riemann surfaces In this case, unlike Problems 6 and 7, successive circuits never bring us back to
the original branch. 7    ' ' ‘ LIMITS '23. (a) If re) = 22, provethat lim f(2) = 22.
I zb'zn 3 i
r . . . _ ' Zr2 29":20 )
(6) Find £13210 f(z) 1f fig) 7 {0 z 2.20 . . \ (a) We must show that, given any e>0 we can ﬁnd 8 (depending in general on e) such that
_z2—zg! < (whenever 0 < Iz—zol < 6. 1 If 8 E1, then 0 < 12—20! < 8 implies that' I .
z2_zg. = lz—zolla+ao <, BIrz—zo+220 < 8{Iz—zo+229i} < 6(1+2zo) Take 3 as 1 or e/(llZIZBI), whichever. is smaller.' Then we have 122—zgl‘ < 6 whenever
z—zo < 6,_ and the required result is proved. ' ' ‘ (b) There is no diii’erence between this problem and that in 'part. (in), since in both cases we exclude
‘zz‘zo from consideration. Hence 'lim f(z) = zﬁ.‘ Note that the limit of . ﬂz) as z—>zn has zuozn _ nothing whatsoever to do with the value of. f(z) at so. I 24. Interpret Problem 23 geometrically.  .1 (a) The equation w = f(2) = 22 deﬁnes a transformation or mapping of points of the—z plane into
I points of_ thew plane. _ In particular let us suppose that point an is mapped into wot'23. \. a h ' zplane I “ ' wplane
 u Fig.225 r  r 3‘ Fig.226 In Problem 23(a) We prove thatgiven any 6 >0 'we can'iind 6 >.0'_such that luf— wo '< e
Whenever jz—zOI < a. Geometricaily this means that if_we wish to to be inside a circle of _
radius 6 [see Fig. 226] we must choose 8 (depending on a) so that 2 lies inside a circle of radius 3.
According to Problem 230:1.) this'iscertainly accomplished if 8 is the smaller of. 1 and e/(l + 2]an). (b) In Problem 23(c), w = w.) = 2:3 is the image of z = so. However, in Problem 23(6), 'w = 0 is the
image of z = zo; Except for this, thegeometric interpretation is identical with that given in
part (a). 7 '  ‘ 35'— newlinixiirwulumswu1M.'Maw.'ﬁnihm" 'an‘vayiirﬁﬁ'IWm" "e liFrgi’z. " ' ‘ ' “1.1;: 'ANALYTIC FUNCTIONS ' isthat, in ‘R, an entity satisfy the Cauehy—Riemann equations I Chapter 3 v DnniVATIVEs _ _ _ _ _
‘ If ﬂz) is singleevalued in some region ‘R of the 2! plane. the derivative of f(z) is
deﬁned as ‘ — r ' . '  , z  f(2 + M)  M) A ~_ “2). 2:3 ~ (I) l ' provided that the limit exists independent of the manner in which .Az> 0. In such case I we say that ‘f(z} is diferentiable at z. _ In the deﬁnition (1) we. sometimes ‘use h instead I of A2. Although diﬁetentiability implies continuity, thereverse is‘ not true (see Problem 4). If the derivative f’(z) exists at all points z of a region"R, then re) isjsaia to be analytic in ‘R’ and is referred to as an analytic function in Q or as function analytic in qt; The tenns'regular and holomorph'ic‘are sometimes used as synonyins for analytic.
' A fun'ction'ﬂz) is said to beeanalytic at a. point 20 if there exists aneighborhood ' < 8 at alI‘Boints of which f’(z) exists. ' CAUCHY—RIEMANN EQUATIONS ‘ 'A necesmry conditiOn that ' w 2 Hz): =‘ u(x,_y)'+ Living) ﬁnely/tie in a regionr‘R. my a» an "Jae . .— w _ "
 '—' "    (3) $362}? 763* 755 If the partial derivatives in (2) are continuous ‘in‘R, then; the Cauchj—Riemnnn equations 7 are sufﬁcient conditions that ﬂz)‘ be" analytic. in ‘R’. ' See'Problem 5. . The fnnctions new) and ’v(a:,'y) are sometimes called conjugate medians; Given ' one we can" ﬁnd the other (within an arbitrary'additiv'e constant) so that uii’u = f(z) is 'an'alytic (see PrOblems 7 and 8). HARMONIC. FUNCTIONS  _ If the Second partial derivatives of u 'and v with respect'to a: and y eati'st end are ' ‘Vcontinuous in arregionqt, then we rﬁnd'from (2) that (see Problem 6) ram 621:. _‘  6% rﬂ _ " . ' it follows that undei' these conditions the real? and imaginary parts of an analytie function
satisfy Laplace} equation denoted by ' ‘ ’ ' ‘2 ' 32x1». 699 H rI'az' 6?.
