# Integral by Residue Theorem

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.

• There is also a pole at $z=-i$ – Seth Nov 29 '18 at 0:48
• You forgot the pole $-i$. – Bernard Nov 29 '18 at 0:49
• 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 – Seth Nov 29 '18 at 0:56
• I think you have accidentally asked this question twice – Seth Nov 29 '18 at 0:57
• 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. – 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

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

and

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

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.

• What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you – Safder Nov 29 '18 at 4:43
• @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. – Trevor Gunn Nov 29 '18 at 5:09