06Bhw4solution

06Bhw4solution - $9 Ave/7% W} 356% SECTION 69 1. (a) Write...

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Unformatted text preview: $9 Ave/7% W} 356% SECTION 69 1. (a) Write 1 Hz) . csc =—--—-—-—=——, whet =1 d = , z Sing gm e [2(2) an q(z) smz Since P(0)=1¢0, 4(0)mSiflO=0, and q’(0)=-cosO=1;¢{}, z = 0 must. be a simple poke of cscz, with residue 3:92.. .._. ._1_ :1 q’( ) 1 (b) From Exercise 2, Sec“ 61, we know that ,. 3. 3. I “a; a esez---z+3fiz~i~[(3f)2 5!]: +~~ (O <Ezi< 77:). Since the coefficient of —1- here is I, it fofiows that z m 0 is a simpie pole of cscz, the‘ z residue being 1. 110 2. {0) Write zu-sinhz = p(z) , where m m ' h z 2 ' . zzsmhz (1m 10(2) 2: 5m 2 and q(z) z smhz it Since p(7ri) = m" ¢ 0, q(m’) : 0, and q' (7131‘) m 752 :5 0, it follows that Ra$z;emhz a Mm? Fm z smhz q'(m) 7: 11: Mi ""5" (b) Write t 6:513: : 5.5% Where 17(3) 2 “13(3) and (1(2) = sinhz, It is easy to see that expert) m pew) m , muff—TWCX ii?!" and R68 — ~' 1“. 2"” sth q (m) p( ) z=-m' sinhz Q'Cwm') MM m ) Evidently, then, RES eX-PCZI) + Res. CX-pC’it) : m2 exp(zm)+exp(—mt) : _ZCOS 7”. PM 8111112: WW smhz 2 3. ((1) Write f (E) “-"' "gag, Where ME) 3 z and q(z) : 0032:, 9(2) Observe that It ‘1(”2“"+mr)=0 (n m0,i1,fl,...)n Also, for the stated values of n, 75 717 p[£+””) = "3+ "7‘ i 0 and 61("2" + 127:) m msin(—i— +1175) 2: (“*1)"+1 :5 0" 2 112 4. Let C be the positively oriented circle tel: 2, shown just below. (a) To evaluate the integral L: tanz dz , we write the integrand as tanz = Pig—l, where 10(2): sinz and q(z) = (3052, W) and recall that the zeros of cosz are 2 2 25+ mt (n = 0,i1,i-2,..,). Only two of those zeros, namely 2 3int! 2, are interior to C, and they are the isolated singularities of tanz interior to C. Observe that Res tanzm pot/2) mwl and Res marw:m1n “"12 Mir/2) We” q’(~ Int/2) Hence Jctanzdz m Zm'(—-l - 1) = —4m'. (b) The problem here is to evaluate the integral , dz . To do this, we write the C smh 22 integrand as _ I =EQQ, where p(z)=1 and q(z)flsinh22:. sthZ q(z) Now sinh 2:: = 0 when 22 2 nm’ (n 2 0,i1,fl,r.n), or when 2:331— (n=0,i1,i2,...). Three of these zeros of sinh 22, namely 0 Midi—1g, are inside C and are the isolated singularities of the integrand that need to be considered here" It is straightforward to show that 1:&_11 Res , .— - , ==0 sthz q’(0) ZooshO 2 ’ 113 I “(xi/22L 1 1 1 qwm/z)"m=mm? and Resw:mmimi_=__im=mi mellsinhZZ q’(-~wm'/2) 2cosh(-m€) Zcos(--7r) 2' Thus 1 . Within (3”, the function —-2—,—— has isolated singularities at z 8111:: 2:0 and z=imr (n31,2,...,N). To find the residue at z 2 0, we recall the Laurent series for 0802: that was found in Exercise 2, Sec. 61, and write =_1?+_1_._1.+[m1m_,1_]z+... (O<Izi<7r). 6 z . 