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06Bhw3solution - fifll’fi‘R/f S’C’h’HfIFOHS‘...

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Unformatted text preview: fifll’fi‘R/f S’C’h’HfIFOHS‘ ftp #fi/ #3 SECTION 64 I. (a) Let us write 1 1 1 1 1 2:—-—--w———=--=«-(1----z~l~z"1'"za-inn)=-—-—}.—I~z~-~2:2+~~ 2+2 2 1+2. 2 z The residue at z = 0, which is the coefficient of —1—, is ciearly 1, z (b) We may use the expansion 2:2 2‘s 25 C0$ZEIWZ+EE"EI+ to write (0<lzi<1). (IZK 00) 1 1 1 1 1 zc08[ijzz[lmmm+__.i_.u——nunE-+..u):~:z_ii+i,i3.wu1_.n;§+.n 2m?” 42 z,“ 6! z z (0 <lzl< no) The residue at z = 0, or coefficient of -1—, is now seen to be ”312-. z (0) Observe that zwsinz 1 . 1 z3 z5 z; z4 =— z~smz “—“m z- 2:w-+—~—~~~- :m—nm+m 0<lzl<oci z z( ) z[ [ 3! 5! ‘ 3! 5! ( ) Since the coefficient of E- ' this Laurent series is 0, the residue at z n 0 is 0“ z (d) Write eotz_ 1 cosz 4 “mfg z 2 sum and recall that 2 4 2 4 z z z z (Izl< m) coszmlu§T+mw~ml—_+mmn 4! 2 24 95 Dividing the series for sin: into the one for cosz, we find that 0032 1 z z =-—---—-—m+... sinz z 3 45 (0<lzf<7r:). Thus 4 __ ....._____.... _—.__S.——~.. ————— +m (0<lz!<1t)n cotzwllzzsw 11111 2 z4z345 Note that the condition of validity for this series is due to the fact that Sims 2 0 when z = my: (:2 m 0,il,:t2,...). It is new evident that 60:2 has residue “2%" at z 2 0. z (e) Recall that 3 5 ’ — E... E... no smhz-z+3i+sg+ (fzk ) and 1 2 Wml+z+z +m GER”)- 1*2 There is a Laurent series for the function sinhz 1 . 1 “2*(w—1mzz)?(smhz)(1~zz) that is valid for 0 <lzi<1w To find it, we first multiply the Maclaurin series for sinhz I d w an 1~z (sinhz)[ 1 )=[z+£z3+u}_z5+.“)(1+zz+z4+n.) {”22 6 120 1 :z+-~ 3-i-———~~-5 H 6" 120g 3 1 s z +~— + 62 £54.... wz+~z +~~ (0<izl<1)n 96 We then see that weight. 0 H 1 24(1-22) 23 6 z ( <Z< )“ sinh z 7 This shows that the residue of m......._,, at z m 9 is —-. z4(1m 2:2) 6 2. In each part, C denotes the positively oriented circle lat: 3. (a) To evaluate JG 6x138?) dz, we need the residue of the integrand at z = O. From z the Laurent series “PP-2L}— ...32. 512:: Mimi: 1 z 32 “32(1 11+2! 3t+ “2,2 1¥z+21 “gt“ (0<iZE<°°), we see that the required residue is -1. Thus j exp?” dz : 2m(-1) m —27ri. C 2 (6) Likewise, to evaluate the integral L 2:2 exp(~1~)dz, we must find the residue of the z integrand at z = 0. The Laurent series which is valid for 0 (izk on, tetls us that the needed residue is 3— Hence 2 1 m . 1 um‘ L" expiridwmizi“? 4. Let C denote the circle lzim 1, taken counterclockwise. (a) The Maclaurin series e «- -22"' --—- (lzl< on) enables us to write 71:0 ’2! M n M 1 1 ex z+w 3-:sz emdzm e"z 5—d = -~— ”ex (—Jd ‘ L p( 2. Le: C We . z 2 . z p z. z (1;) Referring to the Maclauxin series for e1 once again, let us write ,, 1) ,, °" 1 1 '° 1 “k z ex - =z --—~—-= " n=0,l,2,m. pt: 25k! z" é? ( ) Now the l in this series occurs when n Mk 2: ml, or k=n+1. So, by the residue 2 theorem, ,, 1 ' . 