Т.А. Леонтьева, В.С. Панферов, В.С. Серов - Задачи по теории функций комплексного переменного с решениями (1118152), страница 9
Текст из файла (страница 9)
qTo MO)KHO CKa3aTb oCXO.ll:HMOCTH 6eCKOHeqHoro IIpOH3Be,n:eHH.sI Il~=l4.42. IIycTb c - KOHCTaHTa, c*(1 + a,,z)?-1, -2, .... .lJ:OKa3aTb, qToTI~= 1 (1- _z_) e'"a6comoTHO cxo,n:11Tc.si: npH lzl < oo.c+n4.43. IIoKa3aTb, qTo np11 I z I > 1 6ecKoHeqHoe rrpomBe.n:eHHerr~..[r-( J-;;-rZ _,,] CXOAITTC•4.44.
HaifTH o6JiaCTH cxo1111MoCTH 6ecKoHeqHhIX npo113se.n:ett11H::1)n-(1-z")· 2)11=14) TI~_, cos~;'n-"""'s)22(1-)·_!_n2 ' 23)n- (I+~)e-''"·n11=1'Il~= 1 ( sin;;~n)} 6) n~_,c1 + z2").(-1)"+14.45. ,UoKa3aTb, qTQ npoH3se.n:ett11e TI~=' ( 1+ ~)CXOAHTC.sInpH Re z > Yi ' a6comOTHO CXOAHTC.sI IIpH Re z > 1.n·Jn~4.46 .
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IloKa3aTh, qTO rrptt a,~=f: -1, -2, ...TI~-, nca + fJ + n) = rca + o rCfJ+ 1).- (a+n)(fJ+n)f(a+(J+n)4.51. ,I(oKa3aTh, qTO cne.uy10rutte 6ecKoHeqHhie rrpomse,neHM5ICXO.U5ITC5I paBHOMepHo Ha yKa3aHHhIX o6nacrnx:1)n-n=l(1-z2) ' lzl $2) TI~= 1 (1 +~) e-'R < =·'n-.,1",I z IS R < =;23) IT~=/l+z"),lzl$r<l;4) IT~=l(l+z "),lzl$r<l.4.52 .
.UoKa3aTh, qTO ecntt p5I.U L~=il J,, (z) I cxo,nHTC5IMepHoHaMHO:>KecTBeZ,TOM 6ecKOHeqHoepaBHO-rrpott3Be,neHtteIT~=I (1 + f,, (z)) CXO.D;MTC5I paBHOMepHO Ha MHO:>KeCTBe Z.4.53. IIycTh 6eCKOHeqHoe rrpOH3Be,UeHHe IT~=l (1 + J,, (z)) CXO,UHTC5I paBHOMepHo Ha MHO:>KecTBe Z. Cne.uyeT JIH oTcIO,na, 'ITO p5I.UI~=I J,, (z) cxo.n;HTC5I paBHOMepHo Ha MHO:>KecTBe Z?4.54. IIycTh p5I.U I,~=I J,, (z) cxo,n;ttTC5I pasHoMeptto Ha MHO:>KecTBeZ.
CJie,nyeT JIM OTCIO.Ua, qTo Ha MHO:>KecTBe Z cxo;uffc5I paBHOMepHo6eCKOHeqHOe rrpOH3Be,UeHMe IT~=I (1 + f,, (z))?4.55 . .LJ:oKa3aTh, qTo cne,n;y10m11e 6ecKoHeqHhie npoH3Be,n;eHH5I rrpe.n;CTaBJI5IIOT co6oli HenpephIBHhie cpyttKUMM B YKa3aHHhIX o6nacrnx:1) IT~=I (1 + z" ), Iz I< 1;2)IT~ (1+..£)e-~11=lIIIz I<IT ,,= cos2:.,ll~3)'lzl<'1l710000 ,z!l Z·'z "#- n(2k + l);2TT/aea 4544)IT==11n · z I z; I<oo,z;:;t ktr;-sm-,1zn5) IT~= 1 (1 + c-1r ;) I z I<7)n=n=8)n=,(1+~).lzl<2.2"6)z :;t c-1r+ 1 n;,ch...£,2"Iz I<00 ,fl=Iz :;ti 2"-cos...£,2"I z I<00 ,11=1z :;t 2 1c2k + l)n;1(2k + 1)n;11-11=4.56.
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