ﬁ‘i‘yy— 0 01‘ ' Vzir=0_ 'Where ‘V2 E'jTﬁ'ri'W . " ' (4) The.opefator V3 is often celled the Lazilacium 7 63' , ." in
Jed, me. as f st :7) ‘8) i 9) mainaimwmiknwobIimwwseamiwéawmismamr"aimsmas ‘ immeaéweaaawm; {AWWM ‘ HIM , I "meg; m mum ‘ a. may." 4“. wr Qtym. mun GHAP.'3j COMPLEX DIFFERENTIATION, THE 'OAUCHYRIEMA—NN EQUATIONS 65 It is emphasized that dz and the are not the limits of A2 and Aw as. 42+ 0, since these
limits are zero whereasdz and dw are not necessarily zero. Instead, given dz we determine
dw‘froin (8), Le. dw is a dependent variable determined from the independent, variable dz
for a giiren z.  ‘ 7 ‘ _ . ' 'It is useful to think of d/dz as being 'an operator which when Operating on w =rf(z)
leads to decide = m). " ' ' , . RULES FOR: DIFFERENTIATION _ If f(z), 9(2) and Ma) are analytic functions of z, the following differentiation rules
(identical with those of elementary calculus) are valid. ' ' ' 1. diz{f(#)+y(z)}5 = grewggazi = ﬂ“) +912) 727.“ = .—; frtz) _ 91?) _ 3.l%{c}°(z)}' eggﬂz) r; cfr(z) where o isany'constant
' 4., gnaw} =_ reggae” geignz) =. re) 912). ,+ 9(2) Hz)” gemz) — 1(2) m .‘ I d L I. 5 £{ﬂﬂ ‘_ _ 9(2)???) 51163399)
' dz I 9‘2) — ' " J99)? 7=_ ifsgtzhéo ' '
'716 If w=If('§)_,fvhe're'§=g(z) then _ ‘ " 7' r _' r I I
 = =. ﬂag =_rfr{g(§)jgf(z)e 1 I I I ‘ (10) . Similarly, ;_ 17(5)" whereII; 2 9(a) and '1; = 'h(z), then. _ I _ : . awz; dw.d;.i’1 L " i  *‘dgﬁg dz? 7 '. , (117) h . ~ ' Theresults (10): and. (IIj'aresoften .called' chain molesfor diiferentiation' Of  composite functions. 'I  .' I ' . I 13   If w‘=_f(z),ithenr a: f" (to); div/dz
'  ' dw. ‘ 1 . _ .  "'5 " daem , _ s. ' If z: ﬁt) and wzga)’ where‘t is {panmeta, then 7 .
‘ ' £11 __'I dw/dt _ g’(t) "
dz 5 dzIdt _' _
._ Siniilar rules canbe formulated for differentials.  FOEexample, i
I diﬂzi‘i‘gfzii = aft?) the.) 4— 79de + 9'13)!” 7: {7:13) 45912)} dz
' ‘ d{f(z)79(;z)}r = f(z)'fig(3),+Q(zi;df(z) = {f(z)59‘5(z}: it Starlaid? . and are/aw are relatéd by _ a
' ' _ (12) . as)  DERIVATIVES OE ELEMENTARY. ﬁﬁﬁCTIONS_ In the'following We assume that 'the'functions are deﬁned asin’ Chapter2. _‘ In the _cases where functions have branches; 'i.e." are multiéiralued,‘ the ‘of the‘ffunetion
"on the right is chosen so as to correspond to the branch of the functiOn Onthe left. ' Note ' that the results areidentical with those of. elementary calculus; ” '  ' bERIvATIvEs AP. 3] COMPLEX DIFFERENTIATION, THE CAUCHY—RIEMANN EQUATIONS 71 ‘_ Solved Problems 1, Using 'the deﬁnition, ﬁnd the derivative of w = f(2) = 23—2z at the point where. _'
(a) z .—: 20, (b) 2 = —'1. ‘ (a) ‘By deﬁnition, the. derivative at z = 20 in min '= am New) * M = I... . “#0 A5 Mao A2
= m. A2490 . ' 'Az
: 33% + 3zoAz (Az)2 ._ = _ 2 ‘ Az—vo
In general, f’(z) é 3z2—2 for an 2.
