Holomorphic 1/z

I'm going to do the following integral over three different paths: ∮1/z dz

 First Path

Around the unit circle in the complex plane. This seems to be the easiest.

z = e^(iθ, dz = ie^(iθ dθ

∮1/z dz = ∫ ie^(iθ/e^(iθ dθ, [0,2π]
= ∫ i dθ, [0,2π]
= 2πi

 The Second Path

Around the oval of real radius 2 and imaginary radius 1.

To parametrize this oval:

z = 2cos(θ+isin(θ
= (3/2 cos(θ+3/2 isin(θ) + (1/2 cos(θ - 1/2 isin(θ)
= 3/2 e^(iθ + 1/2 e^(-iθ

dz = 3/2 ie^(iθ - 1/2 ie^(-iθ dθ

∮1/z dz = ∫ [3/2 ie^(iθ - 1/2 ie^(-iθ] / [3/2 e^(iθ + 1/2 e^(-iθ] dθ, [0,2π]
= ∫ 3/2 ie^(iθ / [3/2 e^(iθ + 1/2 e^(-iθ] dθ + ∫ -1/2 ie^(-iθ / [3/2 e^(iθ + 1/2 e^(-iθ] dθ, [0,2π]
= ∫ 3/2 ie^(2iθ / [3/2 e^(2iθ + 1/2] dθ + ∫ -1/2 ie^(-2iθ / [3/2 + 1/2 e^(-2iθ] dθ, [0,2π]
= 1/2 ln(3/2 e^(2iθ + 1/2) + 1/2 ln(3/2 + 1/2 e^(-2iθ), [0,2π]
= 1/2 ln(2) - 1/2 ln(2) + 1/2 ln(2) - 1/2 ln(2)
= 0

 The Third Path

Around the unit square on the complex plane.

Going to split this into four paths, one around each side.

z = t+i, [1,-1], dz = dt
z = -1+it, [1, -1], dz = idt
z = t-i, [-1,1], dz = dt
z = 1+it, [-1,1], dz = idt

∮1/z dz = ∫ 1/(t+i) dt [1,-1] + ∫ i/(-1+it) dt [1,-1] + ∫ 1/(t-i) dt [-1,1] + ∫ i/(1+it) dt [-1,1]
= ln(t+i) [1,-1] + ln(-1+it) [1,-1] + ln(t-i) [-1,1] + ln(1+it) [-1,1]
= ln(1+i) - ln(-1+i) + ln(-1+i) - ln(-1-i) + ln(-1-i) - ln(1-i) + ln(1-i) - ln(1+i)
= 0

What am I doing wrong? I don't understand why 1/z is holomorphic when the path of integration is not a circle.

Try doing (1) using cos(theta) + isin(theta).

Common mistake. [well... not really a mistake... more of an oversight]

Hadriel

∮1/z dz = ∫ [-sin(θ + icos(θ]/[cos(θ + isin(θ] dθ, [0,2π]
= ln(cos(θ + isin(θ), [0,2π]
= ln(e^(iθ), [0,2π]
= iθ, [0,2π]
= 2πi

Okay. So, how would I correct for this in the other two paths? Let's just say that in general:

∮1/z dz = ln(z), [0,2π],

where z has a parametrization in the form of z(e^(iθ). Since, the parametrization for the second path is:

z = 2cos(θ+isin(θ
= 3/2 e^(iθ + 1/2 e^(-iθ

What am I supposed to do to cancel out the natural log in the following?

ln(3/2 e^(iθ + 1/2 e^(-iθ), [0,2π]

Using sines and cosines isn't any more revealing, and path 3 is un-parametrizable with in the form of e^(iθ, so how do I proceed?

for one thing, in method 3, you need to make sure your paths are going in the correct direction
(3rd and 4th segments should be -t's, or combination of any 2 that make you go full circle)

@below: k then

My integral spans the full box in the counterclockwise orientation. Instead of reversing t, I reversed the domain for the lines. [-1,1] should be interpreted as going from -1 to 1 (since I couldn't actually type a and b on the integral itself).

