ECE3337NOTES4_20081015125739

ECE3337NOTES4_20081015125739 - .Jmfin’bw 1H; 4 ‘3‘?...

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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: ; co-srz = .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- e-iz = 2i'sin z,_ (2). eiz+ e4; ="2‘egs-z Adding (1) and (-2): _ 29%: i 230;: _+ 2i sin-z - '9‘.” =HEbaz + {sins - Subtifscfing (I) from (.2); ' '28"h- = 72 cosz —. 2i sin: sud ‘ a“? : cos'z —- isinz _. o ' {[21+z,)%_-3-fikzi+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 {ins-z, 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)(cos-zz + isin 22) -|_- (cos 2:,- jisin-zlxms 22 1+ {sin-22') . -. _ :. _ 2 = cos 2, c0322 .—': sin-21 sinzz'. 10. Prove that the-zeros or. ('-)_ sing and '{bycbsz'zareaufreai anth them, - I — —‘z _. ‘I ‘1‘- 3 V 7 7- ‘ _ I ‘.- ' ‘_‘ ’I' -‘ (a) If sinz z flag-ii...— = 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:, tam-5., 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 7-7 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) larelésJIed-bddifu-ractions, while those-1'on - which f('—z) = f(z_) are Quad-even fumt'iom. Thus sin z; ahd--tan 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 for-the'function Of Problem 19. . r We can. have difi‘erent Riemann surfaces corresponding to Fig. 2-23 or 2-24 of Problem 20. Referring to Fig. 2-23, 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| <, BI-rz—zo+220| < 8{Iz—zo|+|229i} < 6(1+2|zo|) Take 3 as 1 or e/(l-l-ZIZBI), 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 zu-o-zn _ 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.2-25 r - r 3‘ Fig.2-26 In Problem 23(a) We prove that-given 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. 2-26] 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'— newlinixi-irwulumswu1M-.'Maw.'finihm" 'an‘vayiirfifi'I-Wm" "e liF-rgi’z. " ' ‘ ' “1.1;: 'ANALYTIC FUNCTIONS ' is-that, 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, the-reverse 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 u-i-i’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». 69-9 H r-I'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) main-aimwmiknwobI-imwwseamiwéawmismamr"aims-mas ‘ immeaéweaaawm; {AWWM ‘ HIM , I "meg; m mum ‘ a. may." 4“. wr Qty-m. mun GHAP.'3j COMPLEX DIFFERENTIATION, THE 'OAUCHY-RIEMA—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 deter-mined 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)" where-II; 2 9(a) and '1; = 'h(z), then. _ I _ : . awz; dw.d;.i’1 L " i - *‘dgfig dz? -7 '. ,- (117)-- h . ~ ' The-results (10): and. (IIj'aresoften .called' chain moles-for 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 {pan-meta, then 7 . ‘ ' £11 __'I dw/dt _ g’(t) " dz 5 dzIdt _' _ ._ Siniilar rules canbe formulated for differentials. - FOE-example, 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 Starla-id? . and are/aw are relatéd by _ a ' ' _ (12) . as) - DERIVATIVES OE ELEMENTARY. fififiCTIONS-_ In the'following We assume that 'the'functions are defined as-in’ Chapter-2. _‘ 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 non-analytic 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 Ann-50 An: +1.51; Mal) : _]im an-‘iy-I-Az—iAy—(x—iy) 7: Hm Am_-—iAy Ann-.0 - . Am+€Ay As»'oAm+iAy ' . Aux—90 ' _ Asa-40 If Ay=0, the-required liinit is‘lim 5‘5 .= 1. r - MaoAx If Nazi), the reduire'd limit is firefly- : -1. Ash-r0 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 non-analytic anywhere. 3.. If.‘w.= f(z) =, (a) Method 1, using the definition. ' (a) $1:- and (b) determine where f(z) is non-analytic. '- 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—z-Azm—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) _~ (1-2)(1)‘—(1+z)(—1) _ 2 dz_' 1—; ‘ ' _ r (Ir-z)? 