Rouche Theorem（Stein复分析）

Rouche Theorem：

\quadIffandgareholomorphicfunctionsinaregionΩcontainingacircleCanditsinterior,and∣f(z)∣≥∣g(z)∣forz∈C,fandf+ghavethesamenumbersofzerosinsidethecircleC.If\quad f\quad and\quad g\quad are\quad holomorphic\quad functions\quad in\quad a\quad region\quad \Omega\\\quad containing\quad a\quad circle\quad C\quad and\quad its\quad interior,\quad and\quad |f(z)|\ge|g(z)|\quad for\quad z\in C,\quad f\quad and\quad f+g\quad have\quad the\quad same\quad numbers\quad of\\zeros\quad inside\quad the\quad circle\quad C.IffandgareholomorphicfunctionsinaregionΩcontainingacircleCanditsinterior,and∣f(z)∣≥∣g(z)∣forz∈C,fandf+ghavethesamenumbersofzerosinsidethecircleC.

Proof:

\quadLetft(z)=f(z)+t⋅g(z),t∈[0,1],thenf0(z)=f(z)andf1(z)=f(z)+g(z).LetntdenotethenumberofzerosoffcountedwithmultiplicitiesinsidethecircleC.Since∣f(z)∣≥∣g(z)∣forz∈C,ft(z)hasnozerosinsidethecircleC(maximummodulusprinciple),whichmeansthatft(z)hasnopolesinsidethecircleC.Accordingtoargumentprinciple,nt=12πi∫Cft′(z)ft(z)dz.Toobtainthatn0=n1,itsufficestotoshowthatntisinteger−valuedandcontinuous.Thusifntisnotconstant,byintermediatevaluetheoremtheremustbesoment0thatisnotintegral,acontradiction.Nowwecometoprovethatntiscontinuous.Sinceft(z)andft′(z)arebothcontinuousinsidethecircleC,andft(z)doesnotvanishonthecircleC,wecometotheconclusionthatntiscontinuous.Hencen0=n1.Theproofiscomplete.Let\quad f_t(z)=f(z)+t\cdot g(z),t\in [0,1],\quad then\quad f_0(z)=f(z)\quad and\quad f_1(z)=f(z)+g(z).\quad Let\quad n_t\quad denote\quad the\quad number\quad of\\zeros\quad of\quad f\quad counted\quad with\quad multiplicities\quad inside\quad the\quad circle\quad C.\\\quad Since\quad |f(z)|\ge |g(z)|\quad for\quad z\in C,\quad f_t(z)\quad has\quad no\quad zeros\quad inside\quad the\quad circle\quad C(maximum\\ modulus\quad principle),\quad which\quad means\quad that\quad f_t(z)\quad has\quad no\quad poles\\ inside\quad the\quad circle\quad C.\quad According\quad to\quad argument\quad principle,\quad n_t=\frac{1}{2\pi i}\int_C\frac{f^{'}_t(z)}{f_t(z)}dz.\quad To\quad obtain\quad that\quad n_0= n_1,\quad it\quad suffices\quad to\quad to\quad show\quad that\quad n_t\quad is\quad integer-valued\quad and\quad continuous.\quad Thus\quad if\quad n_t\quad is\quad not\quad constant,\quad by\\ intermediate\quad value\quad theorem\quad there\quad must\quad be\quad some\quad n_{t_0}\quad that\\ is\quad not\quad integral,\quad a\quad contradiction.\\\quad Now\quad we\quad come\quad to\quad prove\quad that\quad n_t\quad is\quad continuous.\quad Since\\ f_t(z)\quad and\quad f_t^{'}(z)\quad are\quad both\quad continuous\quad inside\quad the\quad circle\quad C,\\ and\quad f_t(z)\quad does\quad not\quad vanish\quad on\quad the\quad circle\quad C,\quad we\quad come\quad to\\ the\quad conclusion\quad that\quad n_t\quad is\quad continuous.\quad Hence\quad n_0=n_1.\\\quad The\quad proof\quad is\quad complete.Letft(z)=f(z)+t⋅g(z),t∈[0,1],thenf0(z)=f(z)andf1(z)=f(z)+g(z).LetntdenotethenumberofzerosoffcountedwithmultiplicitiesinsidethecircleC.Since∣f(z)∣≥∣g(z)∣forz∈C,ft(z)hasnozerosinsidethecircleC(maximummodulusprinciple),whichmeansthatft(z)hasnopolesinsidethecircleC.Accordingtoargumentprinciple,nt=2πi1∫Cft(z)ft′(z)dz.Toobtainthatn0=n1,itsufficestotoshowthatntisinteger−valuedandcontinuous.Thusifntisnotconstant,byintermediatevaluetheoremtheremustbesoment0thatisnotintegral,acontradiction.Nowwecometoprovethatntiscontinuous.Sinceft(z)andft′(z)arebothcontinuousinsidethecircleC,andft(z)doesnotvanishonthecircleC,wecometotheconclusionthatntiscontinuous.Hencen0=n1.Theproofiscomplete.