You do realise that the moment you revert back to Euler's form, it fails, but if you do the direct value crunching for the trigo terms, you return 0? It's abit like trying to integrate... err... can't rmb that function... arctan(x)?

We need the mathmo here. I can't really explain what's going on... not in a math mood now. It's biology time for me.

Hadriel

I'm not sure what you're trying to say here. Cauchy's formula (the first path integral over the unit circle):

∮1/z dz = 2πi

is indeed correct. The point here is that it should be 2πi all the time because of the deformation theorem, but for whatever reason, I'm getting 0 whenever I'm not integrating over the unit circle.

I believe you're talking about ∇arctan(y/x), the vortex vector field. I don't think ∇arctan(y/x) is applicable here, because z∈ℂ, but x,y∈ℝ. In both situations, the winding number appears partially due to the singularity at (0,0), but note that for z∈ℂ, singularities don't work the same way as they do in ℝ. This can be shown by the following integral:

∮1/z² dz = 0

z = e^(iθ
dz = ie^(iθ dθ

∮1/z² dz = ∫ ie^(iθ / (e^(iθ)^2 dθ, [0,2π]
= ∫ ie^(iθ / e^(2iθ dθ, [0,2π]
= ∫ ie^(-iθ dθ, [0,2π]
= -e^(-iθ [0,2π]
= 0

Even though 1/z² has a singularity of order two at z=0, the result of ∮1/z² dz is still 0, because 1/z² is holomorphic. Which is the other thing that's confusing to me. If z has two singularities, why is it holomorphic? Actually, why is 1/z^n, where n∈ℤ \{1} holomorphic? It almost seems to defy the residue theorem, which suggests that I should just count up the number of singularities and multiply by 2πi to obtain the closed contour integral over the entire Re-Im plane, kinda like in here.

hmm let me try again... you haven't incorporated the idea the time marches on into method 3...?

unfortunately i'm not doing it right either right now and don't really have the time to think about why your parametrization can't just be evaluated at -1 and 1 each time

Ok this is out of my hands... way off my realm. Noah...

Hadriel

Hmm... last last time when I saw this:

Complex analysis is not real analysis. Complex is actually fun, its just another class where you manipulate numbers and variables and see cool things.

I thought you had actually studied some of these types of problems. (Not trying to offend or anything.)

 Feel free to skip this

Yesterday, as I was eating dinner, I realized that my TA walked off with my vector calculus textbook after discussion, thus rendering me unable to complete my homework. To remedy this, I tried to search online for my textbook (I ended up downloading a Spanish version of the textbook and typed it into Google translate to work the problems. :|) In the process of searching, I stumbled upon Marsden and Hoffman's Basic Complex Analysis 3rd Edition.

Previously, I had Stein and Shakarchi on my desktop, but I didn't really look through it, since the topics it discusses weren't in my range of knowledge (or range of interest, for some of them). I haven't studied Fourier's transform, Gamma/Zeta/Theta functions, prime numbers, conformational mappings, elliptic functions, etc. in full detail, and that's pretty much all of Stein and Shakarchi. I only kept the book on my desktop and glimpsed through it when I'm bored, because I didn't really have another book from which to view this subject (and we all know that Wikipedia is crap for math).

Anyways, I've known for a while now that ∮1/z dz = 2πi, even though I never really understood how since I didn't know how to parametrize in complex variables. Yesterday, I briefly looked through the first two chapters of Marsden and Hoffman and asked myself the simple question: how do I show that the deformation theorem holds? By showing that different paths give the same value.

And so I wanted to calculate ∮1/z dz over three different paths, and the latter two unfortunately failed. Maybe I'm just parametrizing wrong, or, like hadriel pointed out, I need to somehow get rid of the log function to extract the actual value.

