# compute this integral using cauchy's formula

$$\oint_{\gamma}\frac{z^2+2z-5}{(4+z^2)^2(z^2+2z+2)}$$ $\gamma = \{z \in \mathbb{C} \:\mid\: |z| = R \},\; R > 0$

I think this integral depends on the values that $R$ takes on. there are four points where the function is not analytic $-1\pm i$ and $\pm2i$ so when $R < \sqrt2$ f is analytic and since $\gamma$ is a smooth, closed and simple curve by cauchy's theorem it should be equal to $0$ and when $\sqrt2 < R < 2$ I thought that by applying cauchy's formula to $f(z)=\frac{z^2+2z-5}{(4+z^2)^2}$ and for some point $w$ and after deriving the formula once I would be able to find it but it's been an hour that I'm struggling to solve this so I decided to ask for help I would appreciate any advices, tips or methods on how to get to the answer.

thanks!

• Do you mean $\mathbb{C}\setminus |z|=R$ rather than $\mathbb{C} \:/\: |z|=R$? The first means $\mathbb{C}$ without $|z|=R$ and the second is a quotient. – DMcMor Apr 4 '17 at 18:03
• $\gamma$ is a circle centered at the origin and a radius of $R$ – the_firehawk Apr 4 '17 at 18:06
• @NickPeterson I meant $\{z\in\mathbb{C}\textbf{ such that}\lvert z\rvert=R\}$ so yean I think that's what I actually meant – the_firehawk Apr 4 '17 at 18:08
• I updated your post with slightly different notation for the set definition. (I usually use \mid for a "such that" line.) – Nick Peterson Apr 4 '17 at 18:09
• @1.2.3 You may want to specify that in the question, so one knows where to help. In any case two or more singularities doesn't change a lot, you just need to compute the residue at each pole and sum them up. For the case where $R > 2$ you can calculate the 4 residues and add them up, or use the residue at infinity (en.wikipedia.org/wiki/Residue_at_infinity), which in your case it should be 0 if I'm not mistaken – Ant Apr 4 '17 at 18:32

The given integral is $$\int_\gamma\dfrac{z^2+2z-5}{(z+2i)^2(z-2i)^2(z+i+1)(z-i+1)}dz$$
So the singularities occur at $z=2i,-2i,-i-1,-1+i$ each of which are poles and $\gamma=\{z:|z|=R\}$. If we choose $R>2$, then all the singularities will fall inside the circle. In that case the value of the integral will be $2\pi i\sum\text{Res}_{z=z_i}f(z)$. If we choose $1<R<2$ then the value of integrals with singularities $2i$ and $-2i$ will be 0.
• thank you for your answer I get the first part but if you don't mind can you please elucidate why when $1<R<2$ (perhaps you meant $\sqrt2<R<2$) the integral is 0 ? – the_firehawk Apr 4 '17 at 18:29
• can you just plot the points on the plane ? The last two singularities are at $(-1,-1),(-1,1)$. So any circle of radius $>1$ and $<2$ will make the singularities at $(0,2)$ and $(0,-2)$ fall outside the circle – Hirak Apr 4 '17 at 18:41