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.
The text passage contains multiple questions related to finding the zeros of a complex function and the results of using different Mathematica functions to solve the problem. It also includes an answer suggesting the use of a specific identity and providing a solution.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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.
user
The text passage contains multiple questions related to finding the zeros of a complex function and the results of using different Mathematica functions to solve the problem. It also includes an answer suggesting the use of a specific identity and providing a solution.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS