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Failed to connect to MySQL: (1203) User howardkn already has more than 'max_user_connections' active connectionsPath: ...!weretis.net!feeder6.news.weretis.net!feeder8.news.weretis.net!usenet.pasdenom.info!from-devjntp Message-ID: <9ti6Y3s5Sm3yeoodalYH2gD64qE@jntp> JNTP-Route: news2.nemoweb.net JNTP-DataType: Article Subject: Re: Puissance complexe References: Newsgroups: fr.sci.maths JNTP-HashClient: Cvk_z6xssoEFmEnJTabxfdphJ7w JNTP-ThreadID: 0n0919F69IreuR1l8nnlTNB_YYY JNTP-Uri: http://news2.nemoweb.net/?DataID=9ti6Y3s5Sm3yeoodalYH2gD64qE@jntp User-Agent: Nemo/0.999a JNTP-OriginServer: news2.nemoweb.net Date: Mon, 20 Dec 21 21:45:50 +0000 Organization: Nemoweb JNTP-Browser: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.110 Safari/537.36 Injection-Info: news2.nemoweb.net; posting-host="5c7cc3dd4d5b02ab9eeda727c1c7bf4fa8cd7732"; logging-data="2021-12-20T21:45:50Z/6408118"; posting-account="1@news2.nemoweb.net"; mail-complaints-to="newsmaster@news2.nemoweb.net" JNTP-ProtocolVersion: 0.21.1 JNTP-Server: PhpNemoServer/0.94.5 MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-JNTP-JsonNewsGateway: 0.96 From: Julien Arlandis Bytes: 2667 Lines: 35 Le 20/12/2021 à 22:41, Samuel DEVULDER a écrit : > Le 20/12/2021 à 22:30, Julien Arlandis a écrit : >> Si je dois évaluer sqrt(1), je vois 2 possibilités : >> >> 1) sqrt(1) = 1^(1/2)           = exp(1/2*ln(1))           = >> exp(1/2*0) >>           = exp(0) >>           = 1 > > Attention ln(1) c'est 0 [mod 2pi*i] donc 0 + 2pi*k*i > > Je reprends: sqrt(1) = 1^(1/2) = exp(1/2*ln(1)) = exp(1/2*2pi*k*i) = > exp(k*pi*i) = +/- 1 suivant la parité de k > >> >> 2) sqrt(1) = 1^(1/2) >>           = (exp(2*i*k*pi))^(1/2) >>           = exp(1/2*ln(exp(2*i*k*pi))) >>           = exp(1/2*2*i*k*pi) >>           = exp(i*k*pi) >>           = 1 ou -1 > > Tout pareil ! > > Le truc est de bien voir que ln(x) est un truc qui retourne un résultat > "mod 2pi*i". > > sam. La seconde approche c'est celle que j'avais utilisé pour calculer que 1^x = exp(2*i*k*pi*x), mais toi tu l'avais contesté en utilisant la première approche, je te cite : "1^x = exp(x*ln(1)) or ln(1)=0, donc 1^x = exp(x*0) = exp(0) = 1" Donc retour à la case départ :)