# Calculate integral using the residue theorem

I want to calculate the integral

$$\int_C {{z^2-2z}\over{(z+1)^2(z^2+4)}}dz$$, where $C=\{z:|z|=4\}$

I want to use the Residue theorem to tackle this integral. Now, $f(z)$ has a pole of degree 2 at $z=-1$, and then zeros at $2i, -2i .$

Now, I take the $Res(f, -1)$, and get $-14\over25$ by using $H'(z)/1!$, and then $Res(f,2i)={{7+i}\over 25}$, and $Res(f,-2i)={{7-i}\over 25}$.

For the final answer, I know that I have to add the residue together and multiply with $2\pi$ to get the value of the integral, but which residue do I choose, and does $|z|=4$ play any role in this final answer?

Thanks for any input.

• What is $f$ you are talking about? If you are talking about the function you integrate then indeed it has a second order pole at $z = -1$ and there are simple poles at $z = \pm 2i$, not zeros. The contour $C = \{z:|z| = 4\}$ only needs to be taken account for when checking which residues to evaluate, not after. So in your case, $C = \{z:|z| = 100\}$ would give you the same result. I did not check the values you get for the residues so cannot tell you what the answer is but if you apply the exact formula of residue theorem then there is no problem. – user88595 Apr 28 '14 at 16:59
• – Mhenni Benghorbal Apr 28 '14 at 18:38

Include the residues of points located inside $C$. In your case, this is all of them.
• Then wouldn't I get $0$ as an answer? Is that right? – Akaichan Apr 28 '14 at 16:33
• @Akaichan Yes, that looks right. If you are unsure about your answer using residues, you can always parametrize your contour and get a real integral that you can evaluate by hand or in Mathematica to check. Here it's easy: just let $z = 4 e^{it}, 0 \leq t \leq 2 \pi$. – jc315 Apr 28 '14 at 16:43