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I have a question I needed to show that $$\lim_{R\to\infty} \int_{C_R} \frac {z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$ For $C_R$ the circle with radius R, center z=0 and positively oriented.

Which I have done using an ML estimate. The next part of the question is to show $$\int_{C} \frac {z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$ where C is the circle radius 5, center z=2, positively oriented in the complex plane.

I'm just not too sure how to use the ML estimate to show the second part of the question, because he hasn't done any examples in class and I haven't been able to find any similar questions online.

Thank you!

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  • $\begingroup$ Have you learned the residue theorem? $\endgroup$
    – Mark Viola
    Commented May 9, 2015 at 5:22
  • $\begingroup$ I provided an answer using the residue theorem. Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented May 9, 2015 at 5:38

2 Answers 2

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There are four poles of $\frac{z^2+4z+7}{(z^2+4)(z^2+2z+2)}$ inside $|z-2|<5$ are at $z=\pm 2i$ and $z=-1\pm i$. The residues are

$$\begin{align} &-\frac{7}{20}-i\frac{13}{40} \cdots \text{at z= 2i}\\ &-\frac{7}{20}+i\frac{13}{40} \cdots \text{at z= -2i}\\ \end{align}$$

and

$$\begin{align} &\frac{7}{20}-i\frac{1}{5} \cdots \text{at z= -1+i}\\ &\frac{7}{20}+i\frac{1}{5} \cdots \text{at z= -1-i} \end{align}$$

Inasmuch as these add to zero, the residue theorem guarantees that

$$\oint_C \frac{z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$

where $C$ is defined by $|z-2|=5$.

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August,

I'm guessing you're after help with this assignment. My suggestion would be to apply the Cauchy-Goursat theorem for domains with one hole. Try defining your domain as the circle $C_R$ with the circle $C$ removed.

Hope that helps.

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