Т.А. Леонтьева, В.С. Панферов, В.С. Серов - Задачи по теории функций комплексного переменного с решениями (1118152), страница 22
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IJpo.nomKHTh aHaJIMTttqecKM B KOMIIJieKCHYIO IIJIOCKOCThcPYHKUMH, 1a.rraHHhie Ha .rrettCTBHTeJihHOH OCH:1) sin x; 2) cos x; 3) tg x; 4) ctg x;5) e ·\6) sh x; 7) ch x; 8) th x; 9) arc tg x; 10) arc sin x.11.4 . .UoKa3aTh TO)l(,Uecrna BC:221) sin z + cos z = 1;2) sin (z 1 ± z2) =sin z 1 cos z2 ±sin z2 cos z 1;3) cos (z1 ± z2) =cos z1 cos z2 ± sin z1 sin z2;4) e:,±:,= e:'e±:, .11.S .
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IIocTpOHTh pHMaHoBy noBepxttoCTh ,UJIH cpyHKUHH Argz.11.17. ,[(oKa3aTb, qTo Bee 3HaqeHHH ¢YHKUHH Ln z .n:moTCH cpopMynofi:: Ln z = ln izl + i Arg z.11.18. IIocTpOY!Th pHMattoBy noBepxHOCTh cpyttKUHH Ln z H BhI.n:enHTh O,UH03HaqHbie BeTBH 3TOH cpyHKUHH.11.19. .[(oKa3aTb, qrn crrpaBe.UJIHBhI cpopMyJinI:d11) -Lnz=-;2) eLn: =z.dzz11.20. .[(oKa3aTb, qTo TQqKH 0 H 00 HBJIHIOTCH M30JIHpOBaHHbIMHToqKaMH BernneHHH cPYHKUHH Ln z.11.21. .[(oKa3aTb, qTo eCJIH CD1, CDz - mo6h1e H3 3HaqeHHH cpyHKUHH Ln z COOTBeTCTBeHHO B TOqKax Zr H Z2, TO HX cyMMa eCTb O,UHOH3 3HaqeHHtt cpyHKu;im Ln (z 1 • z 2).11.22.
IIycTh F(z)z = 1, In 1 = 0.= z ln z, HCXO.UHhIH 3JieMeHT In z 3a,n:att B rnqKez = 0, z = oo HBJIHIOTCH H30JIH-,[(oKa3aTh, qTo ToqKHpoBaHHhIMH ToqKaMH BeTBJlemrn cpyHKUHHF(z)6ecKoHeqHoro rro-pH.n:Ka.11.23. ,[(oKa3aTb,qTo cpyHKUH5Iza = eaLnzeCTb aHaJIHTITTeCKOe npo,UOJI)KeHHe B11.24 . .[(oKa3aTb,qTQ cpyHKUH5I11.25 . .[(oKa3aTh,qrnc cpyttKUHHj{x) =Xa, x > 0, a E R.za attaJIHTH'IHa B o6nacrn C\ {0}.T.wea 11146r.ue 3Ha'!emrn za B o6eHx qacrnx paseHCTBa O.UHH H Te :>Ke.11.26.ITycTh KpHsasrycoe,UHHsreT TO'IKH z 0, z 1 H He rrpoxo.uHTH oo. ITycTh B TO'!Ke z 0 3a,UaH 3JieMeHT f(z) cpyHKUHHZa, TaKOH, 'ITO j(zo) = Zo a .
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