Principal argument summation Let $\text{Arg}$ be a an principal argument in $(-\pi, \pi]$. I know that, for all $z_1,z_2\in\mathbb{C}\setminus \{0\}$, the expression $\text{Arg}(z_1z_2)= \text{Arg} z_1 + \text{Arg} z_2$ doesn't hold in general. But they are if $\Re z_1,\Re z_2&gt;0$. I suppose then the following expression must be true, for all $\Re z_n&gt;0$ and all $k\geq 1$,
$$\text{Arg}\left ( \prod_{n=1}^{k} z_n \right )= \sum_{n=1}^{k}\text{Arg}z_n$$
The question is then, is the following expression true, 
$$\text{Arg}\left ( \prod_{n=1}^{\infty} z_n \right )= \sum_{n=1}^{\infty}\text{Arg}z_n$$
if $\Re z_n&gt;0$, without even knowing about the convergence?
The first formula actually holds, but modulo $(-\pi, \pi]$. I believe you could talk about convergence of the partial sums modulo $(-\pi, \pi]$ for the infinite case. But for that to happen, you would need the series $\text{Arg}(z_n)$ to converge.
Convergence is well defined on $\mathbb{R}$, but we would need to define it on $\mathbb{R}/(-\pi, \pi]$. We can use the coinduced topology, so open sets on the quotient would be the image of the projection of open sets on $\mathbb{R}$. A series would converge on the quotient to a point $x$ if given any open set containing $X$, there is a $N \in \mathbb{Z}$ such as that the partial sums up to $n &gt; N$ are inside that open set. Open balls on the line, except that repeated over each interval of $2\pi$.
A open set on the quotient of center $x$ and radius $\rho$ will be correspondent to the union of open balls of radius $\rho$ and center $x + 2\pi k$, $k \in \mathbb{Z}$, on the real line. For each $z_n$, we can associate the set $X_{z_n}$ = $\{..., \text{Arg}_{-1}(z_n), \text{Arg}_{0}(z_n), \text{Arg}_{1}(z_n), ...\}$ of all the members on the congruence class of $\text{Arg}(z_n)$ modulo $(-\pi, \pi]$. If I take the open ball $S$ of center $0$ and radius $2\pi$ on the quotient, then all the elements of $X$ will be inside $S$ when seen on $R$. For different $z_n$'s, $\text{Arg}_j(z_{n_1})$  will be in the same connected open ball on $\mathbb{R}$ as $\text{Arg}_j(z_{n_2})$. Therefore, for a series to converge on the quotient, it will need to converge on every such open ball on $\mathbb{R}$ to a point $x_j$, and all those $x_j$ must be congruent to $x$ modulo $(-\pi,\pi]$. Then, we would say it converges to $x$.
But there's more. The series of arguments could converge on right side, but the product of numbers on the left may not. It may tend to a certain direction, while the radius of the product grows without bounds. So I would say that a necessary condition for the formula to hold is that the product on $\mathbb{C}$ converges. Actually, I think it would be sufficient and necessary, but that would take some delta and epsilon carrying around to solve.
EDIT: Your first equality doesn't follow without modularity. Think about the case when $z_1 = z_2 = z_3$,  $\Re z_1 &gt; 0$ and $\text{Arg}(z_1) = -\pi/3$. The reason it works with two complex numbers and $\Re z_1, \Re z_2 &gt; 0$, is that in that case it cannot jump over the modularity bounds.
The first formula actually holds, but modulo $(-\pi, \pi]$. I believe you could talk about convergence of the partial sums modulo $(-\pi, \pi]$ for the infinite case. But for that to happen, you would need the series $\text{Arg}(z_n)$ to converge.
Convergence is well defined on $\mathbb{R}$, but we would need to define it on $\mathbb{R}/(-\pi, \pi]$. We can use the coinduced topology, so open sets on the quotient would be the image of the projection of open sets on $\mathbb{R}$. A series would converge on the quotient to a point $x$ if given any open set containing $X$, there is a $N \in \mathbb{Z}$ such as that the partial sums up to $n &gt; N$ are inside that open set. Open balls on the line, except that repeated over each interval of $2\pi$.
A open set on the quotient of center $x$ and radius $\rho$ will be correspondent to the union of open balls of radius $\rho$ and center $x + 2\pi k$, $k \in \mathbb{Z}$, on the real line. For each $z_n$, we can associate the set $X_{z_n}$ = $\{..., \text{Arg}_{-1}(z_n), \text{Arg}_{0}(z_n), \text{Arg}_{1}(z_n), ...\}$ of all the members on the congruence class of $\text{Arg}(z_n)$ modulo $(-\pi, \pi]$. If I take the open ball $S$ of center $0$ and radius $2\pi$ on the quotient, then all the elements of $X$ will be inside $S$ when seen on $R$. For different $z_n$'s, $\text{Arg}_j(z_{n_1})$  will be in the same connected open ball on $\mathbb{R}$ as $\text{Arg}_j(z_{n_2})$. Therefore, for a series to converge on the quotient, it will need to converge on every such open ball on $\mathbb{R}$ to a point $x_j$, and all those $x_j$ must be congruent to $x$ modulo $(-\pi,\pi]$. Then, we would say it converges to $x$.
But there's more. The series of arguments could converge on right side, but the product of numbers on the left may not. It may tend to a certain direction, while the radius of the product grows without bounds. So I would say that a necessary condition for the formula to hold is that the product on $\mathbb{C}$ converges. Actually, I think it would be sufficient and necessary, but that would take some delta and epsilon carrying around to solve.
EDIT: Your first equality doesn't follow without modularity. Think about the case when $z_1 = z_2 = z_3$,  $\Re z_1 &gt; 0$ and $\text{Arg}(z_1) = -\pi/3$. The reason it works with two complex numbers and $\Re z_1, \Re z_2 &gt; 0$, is that in that case it cannot jump over the modularity bounds.