Find the domain on which $\log(z^2+9)$ is analytic. I have the following function
$$f(z)=\log(z^2+9)$$
I need to find the set on which this function is analytic.
So far I know that $\log(z)$ is analytic in $D= \left \{ z \in \mathbb{C} | z \notin (-\infty, 0] \right \}$.
So, I understand that for my function $D= \left \{ z \in \mathbb{C} | z^2+9 \notin (-\infty, 0] \right \} = \left \{ z \in \mathbb{C} | z^2 \notin (-\infty, -9] \right \}$.
How should I continue?
 A: $\left \{ z \in \mathbb{C} | z^2 \notin (-\infty, -9] \right \}$ is a perfectly good specification of the set, so you could reasonably stop here. However, if you'd prefer an interval form, note that $z^2 \in (-\infty, -9]$ means $z$ is purely imaginary and has imaginary part at least $3$ or no more than $-3$. Thus $D$ is the union of two rays:
$$
D = \{z\in \mathbb C | iz\notin (-\infty, -3]\} \cup \{z\in \mathbb C | iz\notin[3,\infty)\} 
$$
A: There is not "the" set on which the function is analytic/holomorphic, perhaps contrary to some introductory situations in which the complex plane is almost-always "slit" along the negative real axis, or other iconic things.
To begin with, for example, (the collection of local pieces of) $\log z$ is holomorphic at points on the negative real axis, perhaps contrary to some (too-naive) knee-jerk slittings-of-the-plane along the negative real axis. In fact, in some neighborhood of every point in $\mathbb C$ except $0$, there is a functions $g(z)$ such that $e^{g(z)}=z$.
The real point is best understood, at least in the long run, in terms of the Monodromy Theorem, which deserves more discussion than I should do here.
In the case at hand, a function $\log(z^2+9)=\log(z+3i)+\log(z+3i)$ is locally defined (in the sense that it exponentiates to $z^2+9$) at any point other than $\pm 3i$. Period.
The implicit meaning of the question is finding a (not "the") maximal open set on which there is such a function. With a heuristic notion of Monodromy, the point is that we have to choose a region so that we cannot (for example) go in a closed path around either $\pm 3i$ (at least without also going around the other), because that would add $\pm 2\pi i$ to one of the logs while not changing the other.
One way to do this is to choose a region that does not include any closed path around either $\pm 3i$, e.g., "slitting" the plane from both of the points out to infinity (perhaps not allowing the two slits to intersect, just to avoid technicalities). The slits need not be straight lines...
EDIT: I'd earlier claimed that a "slit" between the two branch points $\pm 3i$ would suffice, but @Maxim observed that my argument was incorrect, since (I add this because the general idea is certainly of interest) going around both $\pm 3i$ does not cancel out the ambiguities in $\log$, since both are in the same orientation.
(Thanks again to @Maxim for observing my inaccuracy!!!)
