text: How do I find the zeros of this function?
  
  How do I find the zeros of:
  
  1/Gamma[1 - I x]
  
  
  The zeros should be $-i\, n$ with $n>0$
  
  I tried with Solve[1/Gamma[1 - I x]==0,x], but I get the message:
  
  Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.
  
  So I tried with Reduce[1/Gamma[1 - I x]==0,x], but I get False as a result
  
  • Use the identity Gamma[z] Gamma[1 - z] == Pi/Sin[Pi z] with z=I x... Apr 14, 2018 at 20:16
  
  f[x_] = 1/Gamma[1 - I x];
  
  Plot[f[y*I], {y, -5, 5}, PlotRange -> All]
  
  
  FindRoot[f[y*I] == 0, {y, #}] & /@ Range[-5.25, -1.25]
  
  (* {{y -> -5.}, {y -> -4.}, {y -> -3.}, {y -> -2.}, {y -> -1.}} *)
  
  Assuming[{Element[n, Integers], n > 0},
  Limit[f[-m*I], m -> n]]
  
  (* 0 *)
  
  • havent you started from the solution? I mean, in your reasoning, why you know already the argument of your plotted function? I am a bit puzzled, as I thought that the OP want to know the solution for $1/\Gamma(1-iz)=0$ by using MMA functions... Apr 15, 2018 at 12:58
  • @JoséAntonioDíazNavas - Function is a given. x has to be pure imaginary to give a real argument to Gamma. Plot of the function suggests that the roots are near the negative integers. FindRoot verified that starting the search near a negative integer would converge to the integer. However, Gamma is undetermined (ComplexInfinity) at negative integers, so the roots only exist in the Limit. The last Mathematica statement verifies root for all negative integers. Apr 15, 2018 at 14:45
  
  Here is my proposal. The product of two gamma functions $\Gamma(z)$ and $\Gamma(w)$, with arguments satisfying the condition that $z+w$ is an integer, as in our case with $z=ix$ and $w=1-ix$, then:
  
  $$\Gamma(z) \Gamma(1 - z) =\frac{\pi}{\sin(\pi z)}$$
  
  So:
  
  Reduce[Gamma[I x] Sin[I \[Pi] x]/\[Pi] == 0, x]
  `
  
  $$c_1\in \mathbb{Z}\land \left(\left(c_1\leq -1\land \left(x=2 i c_1\lor x=2 i c_1+i\right)\right)\lor \left(i x\notin \mathbb{Z}\land \left(x=2 i c_1\lor x=\frac{2 i \pi c_1+i \pi }{\pi }\right)\right)\right)$$
  
  which includes your expected result.
explanation: The provided text is a detailed discussion about finding the zeros of a specific function, \( \frac{1}{\Gamma(1 - i x)} \). It includes a question about how to find the zeros, an attempt to solve the problem using Mathematica, and a detailed solution provided by another user. The solution involves using a known identity and verifying the roots.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: How do I find the zeros of this function?
  
  How do I find the zeros of:
  
  1/Gamma[1 - I x]
  
  
  The zeros should be $-i\, n$ with $n>0$
  
  I tried with Solve[1/Gamma[1 - I x]==0,x], but I get the message:
  
  Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.
  
  So I tried with Reduce[1/Gamma[1 - I x]==0,x], but I get False as a result
  
  • Use the identity Gamma[z] Gamma[1 - z] == Pi/Sin[Pi z] with z=I x... Apr 14, 2018 at 20:16
  
  f[x_] = 1/Gamma[1 - I x];
  
  Plot[f[y*I], {y, -5, 5}, PlotRange -> All]
  
  
  FindRoot[f[y*I] == 0, {y, #}] & /@ Range[-5.25, -1.25]
  
  (* {{y -> -5.}, {y -> -4.}, {y -> -3.}, {y -> -2.}, {y -> -1.}} *)
  
  Assuming[{Element[n, Integers], n > 0},
  Limit[f[-m*I], m -> n]]
  
  (* 0 *)
  
  • havent you started from the solution? I mean, in your reasoning, why you know already the argument of your plotted function? I am a bit puzzled, as I thought that the OP want to know the solution for $1/\Gamma(1-iz)=0$ by using MMA functions... Apr 15, 2018 at 12:58
  • @JoséAntonioDíazNavas - Function is a given. x has to be pure imaginary to give a real argument to Gamma. Plot of the function suggests that the roots are near the negative integers. FindRoot verified that starting the search near a negative integer would converge to the integer. However, Gamma is undetermined (ComplexInfinity) at negative integers, so the roots only exist in the Limit. The last Mathematica statement verifies root for all negative integers. Apr 15, 2018 at 14:45
  
  Here is my proposal. The product of two gamma functions $\Gamma(z)$ and $\Gamma(w)$, with arguments satisfying the condition that $z+w$ is an integer, as in our case with $z=ix$ and $w=1-ix$, then:
  
  $$\Gamma(z) \Gamma(1 - z) =\frac{\pi}{\sin(\pi z)}$$
  
  So:
  
  Reduce[Gamma[I x] Sin[I \[Pi] x]/\[Pi] == 0, x]
  `
  
  $$c_1\in \mathbb{Z}\land \left(\left(c_1\leq -1\land \left(x=2 i c_1\lor x=2 i c_1+i\right)\right)\lor \left(i x\notin \mathbb{Z}\land \left(x=2 i c_1\lor x=\frac{2 i \pi c_1+i \pi }{\pi }\right)\right)\right)$$
  
  which includes your expected result.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed discussion about finding the zeros of a specific function, \( \frac{1}{\Gamma(1 - i x)} \). It includes a question about how to find the zeros, an attempt to solve the problem using Mathematica, and a detailed solution provided by another user. The solution involves using a known identity and verifying the roots.
  
  Label: question-and-answer