# Logarithm of Infinite Product of Complex Numbers

## Theorem

Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers.

The following are equivalent:

$(1): \quad$ The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty z_n$ converges to $z \in \C_{\ne 0}$.
$(2): \quad$ The series $\displaystyle \sum_{n \mathop = 1}^\infty\log z_n$ converges to $\log z + 2 k \pi i$ for some integer $k \in \Z$.

## Proof

### 1 implies 2

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

By Convergence of Series of Complex Numbers by Real and Imaginary Part, it suffices to show that:

$\displaystyle \sum_{n \mathop = 1}^\infty \operatorname{Re} \log z_n = \operatorname{Re} \log z$
$\displaystyle \sum_{n \mathop = 1}^\infty \operatorname{Im} \log z_n = \operatorname{Im} \log z + 2 k \pi$

#### Real Part

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

Then $P_n \to z$.

$\left\vert{P_n}\right\vert \to \left\vert{z}\right\vert$
$\log \left\vert{P_n}\right\vert \to \log \left\vert{z}\right\vert$

By definition of complex natural logarithm and Real Part of Sum of Complex Numbers:

$\displaystyle \log \left\vert{P_n}\right\vert = \sum_{j \mathop = 1}^n \operatorname{Re} \log z_j$

#### Imaginary Part

It remains to show that:

$\displaystyle \sum_{j \mathop = 1}^n \operatorname{Im} \log z_j \to \arg z + 2 k \pi$

for some $k \in \Z$.

Let $\theta = \arg z$ be the argument of $z$.

Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\left({\theta - \pi \,.\,.\, \theta + \pi}\right]$.

$\theta_n \to \theta$

Let $k_n \in \Z$ be such that:

$\displaystyle \sum_{j \mathop = 1}^n \operatorname{Im} \log z_j = \theta_n + 2 k_n \pi$

We show that $k_n$ is eventually constant.

By the Triangle Inequality:

$2 \pi \left\vert{k_{n + 1} - k_n}\right\vert \le \left\vert{\theta_{n + 1} - \theta_n}\right\vert + \left\vert{\operatorname{Im} \log z_{n + 1} }\right\vert$

By Factors in Convergent Product Converge to One, $z_n \to 1$.

By Complex Logarithm is Continuous Outside Branch, $\log z_n \to 0$.

Thus $2 \pi \left\vert{k_{n + 1} - k_n}\right\vert\to 0$.

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

Then:

$\displaystyle \sum_{j \mathop = 1}^n \operatorname{Im} \log z_j \to \theta + 2 k \pi$

$\Box$

### 2 implies 1

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

$\displaystyle \prod_{n \mathop = 1}^\infty z_n = z$

Hence the result.

$\blacksquare$