We have the following problem, which we can't solve.
Let $a \in \mathbb{C}$ be fixed, with real part $1/2$ and imaginary part $\neq 0$. We consider parameters $n \in \mathbb{Z}$ and $k \in \mathbb{Z}_{\ge 0}$. We consider the following function, where $F = {}_2 F_1$ is the hypergeometric function: $$ A^a_{n,k} (t) = \frac{t^{k/2}}{k!} \cdot (a+n)_k \cdot F(a-n , \ a+n+k , \ 1+k , \ t).$$
Problem. We want to see that there exist $0 < t_0 < 1$ and $C>0$ such that $|A^a_{n,k} (t)| \leq C$ for all $t_0 < t < 1$ and all $n \in \mathbb{Z}$ and all $k \in \mathbb{Z}_{\ge 0}$.
Here are some possible approach ideas that we were suggested.
One can rewrite $$ A^a_{n,k} (1-t) = c^a_{n,k} (t) \cdot F(a-n,a+n+k,2a,t) + c^{\overline{a}}_{n,k} (t) \cdot F(\overline{a}-n,\overline{a}+n+k,2\overline{a},t),$$ where $$ c^a_{n,k} (t) := (1-t^{1/2})^k \cdot \frac{(a+n)_k}{k!} \cdot \frac{\Gamma (1+k) \Gamma (1-2a)}{\Gamma (\overline{a}+n+k)\Gamma (\overline{a}-n)}$$ and one can see that $$ |c^a_{n,k} (t)| = (1 - t^{1/2})^k \cdot \left| \frac{\Gamma (2 \overline{a} - 1)}{\Gamma (\overline{a})^2} \right|.$$ So here one needs to bound uniformly an hypergeometric function in the vicinity of $t = 0$.
Returining to the original expression, one can try to use the integral representation of the hypergeometric function, and then deform the contour to travel as much as possible along the locus $$ \left| \frac{1-tz}{1-z^{-1}} \right| = 1.$$ I will not write it down here. Seems to work relatively nicely for a fixed $t$, but as $t \to 1$ it "breaks".
One can try to use a differential equation satisfied by the hypergeometric function. We were explained that introducing a new variable $x = \log (t) - \log (1-t)$, and normalizing $$ Q := \frac{(a+n)_k}{k!} \cdot (e^{x/2}+e^{-x/2})^{-1} \cdot (1-t)^a \cdot t^{k/2+1/2} \cdot A^a_{n,k},$$ we obtain that $Q$ satisfies the differential equation $$ Q^{\prime \prime} (x) + V(x) Q(x) = 0,$$ where $$V(x) := \frac{s^2}{(1+e^{-x})^2} + \frac{s^2 - 1/4 + (n + k/2)^2}{(e^{x/2} + e^{-x/2})^2}- \frac{k^2 e^{-x}}{4(1+e^{-x})}$$ and $s$ is the imaginary part of $a$.
We are interested in the solution that, I think is $\sim c^a_{n,k} (0) \cdot e^{-isx} + c^{\overline{a}}_{n,k} (0) \cdot e^{isx}$ as $x \to +\infty$. But perhaps one can just take the solutions that are $\sim sin (sx)$ and $\sim cos (sx)$ and bound them (then dealing with real solutions).
It seems to me that using techniques from This post I can, using the behaviour of this differential equation, to reduce the estimation of a real solution (let us say now for simplicity that $k$ and $n$ are big enough), to its estimation at a single point $x_0$, where $$ e^{x_0} = -x_{00} + \sqrt{x_{00}^2 + k^2/4s^2},$$ where $$x_{00} := \frac{1}{2} \left( 1 + \frac{4n(n+k)-1}{4s^2} \right).$$ This is the point where $V$ changes sign. I think that knowing estimate at that point I can estimate everything in $(-\infty , x_0)$ and everything at $(x_0 , x_0 + \log 2)$ (the former because the solution is monotone there and $0$ at $-\infty$ and the latter by using a crude estimate). Considering also the point $x_1$ where $V$ starts to be monotone decreasing, I think that I can get an estimate on $(x_0+\log 2 , x_1)$ using the estimate at $x_0+\log 2$ and an estimate on $(x_1 , +\infty)$ using our knowledge at $+\infty$ (those two using the technique of the attached post, the answer of I. Pinelis, or perhaps also the answer of Eremenko).
