Why is $\lim\limits_{r\rightarrow\infty}\int\limits_{C_r}f(z)dz=0$ where $C_r$ is the half circle $\{|z|=r,~\text{Im}(z)\le0\}$

Let $ f(z)=\frac{\cos (z)e^{-2iz}}{z^2+2z+2}$

Why is $ \lim\limits_{r\rightarrow\infty}\int\limits_{C_r}f(z)dz=0$ where $ C_r$ is the half circle $ \{|z|=r,~\text{Im}(z)\le0\}$ ?

I was trying to estimate the modulus of $ f(re^\theta)$ and I’m not even sure if $ |\cos(z)|$ is bounded by $ 1$ . Seems like it’s not necessarily the case: $ |\cos(z)|=|\frac{e^{iz}+e^{-iz}}{2}|={1\over2}|e^{ix-y}+e^{-ix+y}|\le{1\over2}(|e^{-y}|+|e^y|)$

Here $ z=r\cos\theta+ir\sin\theta$ so $ |\cos z|\le{1\over2}(|e^{-r\sin\theta}|+|e^{r\sin\theta}|)$

Thankfully the numerator seems stay rather constant because of the $ |e^{-iz}|=|e^{2r\sin\theta}|$

So $ f(z)$ goes to zero as $ r$ approaches $ \infty$ . Is that sufficient to say $ \lim\limits_{r\rightarrow\infty}\int\limits_{C_r}f(z)dz=0 $ ?