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1、ConciseComplexAnalysisSolutionofExerciseProblemsAiShuXueMarch9,20081Contents1Calculus32CauchyIntegralTheoremandCauchyIntegralFormula143TheoryofSeriesofWeierstrass274RiemannMappingTheorem525Di®erentialGeometryandPicard'sTheorem536AFirstTasteofFunctionTheoryofSeveralComplexVar
2、iables537EllipticFunctions538TheRiemann³-FunctionandThePrimeNumberTheory532ThisisasolutionmanualofselectedexerciseproblemsfromConciseComplexAnalysis,SecondEdition,byGongSheng(WorldScienti¯c,2007).ThisversionsolvestheexerciseproblemsinChapter1-3,exceptthefollowing:Chapter1pro
3、blem37-42;Chapter2problem47,49;Chapter3problem15(xi).1Calculus1.Proof.See,forexample,Munkres[4],x38.2.(1)pProof.j2ij=2,arg(2i)=¼.j1¡ij=2,arg(1¡i)=7¼.j3+4ij=5,arg(3+4i)=arctan4¼0:927324¡¢3(Matlabcommand:atan(4=3)).j¡5+12ij=13,arg(¡5+12i)=arccos¡5¼1:9656(Matlabcommand:13acos(¡
4、5=13)).(2)310862¡3i514nnn+1nProof.(1+3i)=¡26¡18i.=+i.=¡i.(1+i)+(1¡i)=22cos¼.4¡3i554+i17174(3)p¯¯pProof.j¡3i(2¡i)(3+2i)(1+i)j=3130.¯¯(4¡3i)(2¡i)¯¯=p5¢p5=5.(1+i)(1+3i)2¢1023.pp¼11¡iµi4Proof.Letµ=arctan,®=arctan.Then5¡i=26e,1+i=2e4and(5¡i)(1+i)=p5239pi(¼¡4µ)4¡®i¼6762e4.Meanwhil
5、e(5¡i)(1+i)=956¡4i=6762e.Sowemusthave¡4µ=¡®+2k¼,4k2Z,i.e.¼=4arctan1¡arctan1+2k¼,k2Z.Since4arctan1¡arctan1+2¼>0¡¼+2¼=7¼>¼452395239244and4arctan1¡arctan1¡2¼<4¢¼¡2¼=0<¼,wemusthavek=0.Therefore¼=4arctan1¡52392445arctan1.2394.22Proof.Ifz=x+yi,1=x+yi.z2=x2¡y2+2xyi.1+z=1¡x¡y+2yi.z=
6、z¹x2+y2x2+y21¡z(1¡x)2+y2(1¡x)2+y2z2+13232x+xy+x+i(¡y¡xy+y)(x2¡y2+1)2+4x2y2.5.Proof.a=1,b=®+i¯,c=°+i±.So¢=b2¡4ac=(®2¡¯2¡4°)+i(2®¯¡4±)andp¡(®+i¯)§¢z=:26.222Proof.Denoteargzbyµ,thenz+r=reiµ+r=r(eiµ+e¡iµ)=2rcosµ=2Rez,andz¡r=zreiµz2reiµ¡r=r(eiµ¡e¡iµ)=2irsinµ=2iImz.reiµ7.3Proof.If
7、a=reiµ1andb=reiµ2,then12¯¯¯¯¯a¡b¯¯r¡rei(µ2¡µ1)¯¯¯=¯12¯:¯1¡ab¹¯¯1¡r1r2ei(µ2¡µ1)¯Denoteµ¡µbyµ,wecanreducetheproblemtocomparingjr¡reiµj2andj1¡rreiµj2.Note211212jr¡reiµj2=(r¡rcosµ)2+r2sin2µ=r2¡2rrcosµ+r2121221122andj1¡rreiµj2=(1¡rrcosµ)2+r2r2sin2µ=1¡2rrcosµ+r2r2:1212121212¯¯Soj1
8、¡rreiµj2¡jr¡reiµj2=(r2¡1)(r2¡1).Thisobservationshows¯a¡b¯121212¯1¡ab¹¯=1ifa
