I'm working through a Complex Analysis text and am working through the Residue chapter. I am not sure if I am approaching this question correctly.

$$ \int_\gamma \frac{1}{(z-1)^2(z^2+1)}$$

Such that $\gamma$ is a circle of radius of 2 centered at 0. Traversed counter clockwise.

My approach was as follows:

I know that there are poles at $z=1,i,-i$ where $1$ is of order $2$.

First I found the residues at the poles as follows:

$$\operatorname{Res} _{z=1} :\lim_{z\rightarrow 1} \frac{d}{dz} \frac{(z-1)^2}{(z-1)^2(z^2+1)}$$ $$\lim_{z\rightarrow 1} \frac{d}{dz} \frac{1}{(z^2+1)}$$ $$\lim_{z\rightarrow 1} -\frac{2z}{(z^2+1)^2}$$ $$ = -1/2$$

$$\operatorname{Res} _{z=i} :\lim_{z\rightarrow i} \frac{1}{(z-1)^2}$$ $$ = \frac{1}{(i-1)^2}$$

$$\operatorname{Res} _{z=-i} :\lim_{z\rightarrow -i} \frac{1}{(z-1)^2}$$ $$ = \frac{1}{(-i-1)^2}$$

Therefore the integral is $2\pi i$*sum of the Residues.

$$= 2\pi i(\frac{1}{(i-1)^2} - \frac{1}{2} + \frac{1}{(-i-1)^2})$$

Thank you for any guidance. The repeated root is really where I'm wondering if it makes a difference.

  • $\begingroup$ There is also a pole at $z=-i$ $\endgroup$ – Seth Nov 29 '18 at 0:48
  • $\begingroup$ You forgot the pole $-i$. $\endgroup$ – Bernard Nov 29 '18 at 0:49
  • $\begingroup$ I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice $\endgroup$ – Seth Nov 29 '18 at 0:56
  • $\begingroup$ I think you have accidentally asked this question twice $\endgroup$ – Seth Nov 29 '18 at 0:57
  • $\begingroup$ Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer. $\endgroup$ – Safder Nov 29 '18 at 1:21

The residue at $1$ is computed correctly. The residues at $\pm i$ are a bit off. E.g.

$$ \operatorname{Res}_i(f) =\lim_{z \to i} (z - i)f(z) = \lim_{z \to i} \frac{z - i}{(z - 1)^2(z^2 + 1)} = \lim_{z \to i}\frac{1}{(z-1)^2(z + i)} $$

This should simplify to $\frac14$. The residue at $-i$ is also $\frac14$ so the sum of residues is

$$ \underbrace{-\frac12}_{z = 0} + \underbrace{\frac14}_{z = i} + \underbrace{\frac14}_{z = -i} = 0. $$

Alternatively, you can look at the question as

$$ \int_\gamma f = -\int_{- \gamma} f $$

where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $\infty$. Since all the poles in $\mathbb{C}$ are contained in $\{|z| < 2\}$, on $\{|z| > 2\}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So

$$ -\int_{- \gamma} f = -2\pi i\operatorname{Res}_{\infty} (f) = 2\pi i\operatorname{Res}_{0} \left( \frac{1}{z^2} f\left( \frac1z \right) \right). $$


$$ \operatorname{Res}_{0} \left[ \frac{1}{z^2} \frac{1}{\left( \frac1z - 1 \right)^2\left( \frac1{z^2} + 1 \right)} \right] = \operatorname{Res}_{0} \left[ \frac{z^2}{\left( 1 - z \right)^2\left( 1 + z^2 \right)} \right] = 0. $$

As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.

  • $\begingroup$ What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you $\endgroup$ – Safder Nov 29 '18 at 4:43
  • $\begingroup$ @Safder Yes, the residue of $f$ at $\infty$ is the same as the residue of $-\frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information. $\endgroup$ – Trevor Gunn Nov 29 '18 at 5:09

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