114 . This tells us that 2 1 , has a pole of order 3 at z = 0 and that z 3111: Res 21_ ==° z smz 6 As for the points z =1tmt (n = 1,2, ,N), write 1 = p(z) zzsflnz q(z)’ where p(z) =1 and q(z) = 22 sinz. Since Mimic) m1 at 0, inmt) m 0, and q’(:i:mz:) m 1227:! cos mt m (——1)"nz:'r2 ad 0, it follows that l 1 (-I)" ("-1)" "MI-aw il Res 2 _ , z=tmrz smz ("1) n 7: («1)” 1221sz So, by the residue theorem, d .1 N—N JCN‘ z dz=2m[g+22(2 )2]. .22 sin 2 Rewriting this equation in the form ikflimfimfit & ":1 n2 12 41' szzsinz and recalling from Exetcise 7, Sec. 41, that the value of the integral here tends to zero as N tends to infinity, we arrive at the desired summation formuia: 2 z n 12 ~ we 1:: 6. The path C here is the positively oriented boundary of the rectangle with vertices at the points :2 and i2 + i. The problem is to evaluate the integral J- dz W6~m+i 115 The isolated singnlaxities of the integrand are the zeros of the polynomial 4(z) m (212 _1)2 + 3- Setting this polynomial equal to zero and solving for zz, we find that any zero 2: of q(z) has the property 22 =1$wf§i . It is straightforward to find the two square roots of 1+-\/§i and also the two square roots of 1— fit”. These are the four zeros of q(z). Only two of those zeros, «5-H Zozfiemi6= J5 and ~§o=~fiewiaffi=”\f3—+l .fi’ lie inside C. They are shown in the figure below. To find the residues at 2,, and we, , we write the integrand of the integral to be evaluated as ngg' Where M3313“ (1(Z)m(zl—1)1+3, This polynomial 9(a) is, of course, the same q(z) as above; hence q(zo) m 0. Note, too, that p and q are analytic at an and that p(zo) e 0. Finally, it is straightforward to show that q’(z) = 42(22 "1) and hence that q’(zo) 2 4.20053 — 1) m “Q's/6 +6J'i: a 0, We may conclude, then, that 20 is a simple pole of the integrand, with residue 9(20) 2 1 5112,) 446 + (S‘s/5i" Similar results are to be found at the singular point «fl, To be specific, it is easy to see that {IT—Eu) m ~q’(i'"o) = "—q’(zo) = 2J6— + é-fii :5 0, the residue of the integrand at “2}, being P(“fo) : 1 n q’(~:z",) Mme/'2? 116’ 8. Finally, by the residue theorem, “MAE” EM[W£WW+W) bi C(z2 — 132 +3 #245 +6f2‘i 246 + 6451' 2J5 ' We are given that f (z) w 1/ [q(z)}2, where q is analytic at 20, q(zo)=0, and q’(ze)¢0. These conditions on q tell us that g has a zero of order mm} at 20" Hence q(z) : (z - 29 )g(z), where g is a function that is analytic and nonzero at 30; and this enables us to write ¢(z) __ 1 )2, where ¢(z)-— (“any So f has a pole of order 2 at 2:0, and w 23130) [3(30 )]3 . Effie.) m ¢’(ZO) = But, since (1(2) 3 (z - Zo)g(z), we know that q’(z) = (z - zg)g’(z) + 3(2) and q”(z) m (z — zo)g”(z) + 2g’(z)- Then, by setting 2: m 20 in these last two equations, we find that 4’(Ze)=s(20) and q”(zo)=23'(zg)w Consequently, our expression for the residue of f at 20 can