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 non-analytic. 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 4-1] 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» = ream-09) + Hana-m} +' + f(£,;)(bf-2n—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 subdivisions-n. 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. fC-mszzndz = 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 certainly-satisfied if g is analytic in'a region containing curve C}. 98 Agcomvnex INTEGRATION AND-GAUCHY'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“; + 3-2) 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) The-points z_=0 and z = 4+2i on correspond-to 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: + 2-0 +’ i The parametric eguations of C are a: 3:0 2+(89-8i) = 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 , dew-l-ydy + 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 - woo-eds» = = 8_—8i f 4 edit + if T-Zda: . e _r . 108 COMPLEX 'INTEGRATION AND CAUGHY’S THEOREM I _ [CHAPA 21. Evaluate § dz where is any simple closed'curve 'G-and 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 byI-Ceiichy'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) -' _: FIE-429 , '3 . r - r . _ I I ‘ H i 72 .= Augie, o g a <27“. Thus since 2 seeds," 0' Now on I‘, lzr-elré é-orIg—aégei9; i.e.' the right side of (1) becomes . -. - . ' _ I I . V-ria -- '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“Vie-6w 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 indepeIIde-nt of the path joining (Ln-an 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 = x-l-iy = 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 non-inter- secting" curve which is the boundaryrof ai-"egion with; ' finite area equal to - ‘ 1: 7 ' x 3-7.. 3 I 1 3 r ' et/é- +1 (a)(t)2§.+ (any; + (27){e)2§+ s-- ,. _ _ ' V371" 3x75 _ Tar+%+%+"‘) “- 71-1/3 “" T for '1.5 times the area of triangle ABC,- and which has _ .. infinite length (see Problem 91). , I . FIE-M3 '31. ILetI F033,!) and G(x,'y) be 'continuOus and have cootinuous first and second partial denvatlves in I a- simply-cennected region ‘R bounded by a simpler closed curve C. ,- I‘Prove that . ' ' aG _ _fl‘ ' 62_G 32G) gar; fiaG cF(aydx‘79Edy) ‘ IR [F(ax‘~’+ay2 I an: 675*"62; 6 HM“ Let :P-._= Q = in Green's theorem I §Pde+9dy =‘ fl—a—E)dedg A I _ c _ an: 3 Then as required . qt 7 - ' aG aG _ a 1 6G I- a g . c4543???) — ff (aha -—‘-‘{Fay Fed” ‘R. _ I 3%; 3261‘ “'92 '_ it} + (53 at “L 62' WNW” SHDPlementary Problems I LINE INTEGRALS ' I \I ' 2,51 7 r - . ' ' (3:); +11) dIav‘+'(2y met) dy ‘ 'elong' (a) the curve y = x2+ I, (b) the straight line '. 32. zEtraluate ' . I (9.1) I . I _ I I joining (0, 1) anti (2, 5), (c) the straight lines from (0,1) to {0,5} and then from {0, 5) to (2,5), (d) the straight fines fro'm’w, 1) to (2,1) and then from (2, 1) 'to (2,5). ' - ' Ans. (a) 88/3, (5)32, (e) 40. (d) 24 (a) Evaluate f (a:+2y)rdm + (1)4205) tly around the ellipse 0 defined by 9: -"= 4I cos a. y =‘ 3 sin a, c . 0 E a 211’ if C is-deScriIbed'in a counterclockwise direction. (5) What is the answer to (a) if C is described in a clockwise _dire'ction?_ Am. (a) —4IS'1r, {(1) 48m- - 34. Evaluate (xi-agate Iaiong (e) theeparabola y=2e2 from (1,_ 2) to (2,8); -(b) the straight Hues from(1,1) Ito (1,3) m then-from (1,3) to (2)3), (e) the straight line from (1,1).te (2,8). Am. (a) 5t‘--—- 553%. (,6)- fit—3 ‘— 57:; (a) it“ — 82" ~ - .35. Evaluate, f 1212 as; around the square with vertices at (0,0), (1,0), (1,1), (0,1); - Ans. —'-1+i ...
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