Holomorphic 1/z

How do I select a branch cut? I can't even find anything online to visualize complex functions.

got internet back!!!!!!!!

 Spoiler

for example, i can choose

t e [0,8]
z = -1+i(1-t) [0,2]
z = (t-3)-i [2,4]
z = 1+i(t-5) [4,6]
z = (7-t)+i [6,8]

then

∮1/z dz = ∫ -i/(-1+i(1-t)) dt [0,2] + ∫ 1/((t-3)-i) dt [2,4] + ∫ i/(1+i(t-5)) dt [4,6] + ∫ -1/((7-t)+i) dt [6,8]
= {ln(-1-i) - ln(-1+i)} + {ln(1-i) - ln(-1-i)} + {ln(1+i) - ln(1-i)} + {ln(-1+i) - ln(1+i)}
= iπ{5/4 - 3/4} + iπ{7/4-5/4} + iπ{1/4-7/4} + iπ{3/4-1/4} = 0

now then... it's 0. what? well yes. why? branch cut. we crossed it. that's gonna bugger you. up to now, i haven't even declared where my branch cut is. branch cut makes things single valued, and you can't cross it with your integral's path. hence, in method 1, you might see it in a textbook as [0,2π). usually you make a branch cut as convenient as possible. physics people might use [-π,π sometimes, math people kinda don't care as long as you mention it. taking the interval [0,2π implies taking the branch cut along the positive real axis. with our square geometry, it's convenient to cut from 0 to a corner of the square. pick any one. i'll choose the -1,i corner, since that's where t starts. cutting there allows us to define a domain for e^iθ, in this case it's [-π5/4, π3/4). so now take a look at that last line again:

= iπ{5/4 - 3/4} + iπ{7/4-5/4} + iπ{1/4-7/4} + iπ{3/4-1/4}

now that we have taken our branch cut, anything outside of the domain can't be right. so iπ7/4 -> -iπ1/4 and iπ5/4 -> -iπ3/4. (these won't actually matter, but it's good practice to remember they don't really exist with this branch cut). so if changing those doesn't matter, what are we forgetting? we're forgetting we can't cross the branch cut. we START just "below" the branch cut and end just "above" it. that means our starting value of iπ3/4 is wrong and we need to swap it for the clockwise moving way of labeling it: -iπ5/4.

= iπ{-3/4 + 5/4} + iπ{-1/4 + 3/4} + iπ{1/4+1/4} + iπ{3/4-1/4}
=i2π

one more thing... like i said, iπ3/4 is outside of our domain, why did i still use it once? use the fundamental theorem of calculus to evaluate the last integral at its endpoints, since it's just a riemann integral we can do that, but iπ3/4 is outside of our domain, so call that endpoint "a" instead and take the limit as a->iπ3/4 and it converges nicely and all that (just to be sure i don't try to slip one past you, it's the typical justification)

hope this helps

selecting a branch cut is up to you. you'll need one any time you have singularities in the complex plane. branch cuts represent lines you don't cross, ever, when searching for a path to integrate over. they keep everything single valued. and WHAT they look like doesn't matter. for example, in ^ i could have chosen my branch cut to spiral around the origin 20 times, gone through (-1,i), and then done a saw function in any direction out to infinity (long as it doesn't cross my path again). kind of pathological, but it makes more sense when say... you look at normal modes of acoustic waves and you have some strangely placed singularities you need to evaluate the strength of with something like residue theorem.

I haven't really looked at your work yet (since I've just taken two finals and have my physics final tomorrow), but I swear I will get around to looking at it soon. My questions for now are:

1. Does the branch cut always have to be of length 2π?
2. What if the function wraps around itself more than once within this 2π interval? I don't have an example, but I'm sure there exists complex functions that wrap around themselves more than once within [0,2π.
3. Not really related to anything, but I'm looking over the complex argument function, which is really just arctan(v/u) (or is it u/v?). Does it have a winding number of 2π, because it satisfies the Cauchy-Riemann equations, or is it just because of the singularity? Does satisfying the Cauchy-Riemann equations denote that a function has a singularity?
4. Can I imagine the value of the singularity as some sort of height between two branches? (like the distance between two leaves of a helicoid)
5. For what function would I get a value that's not 2πi, and all the constants are 1? (for example, not 2/z)

I haven't really looked at your work yet (since I've just taken two finals and have my physics final tomorrow), but I swear I will get around to looking at it soon. My questions for now are:

1. Does the branch cut always have to be of length 2π?
2. What if the function wraps around itself more than once within this 2π interval? I don't have an example, but I'm sure there exists complex functions that wrap around themselves more than once within [0,2π.
3. Not really related to anything, but I'm looking over the complex argument function, which is really just arctan(v/u) (or is it u/v?). Does it have a winding number of 2π, because it satisfies the Cauchy-Riemann equations, or is it just because of the singularity? Does satisfying the Cauchy-Riemann equations denote that a function has a singularity?
4. Can I imagine the value of the singularity as some sort of height between two branches? (like the distance between two leaves of a helicoid)
5. For what function would I get a value that's not 2πi, and all the constants are 1? (for example, not 2/z)

Gah, /dredging memories. I wish I had a complex book, my prof just went with his own notes.

1. No, a cut has infinite length - it extends from your singularity to any point at infinity and you choose how it gets there. Make use of that.
2. e^ixθ, x > 1, wraps around more than once. You still only choose 1 branch cut, but the interval shrinks. Say you choose re^i2θ in your parametrization... Then run θ from [0,π to keep it single valued. (not sure if help/answer)
3. arctan has an infinite line of singularities every umm... π along the imaginary axis. It wraps around every 2π because sin and cos do... Satisfying the C-R equations implies a function is analytic, nothing about singularities. There isn't really an equation to satisfy that'll imply singularities, you have to notice that you divide by 0 for some value in the plane.
4. Try the height between 2 wraparounds, I think that's applicable... (I never felt the need to attach a visual to the strength of a singularity :f6
5. There may be some pathological example that I don't know about. But usually, it's 2πi. Also, for the wording of this question... I think you want to mean parametrization instead of function? Any function with a singularity can have that point circled by a path, which naturally leads to complex exponentials in the parametrization. I think you would need a geometry other than the complex plane to produce such results, and even then...

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?

I think where you are messing up with the circles excluding singularities is that you neglect the fact that you never need to cross the branch cut. You can evaluate each integral's endpoints with the same angle as the starting points. Only when you actually go around the singularity and return to the same point do you need to specify how you are dealing with the multivalued nature of complex exponentials. For example,

 Spoiler

path 1 doesn't require any special treatment since you end back at the same place while path 2 requires you to tell what level of this pirated-from-wikipedia helix you are on. I'm sure that's why you're getting 2pi's instead of 0's, you are applying branching concept where unnecessary. Closed paths with no singularities don't present the problem of "what level am I on?" because you are always on the same level and end back in the (truly) same place, so forced single-valuedness is not needed.

as for directionality... uh, i did a counterclockwise path. started in the top left corner and went straight down to bottom left corner for the first integral and so on. and yes, expect a sign change when switching clockwise vs. counter. acts like electrical current in that + and - are just definitions used for consistency and don't actually have reasons for being what they are beyond being what dead white guys used. not sure if addressing real issue here...

Edit: I found a practice book with worked out solutions. Finally.