The difficult case is around a fixed point of a function with derivative one. Irvine Noel Baker, 1932-2001, studied these from the viewpoint of formal power series with complex coefficients, beginning with some $ f(z) = z + a_{m+1} z^{m+1} + \ldots, \; a_{m+1} \neq 0.$ He changed the question to finding those
$$ f_\lambda(z) = z + \lambda a_{m+1} z^{m+1} + \sum_{n = m+2}^\infty b_n(\lambda) z^n$$ which commute with $f.$ For a given $f = f_0,$ there may or may not be any other $f_\lambda$ such that the power series is convergent near $z=0.$ The big theorem, with one case taken care of by his student Liverpool, is that the set of $\lambda$ for which $f_\lambda(z)$ converges near $0$ is one of three sets: (a) $\{ 0 \},$ (b) with some fixed $N \in \mathbb Z,$ the fractions $\{m/N, \; \mbox{all} \; m \in \mathbb Z\},$ or $\mathbb C$ itself. In the final case, where any complex $\lambda$ is allowed, Baker called the function $f$ embeddable, saying that the function is embeddable in a continuous group of analytic iterates.

In case (b) there is some minimal $1/N$th iterate which cannot be further, um, divided. So there may be half-iterates of something without there being any one-third iterates.

My summary would be that Baker makes it quite sensible to talk about an $i$ iterate. The conceptual switch from trying to do half iterates to asking what formal power series commute with a given formal power series makes the whole thing tractable.

Oh, original articles and books posted at

http://zakuski.utsa.edu/~jagy/other.html

EDIT: I found some of my notes from 2010. From what I can make out, the only example that we expect to be really pleasant is the family of linear fractional transformations
$$ f_\lambda(z) = \frac{z}{1 + \lambda z} $$
which all comute with each other, and nothing worse happens than a pole for each one at $z = -1 / \lambda. $ Note the group law $f_\lambda \circ f_\gamma = f_{\lambda + \gamma}$
I felt that all other embeddable families were essentially that, just take some holomorphic $h(z)$ with $h(0) = 0$ and $h'(0) = 1$ and get the very similar
$$ f_\lambda(z) = h^{-1} \left( \frac{h(z)}{1 + \lambda h(z)} \right), $$ with Fatou coordinate
$$ \alpha(z) = \frac{1}{h(z)}. $$
There is a bootstrapping method for solving for the Fatou coordinate $\alpha(z)$ which is probably due to Ecalle. I also noted $ \beta(z) = \frac{- h^2(z)}{h'(z)}$ but I forget what $\beta$ was for. No, here we go, it is an explicit description in KCG on solving for the Fatou coordinate, pages 346-352, *Iterative functional equations* by Marek Kuczma, Bogdan Choczewski and Roman Ger. In general $\beta(z) = 1 / \alpha'(z).$

Note, though, that we have now introduced possible bad behavior when either $h(z)$ or, more likely, $h^{-1}(z)$ are undefined, in short we have probably severely curtailed the region of $\mathbb C$ where things are working well.

Edit toooo: the Fatou coordinate may be defined on only a sector out of the origin, anyway
$$ \alpha(f(z)) = \alpha(z) + 1.$$ Then we get a family (but maybe only in a sector) by
$$ f_\lambda(z) = \alpha^{-1}( \lambda + \alpha(z) ), $$
where $f_1 = f$ in this recipe. So once again, as in the linear fractional transformations, we can plug in $\lambda = i.$

notthe use here. $\endgroup$ – j.c. Jul 27 '11 at 19:42