Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$

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.