Compute $i\oint_{|z|=1}\frac{(1-z^2)^2}{z^2(1+6z+z^2)} \ dz$

Question

I was inspired to try and solve the integral from this post. When I get to computing the residues for $z_2=\sqrt{8}-3$ I run into troubles.

We see that $z_1=0$ is a pole of order $2$ and $z_2=\sqrt{8}-3$ is the only simple pole that is inside of the unit circle. Letting

$$f(z)=\frac{g(z)}{h(z)}=\frac{(1-z^2)^2}{z^2(1+6z+z^2)} $$ we have that

$$\text{Res}_{z_1}(f)=\frac{g'(z_1)}{(2-1)!}=\frac{2(z_1^2-1)\cdot 2z_1}{1}=0.$$

For the other residue, we can use

$$\text{Res}_{z_2}(f)=\frac{g(z_2)}{h'(z_2)}=\frac{(1-z_2^2)^2}{z_2^2(1+6z_2+z_2^2)}=...$$

When the arithmetic becomes complicated I just give up and assume the method is incorrect. Is this just a mean problem or am I missing something important here?

Answer 1

In both cases you are doing a couple mistakes. In the first the function $g(z)$ is actually $$g(z)=\frac{(1-z^2)^2}{1+6z+z^2}$$ and $h(z)=z^2$ whereas in the second case $$g(z)=\frac{(1-z^2)^2}{z^2}$$ and $h(z)=1+6z+z^2$, so $h'(z)=2z+6$

From here the calculation goes as follows: in the first case The residue is $$\text{Res}\left(\frac{g}{f},0\right)=g'(0)=\frac{-4z(1-z^2)}{1+6z+z^2}-\frac{(1-z^2)^2\cdot(6+2z)}{(1+6z+z^2)^2}\ \Big|_{z=0}=-6$$ whereas in the second $$\text{Res}\left(\frac{g}{f},z_2\right)=\frac{g(z_2)}{h'(z_2)}=\frac{(1-z_2^2)^2}{z_2^2(2z_2+6)},$$ and here you plug in your value for $z_2$.

WA tells me the residue at $z_2$ is just $4\sqrt{2}$.

So in total we got that the integral equals

$$i\cdot2\pi i\cdot(4\sqrt{2}-6)=2\pi i\cdot(6-4\sqrt{2}).$$

Answer 2

Your method is incorrect. The correct formula for computing the residue at $0$ (since it is indeed a pole of order $2$) is:$$\frac1{(2-1)!}\lim_{z\to0}\frac{\mathrm d\frac{(1-z^2)^2}{1+6z+z^2}}{\mathrm dz}=\lim_{z\to0}\frac{2 \left(z^2-1\right) \left(z^3+9 z^2+3 z+3\right)}{\left(z^2+6z+1\right)^2}=-6.$$