Today, I realized that I could probably simplify this by factoring out e^(iθ. However, as I show below, this causes some strange behavior (here, log is of base e):

g(θ = log (3/2 e^(iθ + 1/2 e^(-iθ)
= log(e^(iθ (1/2) (3 +e^(-2iθ), [0,2π]
= log(e^(iθ) - log(2) +log(3+e^(-2iθ), [0,2π]
= iθ [0,2π]
= 2πi

While that may seem correct now, notice that if I factor out e^(-iθ instead, I obtain:
g(θ = log (3/2 e^(iθ + 1/2 e^(-iθ)
= log(e^(-iθ (1/2) (3e^(2iθ + 1), [0,2π]
= log(e^(-iθ) - log(2) + log(3e^(2iθ + 1), [0,2π]
= -iθ [0,2π] (if log(3e^(2iθ + 1), [0,2π] = 0)
= -2πi

Actually, I can factor out exponential anything. For example, by factoring out e^(2iθ within the log, I erroneously obtain 4πi instead of 2πi. If factoring them out "any way" gives correct solutions, then the above factorization (bolded) must show that

log(3e^(2iθ + 1), [0,2π] = 4πi

which doesn't really make sense.

Also, putting the following into Mathematica:

log(3/2*e^(i*2pi)+1/2*e^(-i*2pi))-log(3/2*e^(i*0)+1/2*e^(-i*0))

returns 0, when it should return the "correct" solution of 2πi. In fact, putting in the simple identity log(e^(i*2pi))=2πi causes Mathematica to erroneously obtain 0. Also, another problem is that even if my first simplification is correct, I'm unable to guarantee that:

log(3+e^(-2iθ), [0,2π] = 0

I'm still completely lost with the third one. There doesn't seem to be a way to kill the log without an exponential term. Hopefully, Noah knows how to cure this mess.

It feels to me as if the closed loop integral you're trying to do will always return 0 because it is a closed loop integral over a scalar plane... That's what guts tell me.

Actually I'm trying to run away from complex numbers now... trying to focus on some of the rest of the topics for my Math exam... I'm running out of HDD space.

Hadriel

z is a complex variable, so f(z) is not really a scalar plane. It's also a vector plane (a,b), where a is the real part and b is the imaginary part (otherwise known as a+bi). I don't really understand the theory behind this, but even though they're different components, Euler's formula for the expansion of exp(iθ still holds so we don't get a vector divided by a vector when we do this.

I did a bit more searching. Apparently, this is linked to ∇arctan(y/x), so your prediction was correct. I found on Wolfram that the complex argument function, arg, is defined as arg(x+yi)=arctan(y/x). Unfortunately, I have no idea how this can help, but it is interesting to note that the deformation theorem works for ∇arctan(y/x), and 2π is the winding number. Similarly, I should obtain 2πi for ∮1/z dz.

Good luck on your finals, by the way.

You guys are giving me too much credit, I don't know everything!

The second one is just some miscalculation, I assume, as I get it correctly:

Let


then



Now, I looked after the last integral for some time, but could not find it. I ended up using Wolfram Alpha which is not really a satisfying answer, but it should leave you content, I hope.

http://www.wolframalpha.com/input/?i=int...D0+to+2+pi

So the second one is correct. The third one is not that much harder:

Your idea is correct, but your use of complex logarithms does not satisfy me nor the mathematics. Look it up! Also, it seems that you try to go clockwise, instead of counterclockwise. And remember, z is a function, and you can only map from one value to another, not to 4 different values. This would not affect the result here, but your professor will kill you for that, and I am against murder! (In all seriousness though, it is smart to be rigorous, at least when defining functions, as you can more easily see where the errors happen.)



Now, the contour integral will look like this:



The first integral evaluated is this:



The remaining integrals return the same value (which should not be a surprise, as this figure is symmetric!), and is up to the reader to evaluate, as I am too lazy to do it. The value of the contour integral is thus 2 pi i.

Noah

 Spoiler

Another common error is to forget to centre the area you are working with. Try, for example, to integrate the normal circle but with the centre somewhere else, say, for example, at 3 or 7. Then

Interestingly, WolframAlpha says that the integral is θ+i*log(1+3exp(iθ). Then we run into the problem of taking the log the form that I was having trouble with earlier, namely the one I bolded and inferred:

log(3exp(4iπ+1) - log(3exp(0i)+1) = 4πi

However, I'm half-willing to believe that the integral is 0, since it doesn't have any singularities, so it should be holomorphic. But then again, that would mean that 1/(1+3exp(-iθ) is also holomorphic, and you could have simplified it the other way and obtained the wrong solution.

