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I'm having trouble with the following problem:

$$ \int \frac{z}{(z^2+1)^2}\ dz $$

I need to integrate around the circle centered at the origin with radius 3/2. When I try to solve using the residue theorem, I find that there are two poles of order 2 and +i and -i. However, when finding the residues, I get infinities as z approaches i and -i. How would I solve this?

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  • $\begingroup$ Slightly off-topic remark: There's no need to use residues in this case, since the integrand has an antiderivative in $\mathbf{C} \setminus \{ \pm i \}$, namely $-\frac12 (z^2+1)^{-1}$. $\endgroup$ – Hans Lundmark Mar 24 '17 at 8:34
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Given $\int_C \dfrac{z}{(z^2+1)^2}\ dz$ where $C:|z|=\dfrac{3}{2}$.

This equals $\int \dfrac{z}{(z+i)^2(z-i)^2}\ dz=\int \dfrac{f(z)}{(z-i)^2} dz$ where $f(z)=\dfrac{z}{(z+i)^2}$

In the above integral representation $z=i$ is a pole of order 2 and $f$ is analytic at $z=i$. So $\text{Res}_{z=i} \:f(z)=\dfrac{f'(z)}{1!}$. So the value of the integral is $2\pi i\text{Res}_{z=i}$. Now do the other part similarly with $\int \dfrac{g(z)}{(z+i)^2}dz$ where $g(z)=\dfrac{z}{(z-i)^2}$ and obtain the residue similarly as abovr. The actual value of the integral will be addition of respective integrals.

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  • $\begingroup$ The problem I have is that f'(z) is undefined at z = i $\endgroup$ – Evan Mar 24 '17 at 4:11
  • $\begingroup$ $f'(z)=\dfrac{i-z}{(z+i)^3}$ $\endgroup$ – Hirak Mar 24 '17 at 4:14
  • $\begingroup$ Never mind, I realized I took the derivative incorrectly. Thank you $\endgroup$ – Evan Mar 24 '17 at 4:17

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