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Encountered this function in one of my research problems

$$\frac{\Gamma \left(1-\dfrac{i c}{a}-\gamma \right) \Gamma \left(1+\dfrac{i c}{a}+\dfrac N 2-\gamma \right)}{\Gamma \left(1+\dfrac{i c}{a}-\gamma \right) \Gamma \left(1-\dfrac{i c}{a}+\dfrac N 2-\gamma \right)}-\frac{\left(a \gamma +b-\dfrac{i c}{2}\right) \left(\dfrac{a \gamma +b-\dfrac{i c}{2}-1}{a \gamma +b+\dfrac{i c}{2}-1}\right)^N}{a \gamma +b+\dfrac{i c}{2}}=0\quad\text{where } \{a,b,c\}\in \mathbb{C} \text{ and } N\in \mathbb{Z^+}$$

I am wondering if it is possible to solve for $\gamma$ analytically.

I solved it numerically for various parameters and the answers are interesting ($\gamma^\prime s$ lie on some sort of nice curve). For example for $$N\to 20,a\to 1.53\, -0.02 i,b\to 0.098\, -0.3 i,c\to -0.3+2 i$$

solutions look like enter image description here

and for another set of random parameters

$$N\to 30,a\to 1.5\, +2 i,b\to -0.5+0.3 i,c\to -3-2 i$$

Solution looks like

enter image description here

Playing with different parameters, I always got most of the solutions on a seemingly nice curve. So, I am wondering if there is a way to find the solution analytically.

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    $\begingroup$ this seems quite hopeless. $\endgroup$ Commented Mar 27, 2021 at 15:59
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    $\begingroup$ I would try using Stirling's formula to get an approximation for large N. $\endgroup$ Commented Mar 27, 2021 at 19:19

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