(b) From (a), or directly, we ﬁnd that if zo=l then f’(—1) _= 3(1)2— 2 = 1. K 2. Show that does not exist anywhere, i'.e. .f(2) = 2 is nonanalytic anywhere.
By deﬁnition, ‘ {inn 2 £130 19+ Ail— ﬂz) if this limit exists independent of the manner in' which A2 = M: + 1' A3; approaches zero. _ —'—.—''——. I ‘
.rThen : limoz+Az,—z : Hmm+w+Ax+zIAy—e+zy
_ dz Az—DD Az Ann50 An: +1.51;
Mal)
: _]im an‘iyIAz—iAy—(x—iy) 7: Hm Am_—iAy
Ann.0  . Am+€Ay As»'oAm+iAy
' . Aux—90 ' _ Asa40
If Ay=0, therequired liinit is‘lim 5‘5 .= 1. r
 MaoAx
If Nazi), the reduire'd limit is ﬁreﬂy : 1.
Ashr0 My  A Then since the limit deﬁends ,on the manner in which Az> 0, the derivative does not exist, i.e. f(z) :5 is nonanalytic anywhere. 3.. If.‘w.= f(z) =,
(a) Method 1, using the deﬁnition. ' (a) $1: and (b) determine where f(z) is nonanalytic. ' 71 4424112) _ 1+2
1—(z+Az) l—z dw _ Hm j(z'+Az)'— ﬂz) = _ _. _lini dz 'Az—DD Az Azao  Az
_ . '. ' 2 _ 2
““ L‘E‘ML—zAzm—z) ‘ (1—2)”. independent of the manner in'which M —> 0. provided 2% 1.
Method 2. using differentiation rules. _
By the quotient 'rule' {See Pfohlem 10(0)} we have if 25* 1,_ __
 a ‘ d! ‘ 
 1 ,.' — 1 + — V — 1 — ' .
1(1” ‘ ( ' 2) w a “Md” ‘2) _~ (12)(1)‘—(1+z)(—1) _ 2
dz_' 1—; ‘ ' _ r (Irz)? H A (1—2)? _ _ (1,4'5)2 _, (b) The function 312) is analytic for all ﬁnite values of 2 except 2: 1 where the deﬁvative'does not exist and the function is nonanalytic. The’point 2.: 1 is a singular point of ﬂz‘). 4. (a) If K2) is analytic at '20,_ prove that it must be continuous at 20.
_.(b') Give an example__to show that the 'c'onverse of (a) is not necessarily true. I.assume'has a ﬁnite length, i.e. C is a 7 , Chapter 4 . COMPLEX LINE INTEGRALS.
' . Let f(z} be continuous at all points
of a curve C {Fig 41] Which We shall i notiﬁable curve. _
 ‘ Subdi'vide G into a parts by means '
of: points. 21, 22, . . ., has chosen arbi...
trarily, and call a=zo,'b in. On each
are joining ate—1 to 21¢ [Where "k goes froml to n] choose a point 5k. 'Form '   Fig M
the sum . , ' _ r 
S» = ream09) + Hanam} +' + f(£,;)(bf2n—1)  . (1)
On Writing as — 'zk—l = AZk, this becomes . ' ‘ I _ '
st = reset—as) =7 fem '  (2) that the largest of the chord Let the number‘of subdivisionsn. increase in such a Way
limit" which does not depend lengths [Azkl' approaches zero. Then the sum Sn approaches a
on the mode of subdivision and we denote this limit by I ﬁe) dz. or I; f(z) dz . v (3) ﬂy line integral of ﬁg) along curve C', orthe deﬁnite
C. In such” case f(z} is said to be integrable slang
of a region ‘R and if ‘C is a curve lying in CR, . called the compleéc line integral or brie
integral of f.(z) from a to b along curve
0. Note that if f(z) is analytic at all points
then ﬂz) 'is certainly integrable along C.— REAL LINE INTEGRALS .
