Existence of $\lim_{n \to \infty} \frac{\prod_{z^n=1}f(z)}{\prod _{z^n=-1}f(z)}$ I have a following problem: Let $f(z)$ be a holomorphic function on a neighborhood of unit disk such that $|f(z)|>1$ for every $|z|=1$. Does there exist the following limit?
$$\lim_{n \to \infty} \frac{\prod_{z^n=1}f(z)}{\prod _{z^n=-1}f(z)}$$
Intuitively, I think if $f$ behaves nicely, such as if $f$ maps $S^1$ to a closed curve..., the limit exists and equals 1 (may be different by $(-1)^n$, for example if we choose $f(z)=2z$). 
In general I don't know how to verify if it's true or not. Someone can give me some conditions such that it's true?
Thanks in advance!
 A: Define
$$ L(\theta) = \int_{0}^{\theta} \frac{f'(e^{it})}{f(e^{it})} ie^{it} \, dt. $$
Then it is easy to check that $L$ is a $C^{\infty}$-function satisfying
$$f(e^{i\theta}) = f(1) e^{L(\theta)}. $$
Thus
$$ \frac{\prod_{z^{n} = 1} f(z)}{\prod_{z^{n} = -1} f(z)} = \exp\left[ \sum_{k=1}^{n} L\left( \frac{2k \pi}{n} \right) - \sum_{k=1}^{n} L\left( \frac{(2k-1) \pi}{n} \right) \right] $$
By the Taylor's theorem, the summation inside the exponential function is written as
$$ \sum_{k=1}^{n} L\left( \frac{2k \pi}{n} \right) - \sum_{k=1}^{n} L\left( \frac{(2k-1) \pi}{n} \right) = \frac{1}{2} \sum_{k=1}^{n} L'\left( \frac{2k \pi}{n} \right) \frac{2\pi}{n} + O\left(\frac{1}{n}\right).$$
(Here, the Big-Oh term is outside the summation.) Thus taking $n \to \infty$, it converges to
$$ \frac{1}{2} \int_{0}^{2\pi} L'(\theta) \, d\theta = \frac{L(2\pi) - L(0)}{2} = \frac{1}{2} \int_{|z| = 1} \frac{f'(z)}{f(z)} \, dz.$$
But we know that 
$$ \frac{1}{2\pi i} \int_{|z| = 1} \frac{f'(z)}{f(z)} \, dz $$
counts the number $N$ of zeros of $f(z)$, with multiplicity, inside the unit disc. Therefore we have
$$ \lim_{n\to\infty} \frac{\prod_{z^{n} = 1} f(z)}{\prod_{z^{n} = -1} f(z)} = e^{iN\pi} = (-1)^{N}. $$
