I have that $f(z)=\frac{e^{3iz}}{z^2+12}$ and what I actually need to show is that the integral below goes to $0$ as $R\rightarrow \infty$. The $\gamma_R$ curve is the semi circle, counterclockwise with radius $R$. Using the triangle inequality for integrals I get.
$$\left|\oint_{\gamma_R}\frac{e^{3iz}}{z^2+12} \ dz\right|\leq\max_{z\in\gamma_R}\left|\frac{e^{3iz}}{z^2+12}\right||\gamma_r|, \tag 1$$
In the book, there is an intermediate step before they proceed with $(1)$, which is
$$|e^{3iz}|\le e^{-3y} \le 1 \tag 2$$ $$|z^2+12|\ge |z|^2-12=R^2-12 \tag 3$$
This then implies that
$$\max_{z\in\gamma_R}\left|\frac{e^{3iz}}{z^2+12}\right||\gamma_r|\le \frac{\pi R}{R^2-12}\rightarrow 0, \quad \text{as} \quad R\rightarrow\infty.$$
I understand everything, except equation $(2)$. Can someone explain that? What is $y$ and why is $|e^{3iz}|\le e^{-3y}$?
You can write $z=x+iy$. Then $e^{3iz}=e^{-3y+3ix}=e^{-3y}e^{3ix}$. Since $|e^{3ix}|=1$, one has that $|e^{3iz}|=e^{-3y}$.