compute usig cauchy integral formula, $ \int_\Gamma \frac{dz}{z^2+1}$

Question

The question is: $$ \int_\Gamma \frac{dz}{z^2+1}$$

Here is how i thought, we can divide it into two contour and then deform continuously the inner curve to a circle around $i$ but my problem is here how should i deform the bigger contour around $-i$ i couldn't do that, because we can't pass through singularity, what do you think? any suggestion would be great,

Thanks

Answer 1

Read the fine print viz. $\oint_\Gamma f(z)dz=2\pi i\sum_{w\in S}\color{blue}{I(\Gamma,\,w)}\color{red}{\operatorname{Res}(f,\,w)}$, with $S$ the set of poles of a holomorphic $f$ the closed rectifiable curve $\Gamma$ encloses, so $w\in S$ has winding number $I(\Gamma,\,w)$. Your contour integral is$$2\pi i\left(\color{blue}{2}\color{red}{\lim_{z\to i}\frac{1}{z+i}}+\color{blue}{1}\color{red}{\lim_{z\to-i}\frac{1}{z-i}}\right)=2\pi i(2/(2i)+1/(-2i))=\pi.$$

Answer 2

Hint.

The result follows from the residue theorem for closed contours.

If you are only allowed to use the Cauchy integral formula with simple closed contours, first rewrite the integral as $$ \int_\Gamma\frac{1}{z^2+1}\,dz=\frac{1}{2i}\left(\int_\Gamma\frac{1}{z-i}dz-\int_\Gamma\frac{1}{z+i}dz\right) $$

Then after you split the contour $\Gamma$, handle $\frac{1}{z-i}$ and $\frac{1}{z+i}$ separately with the Cauchy integral formula.

Answer 3

The "three curves" solution. We use three curves.
The first curve surrounds $-i$ but not $i$. Notice the direction was reversed on the inner loop.

The second and thrid curves surround $i$ but not $-i$.

Add the three together to get the original curve.