Complex integral inside rectangular loop

Question

Consider the integral $$\int {z \over z^2+1} dz$$ on the contour bounding the rectangle $-1<x<1, -2<y<2$ (counterclockwise)

Answer 1

The residue theorem states that

$$\oint_\gamma f(z) dz = 2\pi i\sum\operatorname{Res}(f,a_k)$$

In your case, we have

$$\gamma=R$$ $$f(z)=\frac{z}{z^2+1}$$

As

$$\frac{z}{z^2+1} = \frac{z}{(z+i)(z-i)}$$

we have two poles, one at $z=i$ and the other at $z=-i$. Both of these are inside $R$, so the residue theorem becomes

$$\int_R \frac{z}{z^2+1} dz = 2\pi i\left(\operatorname{Res}(f,i)+\operatorname{Res}(f,-i)\right)$$

The residue of a simple pole at $c$ is

$$\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z)$$

In your case, this becomes

$$\operatorname{Res}(f,i)=\lim_{z\to i}(z-i)f(z)=\lim_{z\to i}\frac{z}{z+i}=\frac12$$ and $$\operatorname{Res}(f,-i)=\lim_{z\to -i}(z+i)f(z)=\lim_{z\to -i}\frac{z}{z-i}=\frac12$$

which gives

$$\int_R \frac{z}{z^2+1} dz = 2\pi i$$