How can I show that the $\lim_{R\to\infty}\int_{C} \frac{e^{iz}dz}{(1+z^2)^2}$ is zero, where $C$ is a semicircle? Where $C$ is the positive semicircle of radius $R$, that is to say, it is the semicircle covering the first and second quadrant, $0 \leq \theta < \pi$
I know I have to use the $ML$ inequality, and I understand most of the steps, but I am having trouble with one particular part of the problem. I know that the points on this semicircle can be parametrized by $Re^{i\theta}$.
Now, in my notes it says that the absolute value of $e^{iz}$, when we plug $Re^{i\theta}$ in for $z$, is $e^{-R\sin(\theta)}$.  When I plug it in I get something different.  If I express $Re^{i\theta}$ as $R(\cos(\theta) + i\sin(\theta))$ I still get something different when I plug it into $e^{iz}$.  I do not understand what is going on here.  

Let $z = R(\cos\theta+i\sin\theta)$. Then
$$e^{iz} = e^{iR(\cos\theta+i\sin\theta)} = e^{iR\cos\theta+(-R)\sin\theta}=e^{-R\sin\theta}e^{iR\cos\theta}.$$
Hence
$$|e^{iz}| = |e^{iR\cos\theta+(-R)\sin\theta}|=|e^{-R\sin\theta}e^{iR\cos\theta}| = e^{-R\sin\theta}.$$ 
