I am unable to get correct answer of this problem and so I am posting here to know I am making a conceptual mistake as I am unable to get a correct answer despite trying 3 times.
Question is - prove that poles of function are integer k$\geq$ n given a >15.
Function is $$-\frac { {\Gamma(nz) }^{a+3} × {\Gamma(n- nz +1) }^3 {\Gamma(nz+2n+1) }^3 } { {\Gamma(nz+n+1)^{a+3}} } $$ .
I tried to used the fact that $\Gamma(s) $ has poles has 0,-1, -2 ,... and then converting it to ns+0 , ns+1,... in the denominator of $\Gamma(ns) $ and similarly for other terms in Gamma function but still unable to get the answer. I think I am doing something wrong. Can anybody please help.
I shall be really grateful.
I didn't complete the problem, but I'll show you what I did. You'll find it straightforward to carry on, I think.
Let the given function be $f(z)$. In the denominator, repeatedly use the formula $\Gamma(z+1)=z\Gamma(z)$ to get $$ \Gamma^{a+3}(nz+n+1)=\Gamma^{a+3}(nz)\prod_{k=0}^n(nz+k)^{a+3}\tag1$$
Similarly, in the numerator we get $$ \Gamma^3(nz+2n+1)=\Gamma^3(nz)\prod_{k=0}^{2n}(nz+k)^3\tag2$$ and $$ \Gamma^3(1-nz+n)=\Gamma^3(1-nz)\prod_{k=0}^{n-1}(1-nz+k)^3\tag3$$
Now we use Euler's reflection formula $$\Gamma(z)\Gamma(1-z)=\frac\pi{\sin{\pi z}}$$ together with $(1),(2)\text{ and }(3)$ to get $$ f(z)=-\frac{\pi^3\prod_{k=0}^{2n}(nz+k)^3\prod_{k=0}^{n-1}(1-nz+k)^3}{\sin^3(n\pi z)\prod_{k=0}^n(nz+k)^{a+3}}\tag4$$
At this point, I stopped, because it's hopeless to prove the claim. Clearly, $z=-\frac1n$ is a zero of order $a+3$ of the denominator, and there simply can't be enough zeros of the numerator to cancel them all.
It should be simple to use $(4)$ to discover the location and order of all the poles and zeros, if you think it worthwhile.