Forum de Mathématiques: Maths-Forum

Bonsoir, 2. (E): (z-a)^n = (z-b)^n <=> z-a/z-b = racine n-ième de 1 = cos (2k;)/n) + isin (2k;)/n) = f(k) avec k;);) et k;)n <=> z-a = (z-b).f(k) <=> z(fk - 1) = b.f(k)-a <=> z = (b.fk - a)/(fk-1) k=0,n =>fk = 1=> pas de solution k=1,2...,n-1 => z=(bf -a)/(f -1) = (bf-a).(fbar-1) / (f-1).(fbar -1) =...