$Evluate $\int_{|z|=3}\frac{1}{z^5(z^2+z+1)}\ dz$$

Question

$Evluate $\int_{|z|=3}\frac{1}{z^5(z^2+z+1)}\ dz$$

Is there a straighforward way to compute
\[ \int_{|z|=3}\frac{1}{z^5(z^2+z+1)}\ dz\]
I don't think using partial fraction is the right approach.

Complex Analysis Calculus Integrals

Fredrick

43

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

The poles of the function $f(z)=\frac{1}{z^5(z^2+z+1)}$ are

\[-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i, \text{and} 0.\].
Since all poles are inside $|z|=2$, by the residue theorem we have

$$\tag 1 \int_{\{|z|=3\}} \frac{1}{z^5(z^2+z+1)}\, dz=\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz$$
for any $R>2.$

But
\[|\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz| \leq 2\pi R \frac{1}{R^5(R^2+R+1)} \rightarrow 0,\]
as $R \rightarrow \infty$. Thus

\[ \int_{\{|z|=3\}} \frac{1}{z^5(z^2+z+1)}\, dz=\lim_{R \rightarrow \infty}\int_{\{|z|=R\}} \frac{1}{z^5(z^2+z+1)}\, dz=0.\]

Erdos

4.7K

$Evluate $\int_{|z|=3}\frac{1}{z^5(z^2+z+1)}\ dz$$

Answer

Related Questions

Search