1
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.