Simplifying this expression of gamma functions I was given a complex function $f(z)$, $z \in \mathbb{C}$, defined as an infinite product which I have managed to reduce to the form below. (I have checked numerically that both expressions agree for some values of $z$ and $N$ very large).
In this exercise, I need to show that for any $z \in \mathbb{C}$ the function 
$$
f(z) = \text{Lim}_{N\rightarrow\infty} \frac{1}{(N^2)!}\frac{\Gamma(N+1-i\sqrt{z})}{\Gamma(1-i\sqrt{z})}\frac{\Gamma(N+1+i\sqrt{z})}{\Gamma(1+i\sqrt{z})} \quad (N\in\mathbb{Z}),
$$
is entire and to find an explicit form of it. So the question is, How can I simplify the gammas above?
Thanks!
 A: Increasing $N$ from $k$ to $k+1$ multiplies the function of $N$ whose $N\to\infty$ limit is sought by $\frac{N^2+z}{\prod_{j=1}^{2N+1}(N^2+j)}$, which $\to0$ as $N\to\infty$. So $f=0$.
A: Rewrite your expression as
$$f=\frac{1}{\Gamma \left(N^2+1\right)}\frac{\Gamma (N+a)}{\Gamma (a)}\frac{\Gamma (N+b)}{\Gamma (b)}$$ Take logarithms, apply Stirling approximation and continue with Taylor series.
You should arrive at
$$\log(f)=-\log (N) \left(2 N^2-2 N+(2 -a-b)\right)-\log \left({\Gamma (a) \Gamma (b)}\right)+(N-2) N+$$ $$\frac{1}{2} \log (2 \pi )+\frac{3 a^2-3 a+3 b^2-3 b+1}{6 N}+O\left(\frac{1}{N^2}\right)$$
Replacing $a$ and $b$ by their values
$$\log(f)=-\Big[2 N(N-1) \log (N)-N(N-2) \Big]+\log \left(\frac{\sqrt{2} \sinh \left(\pi  \sqrt{z}\right)}{\sqrt{\pi z }   }\right)+\frac{1-6 z}{6 N}$$ $$+O\left(\frac{1}{N^2}\right)$$
So $\log(f)\to -\infty$ and then $f\to 0$.
Edit
Using @J.G's approach, using
$$K_N=\frac{1}{\Gamma \left(N^2+1\right)}\frac{\Gamma (N+a)}{\Gamma (a)}\frac{\Gamma (N+b)}{\Gamma (b)}$$ we have
$$\frac{K_{N+1}}{K_N}=\frac{(a+N) (b+N) \Gamma \left(N^2+1\right)}{\Gamma (N^2+2N +2)}$$
$$\log\left(\frac{K_{N+1}}{K_N} \right)=-4 N \log (N)-2+\frac{3(a+b)-{5}}{3N}+\frac{4-3 (a^2+ b^2)}{6
   N^2}+O\left(\frac{1}{N^3}\right)$$ that is to say
$$\log\left(\frac{K_{N+1}}{K_N} \right)=-4 N \log (N)-2+\frac{1}{3 N}+\frac{3z-1}{3N^2}+O\left(\frac{1}{N^3}\right)$$ that is to say
$$\frac{K_{N+1}}{K_N} \sim \frac 1{e^2\,N^{4N}}$$
