I am trying to find the solution to the equation- $$\Gamma(z)=i$$ I have tried doing it the following way- LHS is- $$\displaystyle \int_{0}^{\infty}t^ze^{-t}\ dt$$ Taking $z=a+ib$, we get- $$\displaystyle \int_{0}^{\infty}t^{a+ib}e^{-t}\ dt$$ or $$\displaystyle \int_{0}^{\infty}t^{a}t^{ib}e^{-t}\ dt$$ Using Euler's formula- $e^{i\theta}=\text{cos}\ \theta +\ i\ \text{sin}\ \theta$, we have- $$\displaystyle \int_{0}^{\infty}t^{a}(\text{cos}\ (b\ \text{ln}t) +\ i\ \text{sin}\ (b\ \text{ln}t)) e^{-t}\ dt$$ After this point, I am not being able to solve this integral. I have tried graphing it on desmos, but it doesn't seem like this question has any solution. How can I approach further in this?

  • $\begingroup$ In india we call $i$ as iota? Is that wrong? $\endgroup$
    – Physiker
    May 29, 2020 at 10:24
  • $\begingroup$ If you want numerical solutions you can plug it in wolfram alpha $\endgroup$
    – Mason
    May 29, 2020 at 10:27
  • $\begingroup$ Or inverting Gamma, see here. $\endgroup$ May 29, 2020 at 10:33
  • 1
    $\begingroup$ @SarthakGirdhar Iota is a Greek letter along with pi, alpha, beta etc. It is the ancestor of the Latin / English I in the same way that alpha is the ancestor of A. In its upper case form it looks just like I but again that is similar to alpha whose upper case looks like A. The local case of iota $\iota$ looks a little different in that it lacks the dot. It is not very often used in maths but, if it was, it would be regarded as a distinct symbol. $\endgroup$
    – badjohn
    May 29, 2020 at 10:58
  • $\begingroup$ I guess the point is that, US non-engineering they write $i$ for $\sqrt{-1}$. In (some) engineering they write $j$. And some places (India I guess) they write $\iota$. So should we "correct" $\iota$ to $i$ on math.se? $\endgroup$
    – GEdgar
    May 29, 2020 at 16:21

1 Answer 1


To find the zero of function $$f(z)=\Gamma(z)-i$$ I used Newton method with (why not ?) $z_0=1+i$. The iterates are $$\left( \begin{array}{cc} n & z_n\\ 0 & 1.00000+1.00000\,i \\ 1 & 2.56016+2.59498\,i\\ 2 & 2.24930+1.16218\,i \\ 3 & 2.72561+2.04341 \,i \\ 4 & 2.78276+1.60761\,i \\ 5 & 2.84247+1.68952\,i \\ 6 & 2.84550+1.68427 \,i \\ 7 & 2.84550+1.68429 \,i \end{array} \right)$$

Using @Robjohn's approximation, we should obtain $$z\sim e^{1+W(t)}+\frac 12\qquad \text{where} \qquad t=\frac 1 e \log \left(\frac{i}{\sqrt{2 \pi }}\right)$$ which, numerically, is $2.84071 +1.69496\, i$

Using it as $z_0$, Newton iterates are $$\left( \begin{array}{cc} n & z_n \\ 0 & 2.84071+1.69496\, i \\ 1 & 2.84547+1.68422\, i \\ 2 & 2.84550+1.68429\, i \end{array} \right)$$

Take care : this is only one of the many possible solutions (as usual when dealing with complex numbers).

Warning : In the previous edit of this answer, I had a terrible numerical mistake using @robjohn's superb approximation.


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