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## Homomorphism that preserves exponentiation [closed]

Can you give an example of a function o: C->C (complex numbers) such that o(a + b) = o(a) + o(b) and o(a^s) = o(a)^o(s), other than the trivial one. It might also preserve multiplication - and if so it must be identity on the integers since o(a^2) has to be o(a)^2. What is this type of morphism called?

Thank you!

-Dan

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What does this have to do with algebraic geometry? and why are you looking for such a function in the first place? – Yemon Choi Sep 29 2010 at 0:30
Without having thought much about the problem, it isn't clear to me what your exponentiation property means. In general, a^s and o(a)^o(s) will depend on choosing branches of the logarithm (when a and o(a) are nonzero). So for instance, it could be interpreted as saying that for all nonzero $a$, $o(a)$ is nonzero, and for each logarithm $\log_1$ defined at $a$, there exists a logarithm $\log_2$ defined at $o(a)$ such that $e^{b\log_1(a)}=e^{o(b)\log_2(o(a))}$. (I.e., one must consider different preimage points of $a$ and $o(a)$ under $\exp$.) Is this what you mean, and if not then what? – Jonas Meyer Sep 29 2010 at 0:45
One could consider this about endomorphisms of a ring A with exponentiation, with varying levels of distribution of one over the other (say only assume the usual exponentiation rules e.g. a^(s+t) = a^s.a^t, or also that there is a binomial theorem (a+b)^s = ... – David Roberts Sep 29 2010 at 3:04
Silly observations: $o(0+0) = o(0) + o(0)$ implies $o(0) = 0$. $o(a^0) = o(a)^0$ implies $o(1) = 1$. Then additivity implies $o$ takes rationals to rationals, and using squares to get imaginary arguments, it acts as identity on some countable dense subset of the complex numbers. Even assuming you could make a meaningful interpretation of the complex exponentiation, there is no name for this type of morphism, because no one cares about it. I'm closing the question. – S. Carnahan Sep 29 2010 at 8:11