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Let $a_1,ldots a_nin mathbb{C}$, $|a_j|<1$
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?

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 $1 + a_j = r_je^{iphi_j}$ 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 $2pi i mathbb{Z} = { 2 pi i k mid k in mathbb{Z} }$. 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 $mathrm{Log}left(prod_{k=1}^2 (1+z_k)right) = sum_{k=1}^2 mathrm{Log}(1+z_k) + 2pi i l$ 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 $f equiv f(0,0) = -2pi i k$.

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 $a_k = A_k-1$.