1 jz exp —- dz=2m (n:0,l,2,...). C :5 (n+1)! The final result in part (a) thus reduces to j; exp(z+—:— —--)dz ZniZ-u—mw n=on’(n+1)' 99 5. We are given two polynomials ’ P(z)=ao+a1z+azzz+m+aflz" (aflqeO) and 9(2) : b0 +blz+ (7252 +--«+ Iowa“ (13," a o), where m 2 n+2. It is straightforward to show that 1IP(1/z): 2"”2 +azz’" ”+022”4 +~n~+ a:z,,z""""'"2 z QUIZ)“ buz'" +btzm‘l+bgz’""z+m+bm (#0)" Observe that the numerator here is, in fact, a polynomial since mmn - 2 2 0. Also, since b," ¢0, the quotient of these polynomials is represented by a series of the form do +diz+ c1127" +w-. That is, 1 P(l/z) 2 __4__.__.m:d +d + H, - £2 Q(l/z) o lz dzz + (O<Izl<R2), and we see that t—~z12 Eur-mix: has residueO 2:0. Suppose now that all of the zeros of Q(z) lie inside a simple closed contour C, and assume that C is positively oriented. Since P(z)/ Q(z) is analytic everywhere in the finite plane except at the zeros of Q(z), it follows from the theorem in Sec. 64 and the residue just obtained that Hz) 1 Pale) ~ ... jCQ(z—-)~dz= 2m" Res[? 90 To] am one. If C is negatively oriented, this result is still true since then Ifgld dz:""IC P(—z-—)dz—0 CQQE) CQCE) SECTION 65 1. (a) From the expansion 2 3 gz=1+£+i+mfi+ut (izi<00), 1! 2! 3! we see that zexp(l):z+l+~l—il+im1w+m (0<IZI<°°)‘ z 2! z 3122 100 . . . l . The prmcrpal part of zexp(~) at the Isolated singular point z z 0 is, then, 2. 1 l l 1 2! z 3! z and z n 0 is an essential singular point of that function. 2 (b) The isolated singular point of is at z x —-L Since the principal part at z 2 m1 1 + z involves powers of 2 +1, we begin by observing that 2:2 =(z+1)2~2z~imtz+i)2-2(z+1)+i This enables us to write 2 +12% + z zfiwxw+nwg+$n 1+2: z+1 z+1 Since the principal part is —L~, the point z = --l is a (simple) pole. 1 2+1 1 (c) The point 2: =0 is the isolated singular point of 51-25», and we can write 2 sinz 1 :3 zs _. z2 Z4 z "E(Z“§+ET"'H]-IW'€T+"§TM'H (0<]ZE<°°). The principal part here is evidently O, and so 2: L" 0 is a removable singular point of the function EEK“ z (d) The isolated singular point of 93% is z = 0‘ Since 2. cosz 1 2:2 2:4 1 z 23 -~——-m~1——-—+--—-—~« =-~--+-~mn 0<| l<oo, .. zz(2!4!)z2!4l (Z) . . . 1 . . . (2052 the plinc1pal part IS w. This means that sz lsa(smtiple) pole of ——--—-—. z z I ——1 l . = , we find that the rinci al art of at its isolated singular point z = 2 is simply the function itself. That point is evidently a pole (of order 3). (3) Upon writing 101 2. (a) The singular point is z =2“ 0. Since 1.... 2 4 6 3 Highly 1w(1+£m+£m+§m+m] bigwigs.“ z z 2! 4! 6! 212: 4! 63 when O<izl< 00, we have maland BzM§T=m§ (17) Here the singular point is also 2. = 0. Since z z hiifiiijiiiimzs m . 1' 2.3 2! z2 3! z 4! sgz 3 when O<lzl<oo, we have mm3 and Bzmé-Fz—uj— . . exp(2z') . __ . (c) The smgular pomtof (z-~l)2 IS z——I. The Taylor senes _ 2 ... 2 3 .... 3 exp(?.2.')'-'=32”"“92 =ez[1+ 2(2' I) +—~——--—m2 (z I) + 2 (z D 4-] (izk 9°) 1! 2! 3! enables us to write the Laurent series exp(22) __ 2 1 3.. 