## 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 $$\lim\limits_{r\rightarrow\infty}\int\limits_{C_r}f(z)dz=0$$?