I'm trying to solve this question.
Let $C_{r}$ be the half-circle $|z|=R$, $0\leq \arg(z) \leq 2\pi$. Show that
$$\int_{C_{r}}\frac{1}{x^{4}+x^{2}+1}dx$$
Show that the integral approaches 0 as $R \rightarrow \infty$
What I'm confused about is that why $C_{r}$ corresponds to a half circle, because the question says that $0\leq \arg(z) \leq 2\pi$. I feel like that should translate to $C_{r}$ looking like
I was wondering can someone tell me why this is wrong? I'm not sure why $C_{r}$ is a half circle.
Consider the function $f(z)=\frac{1}{z^4+z^2+1}$, now on the upper half of the circle we have that $|f(z)| \leq \frac{1}{R^4-R^2-1}$. Now, we get that $| \int_{C}{f(z)}| \leq \frac{\pi R}{R^4-R^2-1} \to_{R \to \infty} 0$.