1 22 22 (z--I)2 We [(z—-1)2+1! z—1+2a+?f(z 1)+-- (047' km)" 2 2 2 Thus m=2 and B=e 17:26? . l 3. Since f is analytic at zo, it has a Taylor series representation I Z Ir f(z>:f(ze)+fl(,°)(z«zo)+féf9)<z-—ze)2+~- (IznzakRu). Let g be defined by means of the equation _ flz) 3(z M-w. :5th 4-4 102 (a) Suppose that f (20) ¢ 0~ Then flfl=z ;[fmu+ffd(z zfi+£§%%z»%f+u] This shows that g has a simple pole at 20 , with residue f (20). (b) Suppose, on the other hand, that f (2:0) m 0. Then 1 I 0 rr 0 8(Z)$ zwzif if )(z~zo)+i—§~Z(z—zo)2+..,] =ffio)+f2(fe)(z_,20)+m (O<Ez-ZUE<RD).\ Since the principal part of g at 20 is just 0, the point z x 0 is a removable singular point of g. 4. Write the function _ Saaz2 f(z) -~ (z; + a2? (a > 0) as 3 M2) .. 81513252 fig) (2: ~ at“)3 Where (ME) - (z + at)3 " Since the only singularity of ¢(z) is at z = "ai, 45(2) has a Taylor series representation ¢(z)-- ¢(ai)+ ¢(a i(z) —ai)+ ¢”2(a )(z~ ~ai)z + (Exeaii-(Za) about z z ai. Thus f(z):.~ 1.3[¢(ai)+¢(a1)(z ai)+¢”(m)(z of) +- ] (Miranda). (rat) 1 2. :;- 6 4- __ 3 2 ___ - 2 I a 12 8a :5 and ¢"(z)—— 16c13(z2 4azz a )- (z +ai)5 103 Eonsequently, Mann—air, ¢’(ai)m~%, and ¢”(ar)m~i. This enables us to write 1 (Zeal)3 HZ)“ [“ali“§(z~ai)—%(ZWCIE)2sun] (0<lz—ai!<2a). The principal part of fat the point z = of is, then, i/Z a/Z azi SECTION 67 2 1. (a) The function f(z) = 2 +2 99(3) 1 has an isolated singular point at z :1» Writing f(z) 3 “mi. 7. - Z " where 93(2) m 22 +2, and observing that (25(2) is analytic and nonzero at 2 =21, we see that 2 =1 is a pole of order m =1 and that the residue there is B a $0) m 3. (b) If we write 3 2 (15(2) z} m n M: h = my flit) (234.1] I: ( 1)], w ere 915(5) 3 z t 2 we see that z = mi» is a singular point of 1? Since (M25) is analytic and nonzero at that point, f has a pole of order m m 3 there. The residue is an mum) m 3 21 16" (c) The function exp z exp 2 22m! u(z-—m‘)(z+m') has poles of order m :1 at the two points z = inf“ The residue at z a m' is and the one at zx—Iri is 104 U4 2. (a) Write the function m) = :1 (52b 0, 0 < argz < 22:) as Z f(z) m 533, where ¢(z) a z‘” = aim“ (Ezl> 0,0 < argz < 27:)" The function ¢(z) is analytic throughout its domain of definition, indicated in the figure below“ Also, 1 t . —log(—1} ~—(Iui+in') . 7! fl- 1+ I (—1)m(—“1)m m€34 =e" :e‘m mcos-+isin~:~——~¢O, ‘0 a This shows that the function f has a pole of order m z 1 at z m —1, the residue there being 1+‘i B x “"1 z: —-—-_ ¢< ) ‘5 . . . Log: b Wnte the functxon z a W as ( ) f ( ) (2:2 +1); (NZ) Logz . 2) == . where z ..—.. . fl (Z"£)2 ¢( ) (z-H‘)2 From this, it is clear that f(z) has a pole of order m=2 at z =i. Straightforward differentiation then reveals that Res Logz — qb’(i) _ z=i (6+1)1 “ 3 105 (c) Write the function 112 :5 17(2):??? (lz|>0,0<argz<27r) as (ME) 2‘” 2 =m where 2: .-.~.———-—-—-_ f( > (H); M ) (m): Since . _ (z+i)z‘”2 __4zuz lb (‘7‘) 2(z+i)3 and my: __ “mm "J." J...“ H: mm .... 