Logarithm of Infinite Product of Complex Numbers

Theorem
Let $(z_n)$ be a sequence of complex numbers with real part $>-1$.

Then the following are equivalent:


 * The infinite product $\displaystyle\prod_{n=1}^\infty(1+z_n)$ converges to $z\in\C$.


 * There exists an integer $k\in\Z$ such that the series $\displaystyle\sum_{n=1}^\infty\log(1+z_n)$ converges to $\log z+2k\pi i$.

Proof
Let $P_n$ denote the $n$th partial product.

Suppose $(P_n)$ converges in $\C\setminus\{0\}$.

Then $(\log|P_n|)$ converges to $\log|z|$.

We have $\log|P_n|=\sum_{j=1}^n\Re\log(1+z_j)$.

It remains to show that $\sum_{j=1}^n\Im\log(1+z_j)\to\arg z+2k\pi$ for some $k\in\Z$.

Let $\theta=\arg z$ be the argument of $z$ and let $\theta_n=\arg P_n$ be the argument of $P_n$ in the half-open interval $(\theta-\pi,\theta+\pi]$.

By Convergence of Complex Sequence in Polar Form/Corollary, $\theta_n\to\theta$.

Let $k_n\in\Z$ be such that $\sum_{j\leq n}\Im\log(1+z_j)=\theta_n+2k_n\pi$.

By the Triangle Inequality:
 * $2\pi|k_{n+1}-k_n|\leq|\theta_{n+1}-\theta_n|+|\Im\log(1+z_{n+1})|\to0$.

So $k_n$ is eventually constant, say equal to $k$.

Then $\sum_{j=1}^n\Im\log(1+z_j)\to\theta+2k\pi$.

Conversely, suppose $\displaystyle\sum_{n=1}^\infty\log(1+z_n)=\log z+2k\pi i$.

Because $\exp$ is continuous, $\displaystyle\prod_{n=1}^\infty(1+z_n)=z$.

Also see

 * Absolute Convergence of Infinite Product of Complex Numbers