Complex integration over a square contour (part b)

[bjohnson2001]

Homework Statement

Let $\Gamma$ be the square whose sides have length 5, are parallel to the real and imaginary axis, and the center of the square is i. Compute the integral of the following function over $\Gamma$ in the counter-clockwise direction using parametrization. Show all work.

$\frac{z-1}{z+1}$

Homework Equations

u substitution

The Attempt at a Solution

Starting from the lower right point and going counter clockwise:

1) $^{3.5}_{-1.5} \int \frac{i(2.5+iy-1)}{2.5+iy+1} dy$ = -0.5245 + 2.6194i

2) $^{2.5}_{-2.5} \int \frac{x+i3.5-1}{x+i3.5+1} dx$ = -4.4755-2.3806i

4) $^{-2.5}_{2.5} \int \frac{x-i3.5-1}{x-i3.5+1} dx$ = 3.8299-3.9026i

All of these have been confirmed by the quad function in MATLAB. There is a problem when computing the third integral from 3.5i to -1.5i, crossing from quadrant II into quadrant III. My method seems to be the same as before but my answer is $-4\pi i$ more than what it should be.

3) $^{-1.5}_{3.5} \int \frac{z-1}{z+1} dz$ where $\stackrel{z(y)=-2.5+iy}{z'(y)=i}$

$\int \frac{i(-2.5+iy-1)}{-2.5+iy+1} dy$ where $\stackrel{u=-2.5+iy+1}{du=idy}$

= $\int \frac{u-2}{u} du$

= $\int 1- \frac{2}{u} du$

= $u-2*ln(u) = (-2.5-iy + 1)-2ln(-2.5+iy+1)^{-1.5}_{3.5}$ = 1.1701+3.6638*i

This should be 1.1701-8.9026*i

Can you see what the problem is?

 

[mighty2000]

Thank you for your post and for sharing your attempt at solving this problem. After reviewing your work, I noticed that there may be a small error in your parametrization for the third integral. When crossing from quadrant II to quadrant III, the parametrization should change from z(x) = -2.5 + ix to z(y) = ix - 2.5i. This will result in a change in the calculation of the derivative, as well as a change in the limits of integration.

Therefore, the correct parametrization for the third integral would be:

3) ^{-1.5}_{3.5} \int \frac{z-1}{z+1} dz where \stackrel{z(y)=iy-2.5i}{z'(y)=i}

\int \frac{i(iy-2.5i-1)}{iy-2.5i+1} dy where \stackrel{u=iy-2.5i+1}{du=idy}

= \int \frac{iu-2i}{iu} du

= \int 1 - \frac{2i}{u} du

= u-2ln(u)

= (iy-2.5i+1) - 2ln(iy-2.5i+1)^{-1.5}_{3.5} = 1.1701-8.9026i

I hope this helps clarify the issue and that you are able to obtain the correct answer now. Please let me know if you have any further questions or if you need any additional assistance.