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I found the approximation:
$$left| Gamma left(x + iy right) right|^{2} approx frac{pi y^{(2x - 1)}}{cosh(pi y)}
$$
for $y gt 2$, within an answer for another question, but I could not figure out where it comes from, nor find it anywhere else. How is this approximation derived and what is the error in the approximation?

From the context in which the approximation was used, $0leq xleq 1$ and $y>2$. I will assume that $ygg x$ and, therefore, $y gg 1$.

From the Stirling approximation,
$$
Gamma(z) approx sqrt{frac{2pi}{z}} left(frac{z}{e}right)^{z},
$$
Since $Gamma(z)$ is a holomorphic function for $mathrm{Re}(z)>0$, $Gamma(bar{z})=overline{Gamma(z)}$. Therefore,
$$
|Gamma(z)|^2 = Gamma(z) Gamma(bar{z}) approx frac{2pi}{sqrt{z bar{z}}}left(frac{z}{e}right)^{z} left(frac{bar{z}}{e}right)^{bar{z}}=frac{2pi}{|z|}left(frac{z}{e}right)^{z} left(frac{bar{z}}{e}right)^{bar{z}}.
$$
Using $z=re^{itheta}$,
$$
|Gamma(z)|^2 approx frac{2pi}{r} (r e^{-1+itheta})^{z} (re^{-1-itheta})^{bar{z}} = 2pi r^{z+bar{z}-1} exp(z(-1+itheta)-bar{z}(1+itheta)).
$$
Using $z + bar{z} = 2x $, $z-bar{z}=2iy$ and $r^2=x^2+y^2$, we get
$$
|Gamma(z)|^2 approx 2pi (x^2+y^2)^{(2x-1)/2} exp(-2x -2ytheta).
$$
Assuming $y gg x$ and $y gg 1$, $x^2+y^2 approx y^2$ and $theta approx pi/2$, leading to
$$
|Gamma(z)|^2 approx 2pi y^{2x-1} exp(-pi y),
$$
in which the term $2x$ in the argument of the exponential vanished because it's negligible compared to $pi y$. Since $2cosh y = e^y + e^{-y} approx e^y$ for large $y$, we have finally
$$
|Gamma(x+iy)|^2 approx frac{pi y^{2x-1}}{cosh(pi y)}.
$$
The plot below shows the accuracy of the approximation as function of $y$ for $x=0$ and $x=1$.

The plot below shows the error, defined as $left| |Gamma|^2_{exact}-|Gamma|^2_{approx.} right|/|Gamma|^2_{exact}$. For $x=1/2$ the formula is exact; the error as function of $y$ seems to be identical for $x=0$ and $x=1$. The error here is plotted for $x=0.1$, $x=0.6$ and $x=1$.

For $x=1$, the error is lesser than $1times 10^{-5}$ ($0.001 %$) for $y>2$, agreeing with the original claim on the approximation. For smaller $x$ the error is larger. Suggestions regarding the estimation of the error analytically will be appreciated.

I had a go at answering the approximation part of my own question, as an extension to @rafa11111's answer.

From the approximation (where $z=x +iy$):

$$
left| Gamma(z) right|^{2} approx frac{pi y^{2x-1}}{cosh(pi y)}
$$

for $0 leq x leq 1$ and $y>>1$, define the error in the approximation as:

$$
delta = left| left| Gamma(z) right|^{2} - frac{pi y^{2x-1}}{cosh(pi y)} right| cdot frac{1}{left| Gamma(z) right|^{2}}
$$

First to find the maximum error w.r.t $x$:

$$
frac{partial delta}{partial x} = sgn(delta)frac{left[ 2 ln(y) pi y^{2x-1} left| Gamma(z) right|^{2} cosh(pi y) - 2 pi y^{2x-1} left| Gamma(z) right| frac{partial left| Gamma(z) right|}{partial x} cosh(pi y)right]}{left| Gamma(z) right|^{4} cosh^{2}(pi y)}
$$
$$
= frac{pi y^{2x-1}}{left| Gamma(z) right|^{2} cosh(pi y )}left[ 2ln(y) -2frac{Re(Gamma(z) overline{Gamma'(z)})}{left| Gamma(z) right|} right]
$$
$$
= frac{pi y^{2x-1}}{left| Gamma(z) right|^{2} cosh(pi y )}left[ 2ln(y) -2 Re(psi(z)) right]
$$

where $sgn$ is the Sign Function (which is $=+1$ here) and $psi(z)$ is the Digamma Function. For the derivative of the modulus of a complex function I looked here. Next equate with zero to find the maximum which occurs when:

$$
Re(psi(z)) = ln(y)
$$

which can be numerically calculated and returns $x approx 0.7885$.

Second is to find the derivative of the error w.r.t $y$:

$$
frac{partial delta}{partial y} = sgn(delta)frac{left[ (2x-1) pi y^{2x-2} left| Gamma(z) right|^{2} cosh(pi y) - frac{partial}{partial y}left( pi y^{2x-1} left| Gamma(z) right|^{2} cosh(pi y) right)right]}{left| Gamma(z) right|^{4} cosh^{2}(pi y)}
$$
$$
= frac{pi y^{2x-1}}{left| Gamma(z) right|^{2} cosh(pi y )}left[ frac{2x-1}{y} -2frac{Re(-i Gamma(z) overline{Gamma'(z)})}{left| Gamma(z) right|} - pi tanh(pi y) right]
$$
$$
= frac{pi y^{2x-1}}{left| Gamma(z) right|^{2} cosh(pi y )}left[ frac{2x-1}{y} -2 Im(psi(z)) -pi tanh(pi y) right]
$$

For this to be negative we need:

$$
frac{2x-1}{y} lt 2 Im(psi(z)) + pi tanh(pi y)
$$
$$
y gt frac{2x-1}{2 Im(psi(z)) + pi tanh(pi y)}
$$

which for $y gt gt 1$ approximates further to:

$$
y gt frac{2x-1}{2 Im(psi(z)) + pi }
$$

Of which $Im(psi(z))$ has the same sign as $y$ and so, for $y gt gt 1$ , the RHS is either small (compared to $y$) and positive for $0.5 gt x geq 1$, zero for $x=0.5$, or negative for $0 geq x gt 0.5$. Therefore:

$$
frac{partial delta}{partial y} lt 0
$$

for $y gt gt 1$.

So, if we take $x = 0.7885$ and $y=2$ we get $delta = 0.0042$ which could be taken as the upper bound of the error for which the approximations hold. Then as $y rightarrow infty$, $delta rightarrow 0$ and the approximation becomes more accurate.

NB: I am a little unsure on the partial derivative of $ Gamma(z)$ w.r.t $x$ or $y$. I reasoned that as:

$$
frac{d Gamma(z) }{dz} = int_{0}^{infty} t^{z-1} ln(t) e^{-t} dt = Gamma'(z)
$$

then

$$
frac{partial Gamma(z) }{partial x} = int_{0}^{infty} t^{z-1} ln(t) e^{-t} dt = Gamma'(z)
$$

and

$$
frac{partial Gamma(z)}{partial y} = i int_{0}^{infty} t^{z-1} ln(t) e^{-t} dt = iGamma'(z)
$$