be put in the desired form: ‘1”(30 ) R m *- . gm) [attain]3 (a) To find the residue of the function (:502 2: at z m 0, we write 1 WE)? ’ (:302 z = where (1(2) 7- Sine, Since :1 is entire, (1(0): 0, and q’(0) m 1 e 0, the result in Exercise 7 telis us that Res (:302 z = u q, (0)3 F“ [q (0)] 118 SECTION 72 , we integrate the function f(z) == z2+1 1. To evaluate the integral g! 2:"1 +1 closed contour shown below, where R > 1. Weseethat dx dz + =2 .3, .‘LxN-l fizz-H m where BzRe‘s 11 535.5 1 ="L”] mi” 3:: 3 +1 z=t (z—z)(z+l) Z+i an: 2i Thus R dx d I: 5M :2 _Rx +1 Caz +1 Now if z is apoint; on CR, tzl +1i21szz-m11m R2 —-1; around the simple 121 xzdx We use the simple closed contour {x2 + 1)(x2 + 4) 4. We wish to evaluate the integral J 0 shown below, where R > 2. We must find the residues of the function g x z t ' - f ( ) (124.3002er 4) 3 Its mmple poles z=i and 2:213 They are z2 1 B =Res (z =~—————————---- my... ‘ f ) (42+:sz +4)L 6i and 2 1 =Res (new mm‘ Bl :in f (I:2 +1)(z+2i)]z=2i 3i Thus R 2 2 x dx 3 dz + W: ' £sz + I)(x2 + 4) '[Ca (2:2 +1)(z2 +4) zmwl +32% or RH. xzdx ZZ “(xi —I~l)(.x2 + 4) 3 Ca (:2 -§~1)(z2 + 4) ' If 2: is a point on CR, then fizz +1E2Itz12~1| = R2 w: and I2”1 +4123izF-«4I m R2 "4a Consequently, 2 3 if. I 2 z a": s 2 “R ngugOast-aoo' cfz we +4)! (R ~1)(R~“--4) 1...“: 1 .4 ’ R1 — R2 and we may conclude that (x2+1)(.x2+4) 3' or a?+l)(x2+4) I xzdx 717 xzdx 123 7, In order to show that V? xdx m n: ' (x2+1)(x2+2x+2)“ 5’ «no we introduce the function Z flz) m (z2 +1)(z2 + 22: + 2) and the simple closed contour shown below. Observe that the singularities of f(z) are at i, 20 =-1+i and their conjugates mi, 2‘0 x —1--i in the lower half plane. Also, if R > J5, we see that R j f(x)dx+jc f(z)dz:2m'(Bo +31), -3 R where 1 3 B mRes mflz =----+--—-' 0 m) (21+1)(z-fu)]z= 10 10' and z 1 I BmR 2W "gm-mm" I if”) (z+i)(zz+2z+2):|z=i 10 5‘ Evidently, then, f W “22.4 mid—3M "R(.x2+1)(.x2+2.x+2) 5 c«(z2+1)(zz+2z+2)' Since zdz _ zdz 11:}?2 I 2 2 "J. Mg 2 2"3'0 Cu(z +1)(Z +2z+2) “(Z +1)(z-ZO)(zhfo) (R “IXR-"a/E) as R we no, this means that R 111“ W = - “"40: +1)(x +2x+2) 5 This is the desired result. xdx 7t 124 8. The problem here is to establish the integration formula J- ? 2:31:— using the simple 0 x +1 3J5 closed contour shown below, where R > 1. There is oniy one singularity of the function f(Z) = 3: 1, namely z{3 m em”, that is interior 2 to the closed centour when R > 1“ According to the residue theorem, I :13 +1 :13 4- dz =2m‘Res 1 Q: +1 Caz +1 C: ! z3+l 2% z3+l where the legs of the closed contour are as indicated in the figure. Since C1 has parametric representation 2: = r (0 S r ..