Definition of logarithm on complex plane I meet a problem in understanding logarithm on complex plane.
For example $f(z)=\ln(z+2)+\ln(z-2)$, clearly $f(z)$ has two branch points besides $|z|\to+\infty$
Accoroding to page 17 https://math.mit.edu/classes/18.306/Notes/Branch_Points_B_Cuts.pdf, we can have double dipolar system, that is,
$$z=r_{1}*\exp(i\theta_{1})+2=r_{2}*\exp{(i\theta_{2})}-2\tag{1}$$
and without loss of generality, choose $\theta_{1}\in[0,2\pi)$,$\theta_{2}\in[-\pi,\pi)$.
In this coordinate, how to deal with $g(z)=\ln(z+2)+\ln(z-2)+\ln(-1)$?
My guessing is, acorrding to (1),  $$f(-3)=\ln(-1)+\ln(-5)=\ln|5|\\ f(1)=\ln(3)+\ln(-1)=\ln{|3|}+i\pi\tag{2}$$
therefore (1) has uniquely defined $\ln(-1)=i\pi$, is this understanding correct?
 A: The Product Rule
In (2), you use the product rule to compute $\log(-1)$. However, the product rule doesn't necessarily work with the complex logarithm! For example, suppose we select the branch of the logarithm with the positive real axis deleted, defined with
$$
\log(re^{i\theta}) := \log(r) + i\theta \quad \text{for $\theta \in (0, 2\pi)$.}
$$
Note that $\log(-1) = \log(e^{i\pi}) = i\pi$ and $\log(-i) = \log(e^{3i\pi/2}) = 3i\pi/2$. However,
$$
\log(-1) = i\pi \neq 3i\pi = \log(-i) + \log(-i).
$$
Therefore, you cannot say $f(-3) = \log(5)$ by using the product rule.
Why Branches are Important
Next, notice this line you wrote under equation (1):

and without loss of generality, choose $\theta_1 = [0, 2\pi)$, $\theta_2 \in [-\pi, \pi)$.

I claim that you are losing generality here. In general, the logarithm is multi-valued, but by restricting $\theta_1$ and $\theta_2$, you are making it single-valued. For instance, I defined the logarithm above with $\theta \in (0, 2\pi)$. I could have easily chosen the range of $\theta$ to be $(2\pi, 4\pi)$ or $(\pi / 2, 5\pi / 2)$ or any of infinitely many other intervals. When I make such a choice, I am no longer working with the logarithm, but a branch of the logarithm. In particular, for each of these ranges, $\log(-1)$ takes on a different value!
When you restrict $\theta_1$ and $\theta_2$, you are really defining a function
$$
\log_1(z - 2) = \log_1(r_1e^{i\theta_1}) := \log(r_1) + i\theta_1
\quad \text{for $\theta_1 \in (0, 2\pi)$,}
$$
and another function
$$
\log_2(z + 2) = \log_2(r_2e^{i\theta_2}) := \log(r_2) + i\theta_2
\quad \text{for $\theta_2 \in (-\pi, \pi)$,}
$$
where $r_1, r_2, \theta_1, \theta_2$ are uniquely determined by (1). Altogether, you define $f$ with
$$
f(z) = \log_1(z - 2) + \log_2(z + 2).
$$
Here, we have taken $f(z) = \log(z - 2) + \log(z + 2)$, which is a multi-valued function, and defined it to be single-valued. In order to define this branch of $f$, we needed to define two distinct branches of the logarithm, $\log_1$ and $\log_2$.
log(-1) = ?
Now, how do we compute $\log(-1)$? Well, we can't. As I mentioned above, $\log_1$ and $\log_2$ define two distinct branches of the logarithm. In particular, $\log_1(-1) = i\pi$, while $\log_2(-1)$ isn't even defined! If we write out $g$, we get
$$
g(z) = \log_1(z - 2) + \log_2(z + 2) + \log_{??}(-1),
$$
and we can see the last term is clearly ambiguous.
In reality, this is not a problem. For any branch of the logarithm where $\log(-1)$ is defined, $\log(-1)$ is a constant. So once you pick a branch to work with, you can define $\log(-1) = (2n + 1)i\pi$ for any integer $n$. Just make sure you are explicit regarding the branch you work with!
