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Date: Sat, 03 May 25 15:42:19 +0000
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From: Richard Hachel <r.hachel@tiscali.fr>

Euler pose e^i.π + 1 = 0

Bon, d'accord.

Et si je pose e^(-i)π + 1 = 0

Cela change quoi?

Et si je pose e^i.(-π) + 1 = 0 ?

ou e^(-i)(-π) + 1 = 0 ?

Tout cela reste-t-il invariant?

R.H.