*I am not sure whether this is rather an MO or MSE question but it results from my research, so I put it here.*

In my effort to find (or to disprove the existence of) $k,l,h\in\mathbb{N}$ such that $2^{2(k+h)+l}-3^{h+l}|-(2^{2k}+2)3^{l-1}+2^{2k+l}$ and the quotient of the two being negative, I found that this is equivalent to finding an integer $n\in(-\mathbb{N})$ such that $A_{h,k}\left(-\dfrac{3^{l-1}}{2^{l}}\right)=n$ with $A_{h,k}$ being the Mobius transformation given by $$A_{h,k}=\begin{pmatrix}2^{2k}+2 & 2^{2k} \\ 3^{h+1} & 2^{2(k+h)}\end{pmatrix}.$$ Just for completeness: the matrix $A_{h,k}$ has positive determinant for any $k,h,l$. Since the Gamma function $\Gamma$ has poles on the negative integers, it could be possible to disprove the existence of such an $n$ by choosing a family of closed curves $\lbrace C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)\rbrace$ where $C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)$ contracts to $-\dfrac{3^{l-1}}{2^{l}}$ as $\varepsilon \to 0$ and proving that $$\lim_{\varepsilon\to 0}\oint_{C_{\varepsilon}\left(-\frac{3^{l-1}}{2^{l}}\right)}\Gamma(A_{h,k}(z))\mathrm{d}z=0$$ meaning that there is no pole at the point $-\dfrac{3^{l-1}}{2^{l}}$. And since there is no pole of $\Gamma\circ A_{h,k}$ at $-\dfrac{3^{l-1}}{2^{l}}$ the transformation $A_{h,k}$ does not map $-\dfrac{3^{l-1}}{2^{l}}$ to a negative integer. My problem is now the evaluation of the integral above. A change of variables does not help, since it asks exactly the same question as the initial problem.

Resolving the integral is related to calculating the residues of $\Gamma$. But this is never done using the integral definition of residues but rather the identity $z\Gamma(z)= \Gamma(z+1)$ and some limit considerations when we let $z$ tend to one of the poles of $\Gamma$. See for example the proofwiki proof of the residues. This is based on the simple poles of $\Gamma$ and the Laurent series expansion around one of them, e.g., $n$ as defined above. If $\Gamma\circ A_{h,k}$ has no pole at $-\frac{3^{l-1}}{2^{l}}$, then it is holomorphic in an environment of $-\frac{3^{l-1}}{2^{l}}$ and so the coefficients $a_i$ of its Laurent expansion around $-\frac{3^{l-1}}{2^{l}}$ satisfies $a_{-1}=0$. Therefore, I should obtain

$$\lim_{z\to -\frac{3^{l-1}}{2^{l}}} \left(z+\frac{3^{l-1}}{2^{l}}\right)(\Gamma\circ A_{h,k})(z)=0.$$

Is the reasoning until this point correct? Here I get stuck, since the technique, which is used for the $\Gamma$ function to obtain the residue at $n$ does not work here, since $\frac{3^{l-1}}{2^{l}}$ is not an integer. How could I proceed here?

**This seems like a general problem/approach**

Whenever a quotient of can be expressed as the action of some "reasonable" Mobius transform $A$ on a rational number $q$ one could either look at $\Gamma\circ A$ or $\Gamma\circ (-A)$ to check for negative or positive solutions, respectively, to the initial problem using the approach discussed above **IF** it even works out already in my example. Thoughts on this idea or references where this is already discussed are highly appreciated as well.