<_ R), I dz 3:?" dr ' Cl 5+1 9 r3+1, and, since ~C2 can be represented by z : mm”3 (0 S r S. R), R than R L- dz m”chz3d:1=~JM——mmeizmaj d!" 223+: “(re-WHMI" or344" Furthermore, 1 1 1 Res 2 -—- 2 2:70 ZS +1 3612)!” Consequently, R v dr Zm‘ dz 1__ 121143 3 - _ ( e r3 36121!” C 23+]. But dz 1 211'}? S —---—--- no Liza“ R3m1 3 ’ioaSR") This gives as the desired result, with the variable of integration r instead of .x: _. .— ra +1 3(ei2m'3 w(franc/:3 ( g-sam) _ 3(ei21fl3 wean/'3) : 35inwfil3) 3J3“ 7 dr 2m 2m“ 7: 2;: G 129 This enables us to see that lq(z)l 2 (R-wlzol)4 when 2 is on CR. Thus 1 < mg [91(2)]2 ‘— (Rel-Zn!)a for such points, and we arrive at the inequality 7: I 1 2 dz 5 “Emma, =WJ:Z____S_, Ca [61(2)] (R-lzoi) [ 4591) R which tells us that the value of this integral does, indeed, tend to 0 as R tends to co. Consequently, w "_ - 2 p_VNJ_?m£2mmT=_’E§,Im Wfl_ w[(x ma) +1} 4A 20 But the integrand here is even, and Im -a+i(2a2+3) Mm“? wa+i(2a2+3) _-JA+a_iJAma go mum 1??- s/m " So, the desired result is °° dx “(xi-“afar = edgy [(2“2+3“A+“+“A‘“al' 9 where A = V02 +1. SECTION 74 cosx dx (x2 + azXJt:2 + £22) “mg—w (ZZ + alxz‘l +172) inside the simple closed contour shown below, where R > a, The other singularities are, of 1. The problem here is to evaluate the integral I , where a > b > 0. To do this, we introduce the function f(z) = , whose singularities of and bi lie coarse, in the lower half plane. 130 According to the residue theorem, R . fab: . wmwm——-~m+ “ = ' £sz +az)(x2+b2) iflae dz 2m(B‘+BZ)’ where B 31233 HI: :—-_'£I:—-—'——* :—-————e—:n—— 1 F“ UTE) ] (z+ai)(zz+b2)]mi 2a(b2v-a2)i and Ez _& B :ResL (z e“ :___——L—.——— x ‘3 _, 2 I?“ f ) 1 (£2 +a2)(z+bi):lzahi 2b(a2 That is, R . emdx *— fl ewb emu . fz £(x2+a2)(x2+b2)ma2mbi a )wé‘lf(Z)e dz: or R -1, .- cosxdx 7!: [a e ") “Mm-"m --~ “Re f(z)e"dz L(x2+a2)(x2+b) (12wa b a 6’; Now,ifzis apoint on CR, Ef(z)l$MR where MR 2 1 (R2 ma2)(R2 *b2) and ieizim 8" 51. Hence 75R W122 maZXRZ fibz) -~> 0 as R m} cm“ LR f(z)e‘zdzl S M Rer = Re ICE f(z)e"zdz| S So it follows that (a>b>0). on 2. This problem is to evaluate the integral [63361: air, where £120" The function 0 x + 1 . . . . f (z) = 22 +1 has the smgulantles it; and so we may integrate around the simpie closed contour shown below, where R > 1" 131 We start with R fax 3 . dx+ ““d 22 ' ixhrl iflzk 2: 15:13, Where in: em: emu BxRes[f(z)e ]=——: m _. == 2 I Fl 21 Hence R efar ‘ LX2+1dx=fle "6’;ka de, or fCOdedxm flea-a _Re‘[f(z)eiazdz _R at1 +1 CR ' Since If(z)is MR where MR 2 21 , R --1 we know that - v 717R Re “dz S “(1 S ; iftzk f(z)e 2. R2 _1 and so I gfldx = 72'2"“. m x +1 That is, ” cosax 7?: dx = m "“ a 2 0 ‘ sz +1 2 e ( ) D . “ xsin2x . . . To evaluate the mtegral I 75—h, we first Introduce the functton x Z Z z2+3 (zmzlxzm'z'fl' f(Z) = where 2.1 m «Iii. The point 21 lies above the x axis, and 2'1 lies below it. If we write f(z)ei2z = 45(2) where ¢(z)=26xp(i2z) z-q 3mg ’ 132 we see that 2:1 is a simple pole of the function _f(z)e‘2‘ and that the corresponding residue is . ' F- ‘ amenawzw' Now consider the simple closed contour shown in the figure below, where R > V? Integrating firs)?