If P(:c,y) and Q(a:,y) are real functions of a: and 3; continuous at all points of curve‘C,
the real line integral of de + Qdy along curve C can be deﬁned in a manner similar to that given above and is denoted by
£[P(x,y)dx + Mandel ‘or 1;,de + Qdy  ' (4) the second notation being used for brevity. If C is smooth and has parametric equations
:6 = ¢(t), y 2 Mt) where t1 E t é t2, the value of ('4) yis given by
; mm), ¢(t)}¢'(t) dt + ewe). Home sit} Suitable modiﬁcations can be made if C is piecewisesmooth (see Problem 1), 92 I .GHAP. 4] COMPLEX INTEGRATION AND CAUCHY’S THEOREM 93 CONNECTION BETWEEN REAL AND COMPLEX LINE INTEGRALS
If f(2) = n(:c,y) +iv(x,y) = u + it: the complex line integral (3) can be expressed
in terms of real line integrals as ' ' ' ' 3 Lima : (nativxdxﬁdyy PROPERTIES OF INTEGRALS _ .
If ﬁg) and 9(2) are integrable along 0', then ‘1. fCmszzndz = fcﬂzwz +3'y(z)dz '
‘2.’ 1A f(z)dz = A]; ﬁre) dz 7 where 3A 2 any constant
3', Innate, : ~—_£Tf(2)dzi ‘ m‘ ' I b I
, . , ‘ ﬂz) dz + f(z)’dz_l ‘ where points a,‘b,mare on C.‘ uh
‘q’
A‘
:3,
§
u '§ ML‘ _ pier variable a; = u+£v. Suppose 
‘ _, ' iii the { plane and that the derivative
91;) i3 continuous on 0’. Then ‘ r ' ; ‘  '  cwz r= fee». 91:) d: a  a “ (6).  Theseeonditions are certainlysatisﬁed if g is analytic in'a region containing curve C}. 98 Agcomvnex INTEGRATION ANDGAUCHY'S THEOREM [can i I Solved Problem LINE INTEGRALS (2g + 9:2)dx + (3x —— y) dy: along: (a) the parabola x1 2t,
'3) and then froiu,(2,3) to (2A); a parabola correspond'to t=0 and t=1 respeetively._ Then the = _ 3312 . l r _
= I (24?.2 + 12 — 2t3 ' 6t) dt
0 straight line from (0,3) to (2,3), g = 3, ﬂy 2 0 antithe line integral equals 2 fwd“; + 32) 44/3 
he: 2,ld¢=0‘ and the line integral equals: (6 —. y) all! 5/2 I 4
9:3
3) and (2,4) is 2y—a: = 6. Solving for as. we have 3': 231—6. . . 4 ' ‘ .
{3(2y  6).  y} dy '= J; (81;2 — 393; + 54) cly 97/6  (2.4)
1. Evaiuate
‘ (0.3) _ _
'y = #443; (b) straight lines from (0,3) to (2,
straight line from (0,3) to (2,4).
(0.) The points (0,3) and (2.4) on the
given integral equals '
I ‘{2(t2+3) + (202nm: + {3(2c) — «2+3» am
we  ((1) Along the
2 .
,1 {6+22)dw + (Sm—3w =
. _ 3:0 ‘ .
Along the straight line from (2, 3} to (2,4),
4 .
J (2‘y'+4)0 + {6—y)dy =
u=3  ' .
'frhen the required value = 44/3 + 512 = 103/6.
(c) An equation for the line joining (0,
Then the line integral equals
. r 4 . . r
f any + (2y:—s.)2}2dy +..
I U= .  _ " f r
The result can also be obtained by using 2; =n%(:b + 6).
2. Evaluate f zdz from z = 0 to
r v c m z = 4 +25 along the curve _"C givenby r (a) z :1”? fit, (b) the line from 2 =_ 0 to z = 2i and then the line from 2 = 21'. to 'z 2: 14+72i. ' ' (a) Thepoints z_=0 and z = 4+2i on correspondto t=0 and t= integral equals ‘ . If}.
