Let C be a rectangle with corners $\pm 2\pm 3i$ in the anticlockwise direction. Find $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$
I tried:
Split the contour in to 4 parts: $C_1,C_2,C_3,C_4$.
Let's say $C_1$ is the the bottom side of the rectangle. So,
$C_1: f(t) =t-3i, -2\le t\le 2$
Then I find $$\int_{-2}^{2}\frac{1}{t-3i}+\frac{1}{t-3i-1}dt$$
Similarly, I would find the integrals over $C_2, C_3$ and $C_4$. Then, the final answer is the sum of $C_1,C_2,C_3,C_4$.
Is this correct?
Best Answer
From the comments, it appears that the OP is receptive to seeing how Cauchy's Integral Formula can be used to evaluate the integral of interest.
Let $C$ be a closed rectifiable contour with winding number $1$ about a point $z_0\in \mathbb{C}$. Cauchy's Integral Formula states that if $f$ is analytic on the open region enclosed by $C$ and continuous on $C, then
$$f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}\,dz \tag 1$$
In $(1)$, set $f(z)=1$, $z_0=0$ and $C$ to be the rectangle defined in the OP. Then from $(1)$ we have
$$1=\frac{1}{2\pi i}\oint_C \frac{1}{z}\,dz$$
whereupon solving for $\oint_C \frac{1}{z}\,dz$ reveals
$$\oint_C \frac{1}{z}\,dz=2\pi i \tag 2$$
Similarly, set $f(z)=1$ and $z_0=1$. Then from $(1)$ we have
$$1=\frac{1}{2\pi i}\oint_C \frac{1}{z-1}\,dz$$
whereupon solving for $\oint_C \frac{1}{z}\,dz$ reveals
$$\oint_C \frac{1}{z-1}\,dz=2\pi i \tag3$$
Putting together $(2)$ and $(3)$ yields
$$\oint_C \left(\frac1z+\frac1{z-1}\right)\,dz=4\pi i$$