I am working with the integral $$\lim_{R\rightarrow \infty} \int_\gamma \frac{e^{i3z}}{z^2+4} dz$$ Where $\gamma$ is the contour $\gamma(t) = Re^{-it}$, $0 \leq t \leq \pi$ (the upper half of a circle in the complex plane centered at the origin and with radius $R$ ).
Apparently it follows that $$\biggl| \frac{e^{-i3z}}{z^2+4}\biggr| = \frac{e^{-3R\sin(t)}}{|z^2+4|}$$
I must be missing something obvious, but I cannot see how the equality in the numerators is deduced. Why is the real part of the contour being disregarded?
Hint: Rewrite the $z$ in the exponent of $e$ into polar form: $z=R(\cos t+i \sin t)$. Then notice that $|e^{-it}|=1$ for any real number $t$.