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by cycomanic
1090 days ago
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just to expand on this the issue is not that e^2πi = 1 but that (e^z)^x =/= e^(zx) when z is complex, because in general z^w . instead it would be e^(x ln(e^z)), where the logarithm of a complex number is a multivalued function. We can compute the principle value of ln(e^2πi)=ln(1) + i(2π + 2πk) (where k is an integer) so therefore (e^2πi)^x = e^(xln(e^2πi)) = e^(x2π(k+1)i) |
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