<s>[INST] Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$ Let $\Gamma(z)$ be the gamma function.
In a paper I'm reading, the author states that $$ |\Gamma(z)| = |\Gamma(a+ib)|  \sim \sqrt{2 \pi} |b|^{a-\frac{1}{2}} e^{-|b|\frac{\pi}{2}}$$ as $|b| \to \infty$.
Can this asymptotic behavior be derived from Stirling's formula?
EDIT:
I think I have something.
Assume that $a,b >0$ and that $b$ is very large.
Then it would seem that
$$ \begin{align}|\Gamma(a+ib)|  &\sim \left|\sqrt{\frac{2 \pi}{a+ib}} \Big(\frac{a+ib}{e}\Big)^{a+ib} \right| \\ &= \sqrt{2 \pi}  \left|(a+ib)^{a-\frac{1}{2}} \right| \left| (a+ib)^{ib} \right| \left|e^{-a-ib} \right| \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} \ \left|\left(\sqrt{a^{2}+b^{2}} e^{i \arg \left(\frac{b}{a}\right)} \right)^{ib} \right| e^{-a} \\ &= \sqrt{2 \pi} \left(\sqrt{a^{2}+b^{2}} \right)^{a- \frac{1}{2}} e^{-b \arg \left(\frac{b}{a} \right)} e^{-a} \\   &\sim \sqrt{2 \pi} \, b^{a - \frac{1}{2}} e^{-b \frac{\pi}{2}} {\color{red}{e^{-a}}}. \end{align}$$
But apparently this is not quite correct.
 [/INST] Assume $|y|\gg|x|$. Note that
$$
\begin{align}
y\arctan\!2(y,x)
&=|y|\arctan\!2(|y|,x)\\[6pt]
&=\frac\pi2|y|-|y|\arctan\!2(x,|y|)\\
&\sim\frac\pi2|y|-x
\end{align}
$$
and
$$
x^2+y^2\sim|y|^2
$$
Therefore,
$$
\begin{align}
\left|\sqrt{2\pi}\,\frac{\color{#C00}{z^{z-1/2}}}{\color{#090}{e^z}}\right|
&=\sqrt{2\pi}\color{#C00}{\left(x^2+y^2\right)^{x/2-1/4}}e^{\color{#090}{-x}\color{#C00}{-y\arctan\!2(y,x)}}\\
&\sim\sqrt{2\pi}\,|y|^{x-1/2}e^{-\frac\pi2|y|}
\end{align}
$$
</s>