Limit of quotient of products $\lim_{n\to \infty}\left|\frac{\prod_{z^n=1}f(z)^n}{\prod_{z^n=-1}f(z)^n}\right|$ I have a following problem: Let $f$ be a holomorphic function such that $|f(z)|>1$ for $|z|=1$. So what can we say about the limit $$\lim_{n\to \infty}\left|\frac{\prod_{z^n=1}f(z)^n}{\prod_{z^n=-1}f(z)^n}\right|?$$
Thanks in advance for any help!
 A: This was fun. Where did you get this problem?
[edit] I was a little too rash in my first estimates, but I think the following is correct: The original sequence can converge or not, but I don't think the limit will be very interesting in any case. To see this, we first prove the next claim (which is the original answer, minus the exponent $n$):
Claim.
Let $f$ be holomorphic on an open set that contains the unit circle $S^1$. Then
$$
\lim_{n \to \infty} 
\biggl\lvert
\frac{\prod_{z^n = 1} f(z)}{\prod_{w^n = -1} f(w)}
\biggr\rvert
= \exp \int_{S^1} \operatorname{Re}\biggl(\frac{f'(z)}{f(z)} \biggr) d\lambda.
$$
Proof:
For $n$ fixed write the $n$-th roots of $1$ and $-1$ as
$$
z_j = e^{2\pi j/n}, \quad
w_j = e^{2\pi j/n+ i\pi/2n}.
$$
For future use, note that $z_j - w_j = z_j(1 - e^{i\pi /2n})$, so $z_j - w_j \to 0$ as $n \to \infty$.
We now take logarithms and calculate that
$$
\log
\biggl\lvert
\frac{\prod_{z^n = 1} f(z)}{\prod_{w^n = -1} f(w)}
\biggr\rvert
=
\sum_{j=1}^n \frac{\log|f(z_j)| - \log|f(w_j)|}{z_j - w_j}(z_j-w_j).
$$
The right-hand side reminds us of the derivative. If $\epsilon > 0$ is given then, since $z_j - w_j \to 0$ as $n \to \infty$, there exists an $n_\epsilon$ such that
$$
\biggl\lvert
\frac{\log|f(z_j)| - \log|f(w_j)|}{z_j - w_j}
- (\log |f(z_j)|)'
\biggr\rvert
\leq \epsilon
$$
for all $n \geq n_\epsilon$. If this is the case, then
$$
\biggl\lvert
\sum_{j=1}^n \frac{\log|f(z_j)| - \log|f(w_j)|}{z_j - w_j}(z_j-w_j)
- \sum_{j=1}^n (\log |f(z_j)|)'(z_j - w_j)
\biggr\rvert
\leq n \epsilon |z_j - w_j|.
$$
for $n \geq n_\epsilon$.
Recall now that $|z_j - w_j| = |z_j(1 - e^{i \pi/2n})| = |e^{i\pi/2n} - 1|$. 
Expanding $e^{i\pi/2n}-1$ as a series we find that
$$
n|z_j - w_j| = 
n \Bigl\lvert \frac{i\pi}{2n} 
+ O\Bigl(\frac{1}{n^2}\Bigr)
\Bigr\rvert
\leq \frac{\pi}{2} + O\Bigl( \frac{1}{n} \Bigr).
$$
This is good enough for our purposes, which are to show that the difference of the two sums above becomes smaller than any given $\epsilon$ as $n \to \infty$: Given $\epsilon$ we make an $\epsilon'$ such that $\epsilon' \frac{\pi}2 < \epsilon$, use that one to estimate the derivative and fudge everything so that in the end $\epsilon' \frac{\pi}{2} + O(1/n) \leq \epsilon$, which can be done by taking $n$ ludicrously big.
We have thus shown that the two big sums converge to the same limit as $n \to \infty$. But the sum involving the derivative is a Riemann sum, so
$$
\lim_{n \to \infty} 
\biggl\lvert
\frac{\prod_{z^n = 1} f(z)}{\prod_{w^n = -1} f(w)}
\biggr\rvert
=
\exp \int_{S^1} (\log|f(z)|)' d\lambda,
$$
where we write $S^1$ for the unit circle and the $\exp$ appears because we calculated the $\log$ of the original sequence. A little manipulation shows that since $f$ is holomorphic we have
$$
(\log|f(z)|)' = \operatorname{Re} \biggl(\frac{f'(z)}{f(z)} \biggr),
$$
which gives the result. 
Remark that this simplification is the only part that requires $f$ to be holomorphic, until now the argument carries through with $f$ only complex valued and once-differentiable.
Now back to the original question. Let's write $A_n = |\prod_{z^n = 1} f(z) / \prod_{w^n = -1} f(w)|$. This is a sequence of nonnegative real numbers that converges to $A = \exp\int_{S^1} \operatorname{Re}(f'(z)/f(z)) d\lambda$ by the claim. I see three cases:
Case 1: $A > 1$. Then all $A_n > 1 + \delta$ for all $n$ big enough and for some fixed $\delta  >0$, so $A_n^n$ goes to infinity.
Case 2: $A < 1$. Then $A_n^n \to 0$ by a similar argument.
Case 3: $A = 1$. I don't know. Maybe $A_n^n \to 1$ also, maybe it oscillates weirdly around $1$, maybe this is somehow trivial?
A: There are the same number of $n^{th}$ roots of 1,  $\omega_1 , ~..., \omega_n $, as there are roots of -1, $\eta_1 , ~..., \eta_n $. With this in mind we can consider the expression as $ lim|\frac{f(\omega _1)}{f(\eta _1)}||\frac{f(\omega _2)}{f(\eta _2)}| \ ... \ |\frac{f(\omega _n)}{f(\eta _n)}|$ Then you can estimate how close $f(\omega _i)$ is to $f(\eta _i)$ using the maximum of $f'(z)$ on the unit circle. As $n$ becomes large they will get very close, but there will be more things to multiply.
The $n^{th}$ roots of unity are defined as 
$\omega_j = \exp(\frac{2 \pi i}{n} j ) $ for $j = 0,1,...,n-1$ These are located along a circle of radius 1, evenly spaced.
The $n^{th}$ roots of -1 are defined as 
$\eta_j = \exp(\frac{\pi i}{n}+ \frac{2 \pi i}{n} j ) $ for $j = 0,1,...,n-1$ These are also located along a circle of radius 1, evenly spaced.
So we have two sets of $n$ points evenly spaced along the unit circle. We can then "pair" up these points with each other, pairing each $\omega$ with its closest $\eta$. Since the $\eta$ are evenly spaces there will always be one within an $\frac{2 \pi i}{n}$ arc along the circle. So as $n$ becomes large each $\omega$ will become closer to its paired $\eta$. We relable the roots so that $\omega _l$ is paired with $\eta _l$ then write out the product as above.
As $n$ gets big ethe distance between each pair shrinks, so $f(\omega _l)$ and $f(\eta_l)$ approach each other. How quickly does this happen? Since $f$ is holomorphic on the (compact) unit circle it has bounded derivative there, so $|f'(z)| < L$ for $|z|=1$. It follows that $|f(\omega_l) - f(\eta_l)| < L|\omega_l - \eta_l|$. Then calculate $|\omega_l - \eta_l|$ and use this to put an upper bound on the product. Through further manipulation and using the fact that $|f(z)| > 1$ I think you get the limit is less than $1$. It is probably equal to $1$ as well.
