Dirk Ferus's Komplexe Analysis [Lecture notes] PDF

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Extra resources for Komplexe Analysis [Lecture notes]

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Z z0 0 Nach Voraussetzung ist 0= f (z)dz − f (z)dz = 1C ∂C f (z)dz + 2C f (z)dz. 2C Daher folgt F (z) − F (z0 ) 1 = z − z0 z − z0 = 1 z − z0 f (ζ)dζ − σz f (ζ)dζ σz0 = 1 z − z0 1 f (ζ)dζ 1C 1 f ((1 − t)z0 + tz)(z − z0 )dt = 0 f ((1 − t)z0 + tz)dt 0 1 (f ((1 − t)z0 + tz) − f (z0 )) dt. = f (z0 ) + 0 Mit der Stetigkeit von f folgt aus dem Schrankensatz 41, dass F holomorph mit F = f ist. Also ist auch f holomorph. Ist f holomorph in G, so verschwindet nach dem Cauchyschen Integralsatz das Integral u ¨ber den Rand eines jeden Dreiecks.

H. wir berechnen 2π 0 iφ it Re := σ(Re ). Daf¨ ur finden wir mit (32) nach Differentiation iReiφ dφ = und damit 1 2π |σ(Reit ) − z|2 |σ(Reit ) − z|2 iσ(Reit )dt = iReiφ dt 2 2 R − |z| R2 − |z|2 2π 0 1 f˜(Reit )dt = 2π 2π f (Reiφ ) 0 R2 − |z|2 dφ. |Reiφ − z|2 Wenn man also die Randwertverteilung von f zun¨achst an z spiegelt und dann u ¨ber den Einheitskreis mittelt, erh¨ alt man den Wert f (z). Wir wollen nun Konsequenzen der Poissonformel f¨ ur harmonische Funktionen untersu2 2 chen, also f¨ ur zweimal stetig differenierbare Funktionen u : G → R mit ∆u = ∂∂xu2 + ∂∂yu2 = 0.

Holomorphe Funktionen sind beliebig oft differenzierbar. Korollar 68 (Umkehrsatz). Sind f : G → C holomorph und z0 ∈ G mit f (z0 ) = 0, so besitzt f lokal bei z0 ein holomorphes Inverses: Es gibt eine offene Umgebung U von z0 in G, so dass gilt: (i) f |U ist injektiv, (ii) f (U ) ist offen, (iii) (f |U )−1 : f (U ) → C ist holomorph. Holomorphe Abbildungen mit holomorphem Inversen nennt man auch biholomorph. Also sind holomorphe Abbildungen mit nicht-verschwindender Ableitung lokal biholomorph.

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Komplexe Analysis [Lecture notes] by Dirk Ferus

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