Logarithm of Infinite Product of Complex Numbers

Theorem
Let $(z_n)$ be a sequence of complex numbers with real part $>0$.

Then the following are equivalent:


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


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

Proof
Suppose $\displaystyle\prod_{n=1}^\infty z_n$ converges to $z$.

By Convergence of Series of Complex Numbers by Real and Imaginary Part, it suffices to show that:
 * $\displaystyle\sum_{n=1}^\infty\Re\log z_n=\Re\log z$
 * $\displaystyle\sum_{n=1}^\infty\Im\log z_n=\Im\log z+2k\pi$

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

Then $P_n\to z$.

By Modulus of Limit, $|P_n|\to|z|$.

By Natural Logarithm Function is Continuous, $\log|P_n|\to\log|z|$.

By definition of complex logarithm and Real Part of Sum of Complex Numbers, $\log|P_n|=\sum_{j=1}^n\Re\log z_j$.

It remains to show that $\sum_{j=1}^n\Im\log z_j\to\arg z+2k\pi$ 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 $\sum_{j\leq n}\Im\log z_j = \theta_n+2k_n\pi$.

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

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

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

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

By Exponential of Series Equals Infinite Product, $\displaystyle\prod_{n=1}^\infty z_n=z$.

Also see

 * Absolute Convergence of Infinite Product of Complex Numbers