- #1
bjohnson2001
- 15
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Homework Statement
Let [itex]\Gamma[/itex] 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 [itex]\Gamma[/itex] in the counter-clockwise direction using parametrization. Show all work.
[itex]\frac{z-1}{z+1}[/itex]
Homework Equations
u substitution
The Attempt at a Solution
Starting from the lower right point and going counter clockwise:
1) [itex]^{3.5}_{-1.5} \int \frac{i(2.5+iy-1)}{2.5+iy+1} dy[/itex] = -0.5245 + 2.6194i
2) [itex]^{2.5}_{-2.5} \int \frac{x+i3.5-1}{x+i3.5+1} dx[/itex] = -4.4755-2.3806i
4) [itex]^{-2.5}_{2.5} \int \frac{x-i3.5-1}{x-i3.5+1} dx[/itex] = 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 [itex]-4\pi i[/itex] more than what it should be.
3) [itex]^{-1.5}_{3.5} \int \frac{z-1}{z+1} dz[/itex] where [itex]\stackrel{z(y)=-2.5+iy}{z'(y)=i}[/itex]
[itex]\int \frac{i(-2.5+iy-1)}{-2.5+iy+1} dy[/itex] where [itex]\stackrel{u=-2.5+iy+1}{du=idy}[/itex]
= [itex]\int \frac{u-2}{u} du[/itex]
= [itex]\int 1- \frac{2}{u} du[/itex]
= [itex]u-2*ln(u)
= (-2.5-iy + 1)-2ln(-2.5+iy+1)^{-1.5}_{3.5}[/itex] = 1.1701+3.6638*i
This should be 1.1701-8.9026*i
Can you see what the problem is?