To construct that branch of log $z$, you just define the half-parabola to be the branch cut. This would mean log $z$ = Log|$z$| + $i\theta$ where $\theta$ equals the value of arg $z$ between $\frac{\pi}{2}$ and $2\pi$ for $z$ in the second, third, or fourth quadrant, the value of arg $z$ between $0$ and $\frac{\pi}{2}$ for $z$ in the first quadrant above the half parabola, and the value of arg $z$ between $2\pi$ and $\frac{5\pi}{2}$ for z in the first quadrant below the half parabola.
Explicitly, you can solve for $\theta$ in terms of $r$ where $z=re^{i\theta}$. By the equation of the half parabola, $\frac{y}{x}=\frac{1}{y}$. Then, $\theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{y})$.
$r=\sqrt{x^2+y^2}=\sqrt{x^2+x}$ , then $x^2+x-r^2=0$. By the quadratic formula,
$$x=\frac{-1\pm\sqrt{1+4r^2}}{2}$$
You choose the positive square root because you want the upper half parabola so,
$$y=\sqrt{\frac{\sqrt{1+4r^2}-1}{2}}$$
Therefore, the entire branch would be defined as
$$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$
Let's agree explicitly that "$\log$" refers to the branch of logarithm defined on $\mathbf{C}\setminus(-\infty, 0]$ whose imaginary part is between $-\pi i$ and $\pi i$, and that "$f(z) \leq a$" means "both $a$ and $f(z)$ are real, and $f(z) \leq a$".
If $f$ is holomorphic in some region $U$, then $\log(f)$ is holomorphic at $z$ in $U$ provided $f(z)$ does not lie on the branch cut, i.e., "$f(z) \leq 0$" is false.
In your example, $f(z) = z^{2} + a^{2}$ is holomorphic on all of $\mathbf{C}$, so $\log f$ is holomorphic on the twice-cut plane, from which the imaginary intervals $i(-\infty, -a]$ and $i[a, \infty)$ have been removed. These intervals are precisely the set of $z$ for which $z^{2} + a^{2} \leq 0$. In fancier language, they constitute the preimage of the cut $(-\infty, 0]$ under $f$.
The issue in (2) is that a small circle $C$ centered at $ib$ (with $|b| > a$) crosses the "cuts" of $\log(f)$. The image of $C$ under $f$ is a closed curve (not a circle) "centered on" the negative real axis, so $f(C)$ crosses the branch cut of $\log$. Locating $-b^{2} + a^{2}$ in the $z$ plane, relative to the branch cuts of $\log(f)$, is immaterial: We're not evaluating $\log(f)$ near $-b^{2} + a^{2}$, we're evaluating $\log$ itself. $\ddot\smile$
To see what's going on with the values of $\log(f)$ "near" each imaginary cut, consider points $z = re^{i\phi}$ with $r > a$ and $\phi$ "close to" $\pm\pi/2$:
$$
z^{2} + a^{2}
= r^{2} e^{2i\phi} + a^{2}
= (r^{2} \cos(2\phi) + a^{2}) + ir^{2} \sin(2\phi).
$$
If $\phi$ is a bit smaller than $\pm\pi/2$, then $\sin(2\phi) > 0$. Consequently, $z^{2} + a^{2}$ has "small" positive imaginary part, so $\log(z^{2} + a^{2})$ has imaginary part close to $\pi i$.
Similarly, if $\phi$ is a bit larger than $\pm\pi/2$, then $\sin(2\phi) < 0$ and $\log(z^{2} + a^{2})$ has imaginary part close to $-\pi i$.
Best Answer
You're right that every non-zero complex number has infinitely many logarithms, but it is possible that exponentiation will make all these values equal. For instance, let $ z= r e^{i \theta},$ then the possible values of $\log(z)$ are $ \log(r) + i \theta + 2 \pi i k$ for $k \in \mathbb{Z}.$ Now you would like $\exp(w\log z) $ to be a well-defined function on $\mathbb{C}$ i.e. you want all the possible values of the logarithm of $z$ to give the same value of $w^z.$ Writing this out, you get:
$$ \exp(w(\log r + i\theta + 2\pi i k )) = \exp ( w \log r + i\theta) \text{ for all } k \iff $$ $$ \exp( w \cdot 2\pi i k) = 1 \text{ for all } k \iff w \cdot 2\pi i k \in 2 \pi i \mathbb{Z} \text{ for all }k \iff w \in \mathbb{Z}.$$ Thus only exponentiating by integers is well defined without taking a branch cut, for example: $z^{1} $ is a well-defined analytic function.