Stein-Complex-Analysis-答案---复分析答案

SOLUTIONS/HINTSTOTHEEXERCISESFROMCOMPLEXANALYSISBYSTEINANDSHAKARCHIROBERTC.RHOADESAbstract.ThiscontainsthesolutionsorhintstomanyoftheexercisesfromtheComplexAnalysisbookbyEliasSteinandRamiShakarchi.IworkedtheseproblemsduringtheSpringof2006whileIwastakingaComplexAnalysiscoursetaughtbyAndreasSeegerattheUniversityofWisconsin-Madison.Iamgratefultohimforhiswonderfullecturesandhelpfulconversationsaboutsomeoftheproblemsdiscussedbelow.Contents1.Chapter1.PreliminariestoComplexAnalysis22.Chapter2.Cauchy’sTheoremandItsApplications83.Chapter3.MeromorphicFunctionsandtheLogarithm94.Chapter4.TheFourierTransform105.Chapter5:EntireFunctions116.Chapter6.TheGammaandZetaFunctions137.Chapter7:TheZetaFunctionandPrimeNumberTheorem178.Chapter8:ConformalMappings209.Chapter9:AnIntroductiontoEllipticFunctions2310.Chapter10:ApplicationsofThetaFunctions25Date:September5,2006.TheauthoristhankfulforanNSFgraduateresearchfellowshipandaNationalPhysicalScienceConsortiumgraduatefellowshipsupportedbytheNSA.12ROBERTC.RHOADES1.Chapter1.PreliminariestoComplexAnalysisExercise1.Describegeometricallythesetsofpointszinthecomplexplanedefinedbythefollowingrelations:(1)|z−z1|=|z−z2|wherez1,z2∈C.(2)1/z=z.(3)Re(z)=3.(4)Re(z)c,(resp.,≥c)wherec∈R.(5)Re(az+b)0wherea,b∈C.(6)|z|=Re(z)+1.(7)Im(z)=cwithc∈R.Solution1.(1)Itisthelineinthecomplexplaneconsistingofallpointsthatareanequaldistancefrombothz1andz2.Equivalentlytheperpendicularbisectorofthesegmentbetweenz1andz2inthecomplexplane.(2)Itistheunitcircle.(3)Itisthelinewhereallthenumbersonthelinehaverealpartequalto3.(4)Inthefirstcaseitistheopenhalfplanewithallnumberswithrealpartgreaterthanc.Inthesecondcaseitistheclosedhalfplanewiththesamecondition.(5)(6)Calculate|z|2=x2+y2=(x+1)2=x2+2x+1.Soweareleftwithy2=2x+1.Thusthecomplexnumbersdefinedbythisrelationisaparabolaopeningtothe“right”.(7)Thisisaline.Exercise2.Leth·,·idenotetheusualinnerproductinR2.Inotherwords,ifZ=(z1,y1)andW=(x2,y2),thenhZ,Wi=x1x2+y1y2.Similarly,wemaydefineaHermitianinnerproduct(·,·)inCby(z,w)=zw.ThetermHermitianisusedtodescribethefactthat(·,·)isnotsymmetric,butrathersatisfiestherelation(z,w)=(w,z)forallz,w∈C.Showthathz,wi=12[(z,w)+(w,z)]=Re(z,w),whereweusetheusualidentificationz=x+iy∈Cwith(x,y)∈R2.Solution2.Thisisastraightforwardcalculation12[(z,w)+(w,z)]=12(zw+wz)=Re(z,w)Re(zw)=12((z1+z2i)(w1+iw2)+(w1+iw2)(z1−iz2))=z1w1+z2w2.Exercise3.Withω=seiφ,wheres≥0andφ∈R,solvetheequationzn=ωinCwherenisanaturalnumber.Howmanysolutionsarethere?SOLUTIONS/HINTSTOTHEEXERCISESFROMCOMPLEXANALYSISBYSTEINANDSHAKARCHI3Solution3.zn=seiφimpliesthatz=s1nei(φn+2πikn),wherek=0,1,···,n−1ands1nistherealnthrootofthepositivenumbers.TherearensolutionsasthereshouldbesincewearefindingtherootsofadegreenpolynomialinthealgebraicallyclosedfieldC.Exercise4.ShowthatitisimpossibletodefineatotalorderingonC.Inotherwords,onecannotfindarelationbetweencomplexnumberssothat:(1)Foranytwocomplexnumbersz,w,oneandonlyoneofthefollowingistrue:zw,wzorz=w.(2)Forallz1,z2,z3∈Ctherelationz1z2impliesz1+z3z2+z3.(3)Moreover,forallz1,z2,z3∈Cwithz30,thenz1z2impliesz1z3z2z3.Solution4.Suppose,foracontradiction,thati0,then−1=i·i0·i=0.Nowweget−i−1·i0.Thereforei−ii+0=i.Butthiscontradictsourassumption.Weobtainasimilarsituationinthecase0i.Sowemusthavei=0.Butthenforallz∈Cwehavez·i=z·0=0Repeatingwehavez=0forallz∈C.Sothisrelationwouldgiveatrivialtotalordering.Exercise5.AsetΩissaidtobepathwiseconnectedifanytwopointsinΩcanbejoinedbya(piecewise-smooth)curveentirelycontainedinΩ.ThepurposeofthisexerciseistoprovethatanopensetΩispathwiseconnectedifandonlyifΩisconnected.(1)SupposefirstthatΩisopenandpathwiseconnected,andthatitcanbewrittenasΩ=Ω1∪Ω2whereΩ1andΩ2aredisjointnon-emptyopensets.Choosetwopointsω1∈Ω1andω2∈Ω2andletγdenoteacurveinΩjoiningω1toω2.Consideraparametrizationz:[0,1]→Ωofthiscurvewithz(0)=ω1andz(1)=ω2,andlett∗=sup0≤t≤1{t:z(s)∈Ω1forall0≤st}.Arriveatacontradictionbyconsideringthepointz(t∗).(2)Conversely,supposethatΩisopenandconnected.Fixapointw∈ΩandletΩ1⊂ΩdenotethesetofallpointsthatcanbejoinedtowbyacurvecontainedinΩ.Also,letΩ2⊂ΩdenotethesetofallpointsthatcannotbejoinedtowbyacurveinΩ.ProvethatbothΩ1andΩ2areopen,disjointandtheirunionisΩ.Finally,sinceΩ1isnon-empty(why?)concludethatΩ=Ω1asdesired.TheproofactuallyshowsthattheregularityandtypeofcurvesweusedtodefinepathwiseconnectednesscanberelaxedwithoutchangingtheequivalencebetweenthetwodefinitionswhenΩisopen.Forinstance,wemaytakeallcurvestobecontinuous,orsimplypolygonallines.Solution5.Followingthefirstpart,assumeforacontradictionthatz(t∗)∈Ω1.SinceΩ1isopenthereexistsaballB(z(t∗),δ)⊂Ω1.Nowbyassumptionz(t∗+)∈Ω2.Thus|z(t∗+)−z(t∗)|δforall0.Butthisisacontradictionsincezissmooth.DefineΩ1andΩ2asintheproblem.FirsttoseethatΩ1isopenletz∈Ω1.ThensinceΩisopenandz∈ΩweknowthatthereexistsaballB(z,δ⊂Ω.WeclaimthatthisballisactuallyinsideofΩ1.IfweprovethisclaimthenwehaveestablishedthatΩ1isopen.Lets∈B(z,δ)andconsiderf:[0,1]→Cgivenbyf(t)=st+z(1−t).Then|f(t)−z|=t|s−z|δ.SotheimageoffiscontainedinB(z,δ)⊂Ω.Byconcatenatingthepathsfromwtozandztosweseethats∈Ω1.FinallywewillprovethatΩ2isalsoopen