Holomorphic 1/z

Okay. I looked at this thread again. From what it seems, branch cuts always need to be length 2π. Here's what my book has to say:



Thus, my first question arises from how Noah was able to obtain 2πi, the correct value, by using an interval of [0,8) for the box problem. From what it looks like, the interval that violates the branch cut more than twice.

I'm also still somewhat confused about the contour integral of the ellipse. I mentioned this earlier, but since nobody addressed it, I've decided to type it out to make the problem more visible:

Noah's solution relies on shifting the exponential of e, based on the following method:



This solution relies on the lemma (can this even be called a lemma?) that ∮-2i/(3e^(2iθ+1) dθ from 0 to 2π is 0. This seems correct, since 2i/(3e^(2iθ+1) is an entire function that has no singularities. However, I present an alternate solution:



Thus,∮6i/(3+e^(-2iθ) dθ from 0 to 2π is 4πi (as confirmed by Wolfram Alpha). However, 6i/(3+e^(-2iθ) is also an entire function with no singularities, so it should seem to me that the value should equal 0. After all, it's in a way "symmetrical" to ∮-2i/(3e^(2iθ+1) dθ, and the latter is equal to 0. Also, I was screwing around with the integral, ∮-2i/(3e^(2iθ+1) dθ. Apparently, if I change it to ∮-2i/(3e^(2iθ+2) dθ, it's still equal to 0. However, -2i/(3e^(2iθ+3) dθ equals some strange complex value, and ∮-2i/(3e^(2iθ+4) dθ equals -πi. I don't really understand what's going on here.

Since I have learned how to (correctly) compute the complex logarithm, I find that all of the above troubles can be avoided:



Also, the contour integral over the circle is:

{log(e^(2πi))}-{log(e^(0i))}
= {log|e^(2πi)|+2πi}-{log|e^(0)|+0}
= {log(1)+2πi}-{log(1)+0}
= 2πi

From the above, it makes sense that any contour, no matter how large, will equal 2πi. The value comes from going around the plane exactly once (2π length) and ending at the same value (causing all of the ln® terms to cancel out). However, this method will also erroneously provide 2πi when the region of integration does not include the singularity.

Looking at your work, modular, I'm confused to why you swapped the direction of the integral. After all, when I did this integral over a circle, I took the counterclockwise (the standard) orientation and obtained positive 2πi. I'm under the impression that only the counterclockwise direction should be giving positive values and the opposite direction should give -2πi.

While working this problem, I realized that the integral ∮dz/z is almost the same as integrating the vector field ∇arctan(b/a)= (-b,a)/(a^2+b^2) over a path, since the complex logarithm when a+bi is expressed as (a,b) is defined as log®+iθ, and iθ happens to be i*tan(b/a). The only difference is that (-b,a) is orthogonal (or reciprocal?) to (a,b), but that shouldn't matter, since arctan and tan are inverse functions, right? This raises the question: what IS the integral measuring? The contour doesn't seem to be over a scalar or a vector field.

My next step was to find the integral of ∮dz/(z-1), which should hypothetically equal 2πi if I include the singularity:
∮dz/(z-1)
= log(z-1) [0,2π

If I parametrize my circle to include the singularity:
z = e^(iθ+1

I end up with 2πi.

If I parametrize my circle to avoid the singularity:
z = e^(iθ-1

∮dz/(z-1)
= log(z-1) [0,2π
= log(e^(iθ-2) [0,2π
= {log|-1|+2πi}-{log|-1|+0|
= 2πi

I erroneously get 2πi when I'm supposed to get 0.

If I run my path right onto the singularity via z=e^(iθ:

∮dz/(z-1)
= log(z-1) [-π,π
= log(e^(iθ-1) [-π/2,3π/2)
= {log|-2|+3πi/2}-{log|-2|-πi/2|
= 2πi

If I run the circle through the singularity but the singularity is my start point:

∮dz/(z-1)
= log(z-1) [0,2π
= log(e^(iθ-1) [0,2π
= {log|0|+2πi}-{log|0|+0|
= 2πi

I know log isn't defined at 0, but we have log|0|-log|0| here:
lim log®-log® as r approaches 0
= lim log(r/r) as r approaches 0
= lim log(1) as r approaches 0
= 0

Not sure what goes on now.

Edit: I've run into a problem trying to calculate complex logs. Specifically, I'm trying to calculate:

∮dz/(z-2)

over the circle z=e^(iθ in the interval [-3π/4,5π/4)

∮dz/(z-2)
= log(z-2)
= log(e^(iθ-2)
= log(1/sqrt(2)-i/sqrt(2)-2)-log(1/sqrt(2)-i/sqrt(2)-2)
= {log|1/sqrt(2)-i/sqrt(2)-2|+arg(1/sqrt(2)-i/sqrt(2)-2)}-{log|1/sqrt(2)-i/sqrt(2)-2|+arg(1/sqrt(2)-i/sqrt(2)-2)}
= arg(1/sqrt(2)-i/sqrt(2)-2) -arg(1/sqrt(2)-i/sqrt(2)-2)

My problem is assigning a value to each of the two arguments. If I assign the first argument function as 5π/4 and the second argument function as -3π/4, then I incorrectly obtain 2πi. My question is this: how do I determine what angle to assign to these without knowing the solution and assigning them from hindsight? Is there a definite, unmistakable procedure for assigning an angle?

The reason I'm dwelling on this problem is that the next problem I'm attempting to do is:
∮dz/(z²+1)

I can do this by either noting that it's holomorphic, or I can do this by factoring:
∮dz/(z+i)-∮dz/(z-i)
= -πi - (-πi)
= 0

My actual work involved integrating a unit circle around each singularity. Of course, the angle would thus be different from taking them together. Also, I need help integrating functions in the form of 1/(e^(niθ+m). Any help please?