Complex Analysis: Log Function I want to approach this problem with maximum understanding of everything that is going on. 
I have the function $F(z)=\log(z^2+4)$, and I want to give a region in which it is analytic. I guess I shouldn't call it a function until I give the region since it's multivalued.
$(1)$ I first start by solving $z^2+4=0 \Rightarrow z^2=-4$. (why do we do this)?
$(2)$ Then, I define $z=re^{i\theta} \Rightarrow r^2e^{2\theta i}=-4$
$(3)$ Thus, $r=2$ and $\theta=\frac{\pi + 2\pi k}{2}$ for $k=0,1$.
$(4)$ Next, I draw the rays eminating from the $0's$ of the function (green).
$(5)$ These are the two rays describing 2 complex numbers such that when I double their angle and add $4$ I end up on the negative x-axis.
$(6)$ Now I choose my region of validity. I take my branch cut to be principle branch of log, which seems intuitive. 
However, can I also choose to make $f$ analytic on $\mathbb{C}$ \ {$z=re^{i\theta}: r>2, \theta=\frac{pi}{2},\frac{3\pi}{2}$}?

 A: The crucial point to remember about a branch cut is that it is chosen to ensure that inside the remaining domain of definition of the function, it must be impossible to find a small curve which "winds around" a branch point.
For the straightforward $\log z$ function, then, as you know, it is enough to take any ray emanating from $0$, and this will prevent any "winding".
In your example, we have two branch points; one at $2i$ and one at $-2i$.
We can avoid any possibility of this winding problem by defining branch cuts as follows:
Take 2 branch cuts, one going from $2i$ to infinity (say upwards along the imaginary axis), and the other going from $-2i$ towards infinity (say downwards along the negative imaginary axis). This ensures the domain is now simply connected with no possibility of winding round a branch point.
The alternative possible approach of taking a single branch cut going directly from $2i$ to $-2i$ does not work, since in this case we have a branch point at infinity. This alternative choice of branch cut would work however in the case of the function $\sqrt{z^2+4}$ which just has two finite branch points.
