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I am interested in complex values of $z$ such that $$ \Gamma (z) =z$$ Clearly, the one trivial value of $z$ is 1. Also, looking at a graph of the gamma function on the real axis, I can tell that there are an infinite amount of negative real solutions and ( I believe) two positive real solutions. For other complex values I am stuck at the moment.

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  • $\begingroup$ Considering the plot of the function, I think the answer is yes. $\endgroup$ – mrs Aug 8 '14 at 11:44
  • $\begingroup$ I think we can find. $\endgroup$ – mrs Aug 8 '14 at 11:46
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    $\begingroup$ For positive integer values, $\Gamma (n) = (n-1)!$ so $\Gamma(1) = 0! = 1$, right? $\endgroup$ – Darth Geek Aug 8 '14 at 11:48
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    $\begingroup$ $-0.789467488-0.650981401i$ is a complex fix-point found with the newton-method. $\endgroup$ – Peter Aug 8 '14 at 12:39
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    $\begingroup$ Note that $\Gamma ({a-bi})$ is the complex conjugate of $\Gamma ({a+bi})$. So, if a+bi is a fixpoint, then a-bi is also one. $\endgroup$ – Peter Aug 8 '14 at 12:58
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Here's a picture of the function $\Gamma(z)-z$ for $\Re z,\Im z\in[-10,10]$. The black dots are zeroes of the function, which correspond to the fixpoints of $\Gamma(z)$. The dot in the center is $z=1$, and the big one on the right is $z=3.56$ identified in the comments. There are an infinite number of almost evenly spaced roots on the left near the poles, but since $\Gamma(z)\to 0$ quite quickly, the $-z$ behaviour washes out the poles and they soon become too small to be detected by the grapher.

                              Gamma fixpoints

The roots, then, can be classified as the special points $z=1$, $z=-0.789\pm0.651i$, one real root near each pole at the negative integers $n\le-2$ (there are no crossings in the region $(-1,0)$ because the function is too large by that point), and a line of zeroes along a vertical curve with the $z=3.56$ solution in the center.

The curve plotted here is $|\Gamma(z)|=|z|$. Of course every fixpoint is on this curve, but the curve comes in several connected components - one long line with the imaginary roots, one cycle containing the two exceptional imaginary roots, $z=1$ and the zero near $z=-2$, and a little circle around each pole at the negative integers starting at $-3$. I don't know if there is a closed form for the equation of the curve of imaginary roots such that $|\Gamma(f(y)+iy)|=|f(y)+iy|$, but I doubt it.

I don't think there are any equations to be had here, so numerical explorations of this sort are the best I can give you.

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