Logarithm of Infinite Product of Complex Numbers

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Let $\sequence {z_n}$ 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 $k \in \Z$.


$(1)$ 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:

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

Real Part

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

Then $P_n \to z$.

By Modulus of Limit:

$\cmod {P_n} \to \cmod z$

By Natural Logarithm Function is Continuous;

$\log \cmod {P_n} \to \log \cmod z$

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

$\displaystyle \log \cmod {P_n} = \sum_{j \mathop = 1}^n \Re \log z_j$

Imaginary Part

It remains to show that:

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

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 $\hointl {\theta - \pi} {\theta + \pi}$.

By the corollary to Convergence of Complex Sequence in Polar Form:

$\theta_n \to \theta$

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

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

We show that $k_n$ is eventually constant.

By the Triangle Inequality:

$2 \pi \cmod {k_{n + 1} - k_n} \le \cmod {\theta_{n + 1} - \theta_n} + \cmod {\Im \log z_{n + 1} }$

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 $2 \pi \cmod {k_{n + 1} - k_n} \to 0$.

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


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


$(2)$ implies $(1)$

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

By Exponential of Series Equals Infinite Product:

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

Hence the result.


Also see