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?

  • 2
    $\begingroup$ It's much simpler to use the Residue Theorem. $\endgroup$ – Robert Israel Feb 25 '17 at 1:27
  • $\begingroup$ Your first integral has a lower limit of $c$ and no upper limit $\endgroup$ – mrnovice Feb 25 '17 at 1:27
  • $\begingroup$ @mrnovice that is standard notation... $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 1:29
  • $\begingroup$ Oh ok, this is a topic beyond me then $\endgroup$ – mrnovice Feb 25 '17 at 1:29
  • 1
    $\begingroup$ The two poles are inside the contour, just use the residue theorem. $\endgroup$ – Zaid Alyafeai Feb 25 '17 at 1:29

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

$$\oint_C \left(\frac1z+\frac1{z-1}\right)\,dz=4\pi i$$

  • $\begingroup$ The integral I'm looking for is $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$, so shouldn't the correct answer be $4\pi i$? Also, do the corners of the rectangle not matter at all when using the cauchy integral formula? $\endgroup$ – sucksatmath Feb 25 '17 at 14:44
  • 1
    $\begingroup$ Yes, you're correct. It's $4\pi i$. The corners don't matter. The contour need only be rectilinear. $\endgroup$ – Mark Viola Feb 25 '17 at 15:45
  • $\begingroup$ @sucksatmath Since you're new to the site, I wanted to you know that after you have enough reputation points you can up vote answers too. -Mark $\endgroup$ – Mark Viola Mar 17 '17 at 17:43

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