“ around the closed contour, we have ' R xerz: I Thus ffimxfimhemmwmjjnwhe "R .x + 3 C3 Now, when 2 is apointon CR, I_f(z)l$ MR, where MR 2 RZR —->0 as R—éoo; and so, by Iimit (1), Sec. 74, fire L f (z)e‘2‘dz = 0. Ram-9 Consequently, since 5 lImJ'Cfl 10(2)?” dz Lflmmfl, we arrive at the result no on xsinx m _ xsinx __ at I x2 +3 dx —- flexp(~—2\/§), or Ix: +3 dxnaexm-Zxfi). .4, 0 135 Now if z is a point on CR, then R Ef(z)l$ MR where ME =: (R2 “MR2 m9) as R ——> 00‘ So, in view of limit ( 1), Sec. 74, S 'ImJC f(z)ei‘dz ~—->0 as Rm) on; LR f (z)e"‘dz and this means that I xasinxdx m£(§__1) or]: .xssinxdx _i[2___1) ennui—+9) Se e"- ' 0(x2+1)(x2+9)_168 e2 a” in.” m sinxdx x2 +4x +5 9. The Cauchy principal value of the integral I can be found with the aid of the 1 z2+4z+5 Using the quadratic formula to solve the equation zz+4z+5m0, we find that f has 1 (Z*ZI)(Z_EI) function f(z)-~ and the simple closed contour shown below, where R>-\/§. singularities at the points zI m ~2 +1? and fl = "-2 -i. Thus f (z) a , where zE is interior to the closed contour and E; is beiow the real axis. The residue theorem teHs us that R v e“dx - m4. '3 a ' if +4x+5 LRszk dz 27:13, where 81.: 651] B 2 Res = --——----—; W: (z ~zl)(z—2"1) (2:i --%”5) and so if sinxdr m Zm'eiz' x2+4x+5 wR 136 01' sinx dx 7—-—— ~£sin2-lm x +4x+5 e r. LR f(z)e"‘dz. Now, if zis apointon CR, then ieizlze’y $1 and 1 lf(z)iS M"R where W = 1 M = W" R (Ia—«Bf Hence 717R .<.. W R m, WWW/5240215 ~> Ln f(z)ei‘dzl S Mam? = llm [CR f(z)ei‘dz and we may conclude that RV”! 2511M612: x +4x+5 J'L‘ . ---~sm2. e (x+1)cosx 2 dx, we shall use x +4x+ 5 10. To find the Cauchy principal value of the improper integral I 3+1 z+1 2:2 +4z+ 5 (z "31)(2 "‘21) same simple closed contour as in Exercise 9. In this case, MZ—vl, and the the function f (z) = , where zi m —2 H, and f: R . (x+1)e“dx E. , w+ =d m2 3, _sz+4x+5 Jcnf(2)e Z m where BER“ (2+1)?z :(zi+1)eiz' m(-—1+i)e"2‘ H (z-"lezw-El) (2"?!) M ' Thus R (x+1)cosx _ _ _ 1.: -g x2 +4x+5 dx m Rgcsz) Laflzk ’ or R (x+1)cosx 7r . . dez-m sn2—c 2 w "d" ix2+4x+5 (3(1 OS) J‘cnflz)‘; 2 Finally, we observe that if z is a point on C R, then R+1 R+l (RuizllXR—dfl!) = (12—4332 ""9 0 as R ""9 °°‘ If(z)[$ MR where MR= ...
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This note was uploaded on 12/27/2011 for the course MATH 5B taught by Professor Rickrugangye during the Fall '07 term at UCSB.

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06Bhw4solution - $9 Ave/7% W} 356% SECTION 69 1. (a) Write...

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