Now, the contour integral will look like this:



The first integral evaluated is this:



The remaining integrals return the same value (which should not be a surprise, as this figure is symmetric!), and is up to the reader to evaluate, as I am too lazy to do it. The value of the contour integral is thus 2 pi i.

This pretty much what I did, without shifting the "index" (aka I parametrized every line starting from (1,1) from 1 to -1 and then from -1 to 1 when I reached (-1,-1). I do see the problem now, however. In the last step, simplification must be grouped before evaluated:

∮1/z dz = ∫ 1/(t+i) dt [1,-1] + ∫ i/(-1+it) dt [1,-1] + ∫ 1/(t-i) dt [-1,1] + ∫ i/(1+it) dt [-1,1]
= ln(t+i) [1,-1] + ln(-1+it) [1,-1] + ln(t-i) [-1,1] + ln(1+it) [-1,1]
= {ln(1+i) - ln(-1+i)} + {ln(-1+i) - ln(-1-i)}+ {ln(-1-i) - ln(1-i)} + {ln(1-i) - ln(1+i)}
= ln[(1+i)/(-1+i)] + ln[(-1+i)/(-1-i)] + ln[(-1-i)/(1-i)] + ln[(1-i)/(1+i)]

And you claim that each one of these grouped logarithms is equivalent to πi/2. Turns out to be -2πi. You're right. I seem to have gone backwards. I'll check my parametrization later.

You guys are giving me too much credit, I don't know everything!

You seem to be beyond us all on math. I've yet to observe a math post that you couldn't answer, on Southperry, at least.

 Spoiler

Interestingly, WolframAlpha says that the integral is θ+i*log(1+3exp(iθ). Then we run into the problem of taking the log the form that I was having trouble with earlier, namely the one I bolded and inferred:

log(3exp(4iπ+1) - log(3exp(0i)+1) = 4πi

but e^(4 pi i) = e^(2 pi i) = e^(0 pi i) = 1, so log(3e^(4 i pi) + 1) = log(3 + 1) = log(3 e^(0 pi i) + 1), and that leaves us with zero?

This is precisely what I did, without shifting the "index" (aka I parametrized every line starting from (1,1) from 1 to -1 and then from -1 to 1 when I reached (-1,-1). I do see the problem now, however. In the last step, simplification had to be grouped:

∮1/z dz = ∫ 1/(t+i) dt [1,-1] + ∫ i/(-1+it) dt [1,-1] + ∫ 1/(t-i) dt [-1,1] + ∫ i/(1+it) dt [-1,1]
= ln(t+i) [1,-1] + ln(-1+it) [1,-1] + ln(t-i) [-1,1] + ln(1+it) [-1,1]
= {ln(1+i) - ln(-1+i)} + {ln(-1+i) - ln(-1-i)}+ {ln(-1-i) - ln(1-i)} + {ln(1-i) - ln(1+i)}
= ln[(1+1)/(-1+i)] + ln[(-1+i)/(-1-i)] + ln[(-1-i)/(1-i)] + ln[(1-i)/(1+i)]

And you claim that each one of these grouped logarithms is equivalent to πi/2.

No, I claim that specific result to be -2 pi i as you went clockwise instead of counterclockwise.

Also, Wolfram Alpha is partly correct, but look up complex logarithms! Some of them have infinitely many solutions.

Noah

Wait.

∮1/z dz = ∫ [3/2 ie^(iθ - 1/2 ie^(-iθ] / [3/2 e^(iθ + 1/2 e^(-iθ] dθ, [0,2π]
= i ∫ [3 e^(iθ - e^(-iθ] / [3 e^(iθ + e^(-iθ] dθ, [0,2π]

When you reached this step, how did you go to the next one? It seems like you somehow dropped the exp(-iθ terms. Going backwards from your work:

i - 2i /(1+3exp(iθ)
= i - 2i /(3exp(iθ+1)
= i (3exp(iθ+1) /(3exp(iθ+1) - 2i /(3exp(iθ+1)
= (3i*exp(iθ +i -2i) /(3exp(iθ+1)
= (3i*exp(iθ -i) /(3exp(iθ+1)
?