t=0 (:2 '+ it) {161:2 + it) 3 ‘ fubexz'tee dt
. u . 2 respectively. Then the line " .—. ,10 — Sits ' r(2t3——it2+t)dt
o . ’ Another Method. ~ The given integral equals Vf"(z—¢g)(de +i
C‘ _ ' equals 2
f . (t2){2t cit) +
téo ‘ ‘7: J: (b) The given line integral equals. L (a:  iyﬂda: + idy) The line from 2:6 to 2:25 dw=0 and the line integral equals _
.  2 I ‘ 2 I
f_o(9)(0}+ydy + if e (owe—mo)
_ ,E 2i to _z = 4+2i is the same as the line from (G, 2) to (4,2) for which
0 and the line integral equals _ The line from 2 =_
" 1;:2, dy = f
3:0 Then the3 required value 4
xda: + 20 +’ i The parametric eguations of C are a: 3:0 2+(898i) = 10—35. dy) f’mdx+ydy + if'wdyr—ydx
 c . c :9, y=t from t=_0 to t=2. Then the'line integral r . 2 7
(t)(dt) + t]; (t2){dt) —— (t)(2tdt) . 4 2 7
(2t3+t)dt _+ if {—tzwt = 10 — 3:13
o , dewlydy + if xdﬁﬂyda:
c ' c ‘ is the same as the line from (0, 0) to (0, 2) for which m=0,  2 7 _
I ydy = 2
. y=0  wooeds» = = 8_—8i f 4
edit + if TZda:
. e _r . 108 COMPLEX 'INTEGRATION AND CAUGHY’S THEOREM I _ [CHAPA 21. Evaluate § dz where is any simple closed'curve 'Gand 2:11; is I. ((1) outside 0, 22. C 2' '— a
(b) inside 0. I (a) If a is outside 'CI,I then. 1(2) = 1/(2— a) is ena'lytie'every
where inside andson Hence byICeiichy'r's theorém, :0.”
Cz ‘1 . (b) Supposela. is inside C and let I‘ be 'a circle of V'radins' a with '
center at 2 =0. so that'II‘ is inside 0 [this can be done since
2 :6. is an interior point]. ' ' " By PrOblem '20, f; z—a =7, fig—far. (1) ' _: FIE429 ,
'3 . r  r . _ I I ‘ H i 72 .= Augie, o g a <27“. Thus since 2 seeds," 0' Now on I‘, lzrelré éorIg—aégei9; i.e.'
the right side of (1) becomes . .  . ' _ I I
. Vria  '21s “6..” z 1“. 0197:7211»; Eele or . II «1:6 which isrthe tequiredvelue. ‘ 7 . Evaluate n = 2, 3, 4, . . . where 2:41 13‘1n31de the sample closed curve C. .
,  c.(z—a)"’ ,  r  
. r '  7 'dz __ dz
As in Problem 21, ct (z_a)n  I F {zﬂay‘
2“Vie6w d0 1'. 2”
._ _. (1 no
" o in awe cunt j; e 1‘ do
__ .; e(l—n)iﬂ 2:: __ 1 I 2“_ nt  _
‘ s“_1_(1—n)zd ‘ muggy—IV " 1] “ ° ‘Where 1; 5* 71.  23. If C is the curve q) = w3m3m2+4x—1 joining points (1,1) and (I2, 3), ﬁnd the ' any path can be choSen.I In particuler let us value. of ' _ _
" (1222 ——' 4iz) dz
a (2, 3). ﬁence Method 1. By Problem 17, the'integral is indepeIIdent of the path joining (Lnan
choose the straight line paths from (1,1) to (2,1) and  then from (2,1) to (2,3). I_
ICase 1. Along‘the path from (1,1) 130 (2,1), #:1. (121:0 .30 that 2 = I441! = 95+}; dz :1 dat
Then the integral equals ‘ ' 2 .
I {12(m+ 132 — 4i(:n + i)} dz: = {4(9: + '03 — 2112: + {)2} F = 20 + 30$
:51 ' 1 ' Case Along the path from (2,1) to (2,3), 95:2, dw=0 so that z = xliy = 2+iy, dz = idy.
Then the integral equals . . . I ' 3  . _ _ _ ' ‘ 3 _ A
I I{12(2+iy)2 —, 4112+ iy)}i dy I: {4(2+iy}3 — 2%(2+iy)2} . 4: "1'16 + 8i
 yr=1I.I_A _ . 1 .._
_ Then adding, the required value = (20 + 302) + (—176+8i) = —156 + 382'. _
Method 2. The given integral eqnals ' 7 "
I2+si ' r ‘ 7  ' 2+3:
. I (1222— 4iz)dz = (423 —. 21:32) . = —156 + 38% I
' 1+i' , . I‘ 1+} _ I _  I .
It is.c1eer that Method '2 is; easier. I ' 112 I . ' ' COMPLEX iNTEGRA'TiON AND CAIUCHY’S'THEORE'M j [CEAPA  I The process marrow be continued" by trisecting sides ‘
‘DE, EF’, FB, BG;IGH, etc., and constructing equilateral
triangles as 'beforcL By repeating the process indeﬁniter
7 [see 'Fig. 4—23] we ‘ohtain a continuous closed noninter
secting" curve which is the boundaryrof ai"egion with; ' ﬁnite ar...
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 Fall '08
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 Vector Space, Analytic function, integral equals

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