What number's factorial is $i$? 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?
 A: 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.
