TITLE: complexification of the vector space $\Bbb R^+$
QUESTION [5 upvotes]: Consider $\Bbb R$ and $\Bbb R^+$ as isomorphic vector spaces, with isomorphism $f(x)=\exp x.$

Is the complexification of $\Bbb R$ isomorphic to the complexification of $\Bbb R^+?$ What is the complexification of $\Bbb R^+?$

I understand that when going from $\Bbb R$ to $\Bbb R^+$ under $f$ we have $a+b:=ab$ and $kb:=b^k.$ Here is a relevant link: vector space.

REPLY [2 votes]: Yes, isomorphic vector spaces have isomorphic complexifications.
If you're asking how to "implement $\mathbf{R}^{+} \otimes \mathbf{C}$" without unwrapping the isomorphism with the usual vector space structure on $\mathbf{R}$, I don't see a way to do this offhand. At first glance it's tempting to think of the complexification as $\mathbf{C}^{\times}$ with complex multiplication as "addition", and exponentiation as "scalar multiplication". The fatal problem is, the complex exponential mapping is not injective (unlike the real exponential mapping), so to make sense of the operations you end up "taking logarithms" (really, passing to the universal cover), i.e., to the usual vector space structure on $\mathbf{C}$.