Math 656 March 10, 2011 Midterm Examination Solutions

Transcription

1 Math 656 March 0, 0 Mdtrm Eamnaton Soltons (4pts Dr th prsson for snh (arcsnh sng th dfnton of snh w n trms of ponntals, and s t to fnd all als of snh (. Plot ths als as ponts n th compl plan. Mak sr or ponts agr wth th prod of th hprbolc sn fncton. w w w w w w = snh w = = 0 w = 0 { } { } { } s s = 0 s = + ( + w = snh = log + ( + / / k k / snh ( = log + ( 4 + = log ± 3 = ln( ± = ± ln( s s ln(+3 / / ln(+3 / Ths agrs wth th prod of hprbolc sn fncton, whch s

2 (5pts Show that mags of rtcal lns ndr transformaton w = cos ar hprbolc = ±. What shap ar mags of horontal lns? What s th angl btwn ths two sts a b of crs? w = cos = cos( + = cos( cos( sn( sn( = cos cosh sn snh. Imag of rtcal lns: = c +, < <, c s fd w = cos c cosh sn c snh Us dntt cos sn cosh snh = : = c c A hprbola (nlss cos c = 0 or sn c = 0 If cos c = 0, th mag s th -as (s fgr. Whn sn c = 0, th mags ar ral: (, or (, +. Imag of horontal lns: = + c, < <, c s fd w = cos cosh c sn snh c Us dntt cos + sn = : + = An llps (or a ln [-,] f snh c = 0 cosh c snh c w = cos / / Ths s a conformal mappng, whch prsrs angls, so mags of horontal and rtcal lns ar orthogonal whrr (cos ' 0, snc th horontal and rtcal lns ar orthogonal.

3 3. (6pts Is th fncton f( = / contnos for all? Is t dffrntabl anwhr? Is t analtc anwhr? Pro or answrs drctl (sng lmts, and rf or answr abot analtct sng ach-rmann qatons. f ( s contnos whrr ts ral and magnar parts ar contnos f ( = = = R f (or, n polar form = = = cos( θ + sn( θ f ( r θ / r θ θ and Im f ar contnos apart from pont = = 0, whr f ( s not n dfnd, so th onl dscontnt s = 0 Now lt s amn dffrntablt: Th smplst wa to show non-analtct n ths cas s to dffrntat n polar coordnats (s homwork #4, bt w can also s th sal artsan rprsntaton: df f ( + h f ( + h ( + h ( + h h h = lm = lm = lm = lm d h 0 h h 0 h h h 0 h ( h h h ( + h Lt's tak th drat along th horontal drctons: h = h = 0 df Im = lm = = d 0 ( + ( ( Now lt's dffrntat along th rtcal drcton: h =, h =, 0 df + + R = lm = = 0 d ( ( ( Ths two prssons ar nr qal (not that =0 s otsd of doman of dfnton Also, lt s chck ach-rmann qatons: ( + ( 4 = = = + ( + ( + Eqal whn ( + ( = = = + ( + ( + ( + ( 4 = = = + ( + ( + Eqal whn ( + ( = = = + ( + ( + Both ach-rmann qatons cannot b satsfd 3 0 = or = or 3 = = 0 Not: t s asr to s polar -R qatons: r = θ /r, r = θ /r (hr =cos(θ, =sn(θ, so r = r =0, whl θ = sn(θ and θ =cos(θ, and th polar -R qatons ar not satsfd

4 4 (6pts Fnd all branch ponts of tanh = log +, and fnd th branch ct for an branch choc for ths fncton (hnt: ths s smplr than ampls w sold n class. How mch dos th fncton jmp across th branch ct(s? + = + Snc log [ log( log( ] t s clar that th branch ponts ar = and = (not thogh that abo logarthmc dntt wll not hold for all branch chocs Not that = s not a branch pont snc plggng n = / t lds + / t log t + = log / t t It s as to fnd a branch of ths prsson that s contnos at t=0 (tak for nstanc th prncpal branch of logarthm of (t+/(t W can fnd a branch of ths mlt-ald fncton sng an of th thr mthods w larnd n class: METHOD : hoos + log = log( + log( log p ( + log p ( + log( log p (+ has a brnch ct (jmp b along (, log p ( has a brnch ct (jmp b along (+, Sbtractng th two cancls th jmp rwhr cpt th ct (,+, whr th fncton jmps b. + METHOD (mappng hoos + + log = log p Prncpal branch has a ct whn >0. Solng for lds =. As ars from 0 to nfnt, + cors ral ntral (,+. Fncton jmps b as ths branch s crossd from abo. METHOD 3 (Indpndnt paramtraton of factors r p( θ + + r + θ θ log = log = ln +, 0 < θ, θ < r r p( θ Fncton jmps b as th branch s crossd from abo BRANH DEFINITION θ θ =0 θ θ =0 θ =0, θ = θ θ = θ θ =+ θ =, θ = θ θ =0 θ θ =0

5 5 (6pts Us paramtraton to show that th followng ntgral s ro or an crcl arond th orgn: t t t t t t + R t d dt dt t ( dt R R 0 R = = + R R = + R R = + R = = 0 Dos t follow that th ntgrand has an ant-drat rwhr n th doman /{0}? Dos t follow that th ntgrand s analtc n ths doman? Eplan or answrs. No, nthr of th abo statmnts s tr n ths cas, snc th ntgral ma wll b non-ro along a non-crclar Jordan contor (on can tr for nstanc a rctanglar closd path d 6 (6pts Wthot rsortng to paramtraton, calclat along two dffrnt contors: + a = an contor from = to = not contanng snglarts of ntgrand Frst, not that th ntgrand has two snglarts at Ths ntgral dos dpnd on th contor choc. If thr ar no snglarts btwn th contor and th ral as (s fgr, w can dform th contor to th ntral of th ral as, along whch th ntgral s ro snc /( + s an odd fncton. W can also s th Fndamntal Thorm of alcls; f th contor dos not ntrsct th branch ct of th ant-drat log( +/ (o had to mak ths commnt for fll crdt w obtan:, / / / / 4 = ( = ± ( = ± = ± + b-cts of log( +/ d + log( log( log( 0 = + = + + = Optonal: ths s not tr f thr s a pol btwn th contor and th ral as (s bottom fgr. In ths cas w can clos th contor along th ral as (whch dos not contrbt to th ntgral accordng to abo, and s th ach Intgral Formla: d d = = = = + + / + Not that ths qals th jmp of th ant-drat across ts branch ct (maroon dashd ln b = crcl of rads arond th orgn n th post drcton Us th ach Intgral Formla: d d + = ( ( = + =

6 7 (5pts Paramtr th ntgral Formla to show that α d (whr α s a ral constant and s th ach Intgral = α cos θ cos( α sn θ dθ =. 0 B ach Intgral Formla w ha = α α 0 d= = Also sng paramtraton = t, w obtan α p( θ θ p( cos + sn cos sn = α θ = α θ α θ = α θ α θ θ dθ dθ dθ dθ α cosθ α cosθ α cosθ { } = cos( α sn θ + sn( α sn θ dθ = sn( α sn θ dθ + cos( α sn θ dθ Now, th ral part of ths ntgral s ro (t s an odd fncton, whl th magnar part has to qal b th ach Intgral Formla, whch gs s th fnal rslt 0 α cos θ cos( α sn θ dθ = Snc th ntgrand s n wth rspct to θ=, w can dd th ntgraton ntral b half, ldng th fnal answr of