1 l [e Z‘fi’lwemzmkz’ U2 . z 1..., Res 2 ’1' mm, (at-ii)2 ‘3 0 3 2 3. (a) We wish to evaluate the integral 3 I 32 «:2 dz, C (z“1)(z +9) where C is the circle I: - 21 = 2, taken in the counterclockwise direction. That circle and the singularities 2: =1, i3t’ of the integrand are shown in the figure just below. Observe that the point z = 1, which is the only singularity inside C, is a simple pole of the integrand and that R 32:3 + 2 3Z3 + 2 1 68W 2 ”1,—- x w. ml (Eh-1X2. +9) z +9 :21 2 According to the residue theorem, then, 323+2.(1) . W :2 ... : ~ Jc(z-1)(zz+9)dz 7” 2 7“ 106 (15) Let us redo part (a) when C is changed to be the positively oriented circle Izl m 4, shown in the figure below. In this case, all three singularities 1:1, $3i of the integrand are interior to C. We already know from part (a) that 323 + 2 1 Res-—*—2-—"— " "”~ z=i (z —1)(z + 9) 2 It is, moreover, Straightfbrwa‘d to show that esmfiflimu 3334.2 _15+49i ma.- (2 m 1x6 + 9) (z -1)(z + 30 W3.- 12 and Res 3z3+2 =__3E'E_2-_.__. WISH—491. 1;"31. (z — 1)(Zz + 9) (Z W 1)(Z m 3i) gun-41' 12 ' The residue theorem now tells us that z —- + (poo-2+9) z ,2 12 12 3+2 . ' m ' chflgmd [1 154-49: 15 491):6m', 4. (a) Let C denote the positively oriented circle 1zE= 2, and note that the integrand of the dz integrai I has singularities at z m 0 and z 3 ~61. (See the figure below.) C2.3(2:~i»4) To find the residue of the integrand at z n O, we recall the expansion and write Mimi; Him 23...: 33am) 42.3 1+(z/4) 423 . 1 Now the coefficrent of «- here occurs when n = 2, and we see that Consequently, 107 (12k: 1) (0 <le< 4). (17) Let us replace the path C in part (a) by the positively oriented circle I: + 2! z 3, centered at -—2 and with radius 3. It is shown below. To find the residue at —4, we write 1 _ 45(2) 1 W"z—~c-—4)’ Where “2):?” This tells us that z = "4 is a simple pole of the integrand and that the residue there is M" 4) = ”-1 l 64“ Consequently, dz . 1 l ———.—.-2 —-~--- = . Cz3(z+4) ”{64 64) 0 108 . coshn‘ d . . . . . 5. Let us evatuate the integral f WTiE, where C 15 the posmvely ouented Circle lem 2. C z(z +1) AH three isolated singularities z : Oil" of the integrand are interior to C" The desired residues are “:1, cosh fiz cosh 7:3] Res 2 m 2 ==° 22(z +1) 2 +1 1:0 cosh 11:2: 6081} 712 1 Res 2 m , = "a at! z(z + 1) 2(2: +16) ,7, 2 and cosh m cosh m5] 1 Res 2 = . 3 '" z=~r z(z +1) 25(2' “ I) We; Consequently, Iwmzni(l+i+ij=4m‘ C z(z +1) 2 2 6. In each part of this problem, C denotes the positively oriented circle Ezlm 3" (a) It is straightfomaxd to show that ., (sin—2)2 1 (1) (3+22)2 If = M, th —— — z w, fig) z(z -1)(2z + 5) en z2 f z z(1-- z)(2 + 52) . . 1 1 . This functlon “I f - has a sunple pole at z m 0, and z z 2 J' M dz = 2751' ReS|:";i'f('£):l L" 2751(2) m 97d“ C z(z "1)(22+5) 2:0 :5 z (b) Likewise, .. : 23(1~32) "I“ (i): z--3 If”) (1+z)(1+2z4)’ the“ 22f z z(z+l)(z4+2)' The function “lg-f (—1-) has a simple pole at z 2 0, and we find here that z z ‘ 3 f i jJ—gj334 dz=2mRes[-1—;~f[i)]=2m(—-—3~)=—3m. C(1+z)(1+2z) F 2 z i 2‘ ...
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