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From: Samuel DEVULDER <samuel_dot_devulder@laposte_dot_net.invalid>
Newsgroups: fr.sci.maths
Subject: Re: Puissance complexe
Date: Mon, 20 Dec 2021 22:41:24 +0100
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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.