[cycomanic]

he writes:

> where ln(e^2πi)=ln(1)=0,

that is incorrect because the logarithm of a complex number is multi-valued. He even cites the correct source on wikipedia, but his argument is incorrect (and because the exponent is 2πi he actually would get a meaningful results I believe).

[mgunyho]

Hmm, this indeed seems to be the case! I think I was confused by ln(1), since 1 is a real number, but the multi-valued complex logarithm of 1 is indeed k 2pi (and now I see the appropriate notation should be "log" instead of "ln"). Perhaps the whole thing could indeed be reduced to "1^x" with an asterisk that 1^x means complex exponentiation. I'll have to update this section in the post.

[cycomanic]

See my other reply down below. That said I agree with the final conclusion that including the i is not a good idea, although I'm a bit on the fence on if the whole thing (I'm convinced by the tau=2pi arguments on the other hand).

I think this really is just a redefinition of sin and cosine (and consequentially the exponential). My feeling is that we now have to deal with different derivative operators for periodic and non-periodic functions and a lot of other weird disconnects, e.g. we don't have a relation between a cycle and the radius (diameter, circumference ...) anymore. I'm not convinced that the (minor IMO) conveniences that gives us is worth it.

[cycomanic]

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)