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e^(pi * i) + 1 = 0

Name: Anonymous 2007-12-13 5:18

Why? I don't understand this at all ><

Name: Anonymous 2007-12-13 15:58

do an analytic continuation from exp x to all complex values; the same with cos x and sin x; then is easy to verify(by power series) that (1) e^(iz) = cos z + isen z; doing z = pi u have the result.
You can think that (1) is heuristic based that both classes of function (exp x  and cos x + isen x) are given by the same functional equation viz:
f( x + y ) = f( x )*f( y );
sorry for my possible lack of analysis :)

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