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.
 A: Let $z=a+ib$.  As $\vert  b \vert \to \infty$, the argument of $z$ does not approach $\pi$.  Thus Stirling's approximation applies, giving
$$\vert \Gamma(z) \vert \sim \left\vert \sqrt{\frac{2\pi}{a+ib}}\left(\frac{a+ib}{e}\right)^{a+ib}\right\vert\sim \sqrt{2\pi}\left\vert(ib)^{-1/2}\left(\frac{a+ib}{e}\right)^{a+ib}\right\vert$$
$$\sim\sqrt{2\pi}\vert b \vert^{-1/2} \left\vert \left(\frac{a+ib}{e}\right)^{a+ib}\right\vert \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a} \vert z^z\vert,$$
The $\vert z^z \vert$ term above satisfies
$$\vert z^z \vert = \vert e^{z \log z} \vert = \vert e^{(a+ib)(\log \vert z \vert + i \mathrm{arg}z)}\vert=e^{a \log \vert z \vert-b \mathrm{arg} z} \sim \vert b \vert^a e^{-b \mathrm{arg} z}$$
because $\vert z \vert \sim \vert b \vert$ as $b \to \infty$.  Thus, it remains to approximate $e^{-b \,\mathrm{arg z}}$ as $\vert b \vert \to \infty$.
Note: An easy mistake at this point is the false implication
$$\mathrm{arg}z \to \pm \pi/2 \quad \Longrightarrow \quad e^{b \mathrm{arg} z} \to e^{\pm b \pi/2}.$$
However, this need not hold, because the exponential is sensitive to non-dominant terms.
We compute
$$\lim_{\vert b \vert \to \infty} e^{-b\, \mathrm{arg} z}e^{-a+\vert b \vert \pi/2}=\mathrm{exp}\lim_{\vert b \vert \to \infty}\left(-b \arctan(b/a)-a+\vert b \vert \pi/2\right)$$
$$=\mathrm{exp}\lim_{\vert b \vert \to \infty}-b\left(\frac{\pm\pi}{2}-\frac{a}{b}+O\left(\vert b \vert^{-3}\right)\right)-a+\vert b \vert \pi/2,$$
in which we take the positive sign if $b \to \infty$ and the negative sign if $b \to -\infty$.  (This is simply the Taylor series to $\arctan(b/a)$ at $\pm \infty$.) Therefore we can replace said $\pm$ with $\vert b \vert/b$, at which point we see that our limit is $1$.
Thus
$$\vert \Gamma(z) \vert \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a}\vert b \vert^a e^{-b \,\mathrm{arg} z} \sim \sqrt{2\pi}\vert b \vert^{-1/2}e^{-a}\vert b \vert^a e^{a-\vert b \vert \pi/2}$$
$$\sim\sqrt{2\pi}\vert b \vert^{a-1/2} e^{-\vert b \vert \pi/2},$$
as claimed.
A: 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,
$$
\begin{align}
\left|\sqrt{2\pi}\,\frac{\color{#C00}{z^{z-1/2}}}{\color{#090}{e^z}}\right|
&=\sqrt{2\pi}\color{#C00}{\left(x^2+y^2\right)^{x/2-1/4}}e^{\color{#090}{-x}\color{#C00}{-y\arctan\!2(y,x)}}\\
&\sim\sqrt{2\pi}\,|y|^{x-1/2}e^{-\frac\pi2|y|}
\end{align}
$$
