$\oint_{\gamma}\frac 1{z^2+1}dz$ on different curves

Question

$\oint_{\gamma}\frac 1{z^2+1}dz$ on the curves: $\gamma_1: |z + i| = 1$, $\gamma_2:|z - i| = 1$, $\gamma_3: |z| = \frac 12$, $\gamma_4: |z - i| = \frac 32$. I would like to use the residue theorem and we see that we have $z_{1,2}=\pm i$. Now it makes it a problem of in which of those circles which poles we have, right? For $\gamma_1$ we will only use $z = -i$ since it is the center of the circle and will always be there, do I have to consider both cases when $|z + i| < 1$ and $|z + i| > 1$? for each of them? Do you have any techniques to faster know if a pole is inside the curve or not? Or actually, how do I check that algebraically?

Answer 1

I'm not sure whi it is a problem: the only poles are the simple ones $\;\pm i\;$ ,so check:

$$\gamma_1: |-i+i|=0<1\;,\;\;|i+i|=2|>1\implies -i\;\text{ is in, $\,i\,$ is out}$$

...and etc. Do the same as above with all the other paths and check whether $\;-i\,,\,\,i\;$ are enclosed by the path...and then use that the given function is analytic anywhere in $\;\Bbb C\setminus\{-i,i\}\;$ .

Answer 2

Hint. You can rewrite $$\oint_{\gamma}\frac{1}{z^2+1}dz= \oint_{\gamma}\frac{1}{(z-i)(z+i)}dz=\\ \frac{1}{2i}\oint_{\gamma}\frac{1}{z-i}dz-\frac{1}{2i}\oint_{\gamma}\frac{1}{z+i}dz$$ and use Cauchy's integral formula $$f^{(n)}(a)=\frac{n!}{2\pi i}\int\limits_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz \tag{1}$$ where $f(z)=1$, depending on whether $z_1=-i$, $z_2=i$ fall inside or outside a particular disk bounded by the relevant $\gamma$.