# 关于解析函数的一个易想象的不等式及其应用(英文)

【摘要】本文关于解析函数给出了一个可作几何理解的不等式,由此易得出一个有关解析函数零点的命题,而代数学基本定理成为它的直接推论．

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Without1055ofgeneralit又aGaussianPlaneCmaybewrittenno介Collstsntintheformanal外iefunetionf(z)definedfor}:}0亡herehol由亡heInequalitysu用den街slnall!f((一a占/。)‘/仇)!0.口Evidentl叭Ine卿ality(2)m盯bestatedinamoregeneralform:Iff(:)15non一eonstantandanalytieinaneighborhoodof:。任C，withf(句)=a笋0，thenf(z0+:)m盯bewrittenina51而larform，asthatof(1)，f(句十习二了(耐十加m十二饥+l十二，(m全l)，(3)andconse卿entlythefollowinginequalityIf(句+(一a占/6)‘/爪)I0eanneverbecomeanabsolute而nimummin二}F(z)1Thisimpliesthatifmin二!F(z)}really树stsand15attainedatz=句任D，thenitmustbethat而n:}F(:)卜}F(:。)}=0.Aeeordingl从wegetthefollowingusefulandwell-knownpropositionasaconsequeneeof(4).proposition2Le亡F(z)beanon一cons艺an亡analytjc允ne七jon加DC，andle亡}F(:)}at亡ajnanabsoluteminjlnuma七名。任D.孙enj亡mustbe亡ha亡呼nIF(“)l一IF(“o)1一0，In环.rds，句mus七beazeroofF(z).Inpartieular，ifF(:)15apolyno而alin2ofde盯een(n之1)，thentheobviousfactthat}F(:)}弓co(}z!科co)i哪liesthat}F(z)!shouldattainmin:}F(:)1ateertain:0任C.Thus饰Proposition2wemusthaveF(z。)=0.This15whatso-calledthewell-knowne劝steneetheoremfirstprovedbyGauss(1799):FTAA盯polyno功勿1equa亡jonF(:)二0ofd卿een(n全1)hasa亡leas亡aroo。:。任C，viz.F(z0)二0.Notethatfortheexa呷leF(:)=ezwehave}e名1=!e忠+‘，卜e劣>0for名=x+乞，任C.ThiSshowsthatthesecondeonditinninProPosition2eannotbeolnitted.Remarkl取callthatintheeomplexanal邓15ali而tprocesssuchasf(z)叶A(z、acC)involvesthatboth2andf(:)eouldtendtotheirlimitsinvariouspossibledireetionsinC.Inpartieular，iff(z)、f(a)笋0(:、a)andifthemodeofp韶s鳍ez、aeouldbe50chosenthat}f(z)}丁}f(a)}，thenweshallhave}f(:){<}f(a)}forallthose2suffielentlyeloset

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