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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

enter image description here

I was wondering can someone tell me why this is wrong? I'm not sure why $C_{r}$ is a half circle.

mathreadler
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user262291
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1 Answers1

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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 $| \int_{C}{f(z)}| \leq \frac{\pi R}{R^4-R^2-1} \to_{R \to \infty} 0$.

ADA
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