$\mathrm{Log}(\prod_{k=1}^n (1+a_k))=\sum_{k=1}^n (\mathrm{Log}(1+a_k))$ Let $a_1,\ldots a_n\in \mathbb{C}$,  $|a_j|<1$
for each $j=1,\ldots n$ such that $\left| \prod_{k=1}^j (1+a_k)-1 \right| < 1$.
Show that
$$\mathrm{Log}\left(\prod_{k=1}^n (1+a_k)\right)=\sum_{k=1}^n (\mathrm{Log}(1+a_k))$$
Can anyone help?
 A: Let $\mathrm{Log}$ denote the (holomorphic) principal branch of the complex logarithm, defined on $G = \{ z \in \mathbb{C} \mid z \notin (-\infty,0] \}$, whose imaginary part lies in the interval $(-\pi,\pi)$. This means that for $z \in (0,\infty)$ the value of $\mathrm{Log}(z)$ is the ordinary real logarithm of $z$.
Let us start with $n = 2$. Since $\lvert a_1 \rvert, \lvert a_2 \rvert < 1$, we see that $1 + a_j$ have positive real parts so that we can write $1 + a_j = r_je^{i\phi_j}$ with $r_j >0$ and $\phi_j \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore $\phi_1 + \phi_2 \ne -\pi$ and we conclude $(1+a_1)(1+a_2) \in G$. Let $D = \{ z \in \mathbb{C} \mid \lvert z \rvert < 1 \}$ and define
$$f : D \times D \to \mathbb{C}, f(z_1,z_2) =  \mathrm{Log}\left(\prod_{k=1}^2 (1+z_k)\right) - \sum_{k=1}^2 \mathrm{Log}(1+z_k) .$$
We have $e^{f(z_1,z_2)} = 1$ for all $z_1,z_2$, hence $f$ must have values in $2\pi i \mathbb{Z} = \{ 2 \pi i k \mid k \in \mathbb{Z} \}$. Since $f$ is continuous and $D \times D$ is connected, also $f(D \times D)$ is connected, and this implies that $f$ must be constant. But $f(0,0) = 0$ which proves that $f \equiv 0$. Thus the claim is true for $n=2$.
Note that it is essential that $\mathrm{Log}$ was chosen as the principal branch, otherwise we would have $\mathrm{Log}\left(\prod_{k=1}^2 (1+z_k)\right) = \sum_{k=1}^2 \mathrm{Log}(1+z_k) + 2\pi i l$ for some fixed $l \ne 0$. In fact, if $\mathrm{Log}$ is a different branch on $G$, then $\mathrm{Log}(1) = 2 \pi i k$ for some fixed $k \ne 0$ and therefore $f \equiv f(0,0) = -2\pi i k$.
We finish the proof by induction. The equation is true for $n=1,2$. Assume it is true for some $n \ge 2$. For $n+1$ we argue as follows.
Let $b =  \prod_{k=1}^n (1+a_k)-1$. Then $\lvert b \rvert < 1$ and $ \prod_{k=1}^{n+1} (1+a_k) = (1+b)(1+a_{n+1})$. Hence
$$\mathrm{Log}\left(\prod_{k=1}^{n+1} (1+a_k)\right) = \mathrm{Log}\left((1+b) (1+a_{n+1})\right) = \mathrm{Log}(1+b) + \mathrm{Log} (1+a_{n+1})$$
$$= \sum_{k=1}^n \mathrm{Log}(1+a_k) + \mathrm{Log} (1+a_{n+1}) = \sum_{k=1}^{n+1} \mathrm{Log}(1+a_k) .$$
ADDED:
We can reformulate the above "theorem" as follows.
Let $U =  \{ z \in \mathbb{C} \mid \lvert z - 1 \rvert < 1 \} \subset G$ be the open disk with center $1$ and radius $1$, and let $\mathrm{Log}$ be a holomorphic branch of the logarithm on $G$. It is uniquely determined by the value $\mathrm{Log}(1)$ which has the form  $\mathrm{Log}(1) = 2 \pi i k$ for some $k \in \mathbb{Z}$.
If $A_1,\dots,A_n \in U$ and $\prod_{k=1}^j A_k \in U$ for $j =1,\dots,n$, then
$$\mathrm{Log}\left(\prod_{k=1}^n A_k \right) = \sum_{k=1}^n \mathrm{Log}A_k - (n-1) \cdot2 \pi i k .$$
This is proved as above. Now set $a_k = A_k-1$.