Also...

Your idea is correct, but your use of complex logarithms does not satisfy me nor the mathematics. Look it up! Also, it seems that you try to go clockwise, instead of counterclockwise. And remember, z is a function, and you can only map from one value to another, not to 4 different values. This would not affect the result here, but your professor will kill you for that, and I am against murder! (In all seriousness though, it is smart to be rigorous, at least when defining functions, as you can more easily see where the errors happen.)

I don't have a professor. I'm not taking a class for this. And I thought z can map to infinitely many values, or so it seems from the Riemann surfaces resembling overlapping spirals.

Also, Wolfram Alpha is partly correct, but look up complex logarithms! Some of them have infinitely many solutions.

I looked it up. I'm wondering: how do I change the base of a complex logarithm? The formula log(a) {base b} = log(b) /log(a) doesn't seem to be working.

Wait.



∮1/z dz = ∫ [3/2 ie^(iθ - 1/2 ie^(-iθ] / [3/2 e^(iθ + 1/2 e^(-iθ] dθ, [0,2π]
= i ∫ [3 e^(iθ - e^(-iθ] / [3 e^(iθ + e^(-iθ] dθ, [0,2π]

When you reached this step, how did you go to the next one? It seems like you somehow dropped the exp(-iθ terms. Going backwards from your work:

And it seems like I forgot at 2 in the exponent, but the result of the integral did not change. I'm a lucky guy!

I don't have a professor. I'm not taking a class for this. And I thought z can map to infinitely many values, or so it seems from the Riemann surfaces resembling overlapping spirals.

That's a shame. Professors love interesting and curious students, and you do seem like one.

Well, yeah for a single theta, there can be several points. You should have some function you integrate over to distinguish the points, though. Usually, it is "time"-based.

I looked it up. I'm wondering: how do I change the base of a complex logarithm? The formula log(a) {base b} = log(b) /log(a) doesn't seem to be working.

It works fine, but you need to be careful due to branching: http://books.google.com/books?id=FUWPyHM...tAPZ5-XjCw

Noah

Argh, I had a nice post all typed out and my internet went down.

Hmm... last last time when I saw this:


I thought you had actually studied some of these types of problems. (Not trying to offend or anything.)

I have, but I hadn't had time to think about the problem. The key concept you are missing out on is branch cuts. Mathematica won't do branch cuts for you automatically (and I don't know how to do them manually either because it's easy enough to ponder what your branch cut will be without actually doing any computations), which explains why you get multiple answers in your Mathematica codes. So anyways, complex exponentials are inherently multivalued. Integrating around a singularity inherently begs the question of where exactly you are in the complex plane, since e^0 = e^i2π = e^i4π etc. A branch cut is an imaginary line you specify, and don't cross. The problem you run into when you integrate method 3 is not only with your parametrization, which you need to be more careful about, but also the fact that you never specify a branch cut. Method 1 sort of implies that you've done a branch cut, by integrating over [0,2π). But you don't even know it.

Ugh, I have Method 3 redone as an example for you at home. When I get my internet back up there I'll post it. Blame comcast. Wikipedia isn't that much help for this concept, if you ask me.

And it seems like I forgot at 2 in the exponent, but the result of the integral did not change. I'm a lucky guy!

What if I did this instead?
[3exp(it)-exp(-it)] /[3exp(it)+exp(-it)]
= [3exp(it)-exp(-it)]exp(-it) /[(3exp(it)+exp(-it))*exp(-it)]
= [3-exp(-2it)] /[3+exp(-2it)]
= - [-3+exp(-2it)] /[3+exp(-2it)]
= - [3+exp(-2it) -6] /[3+exp(-2it)]
= -1 -6/[3+exp(-2it)]

Then I'd end up with -2πi when I integrate. Does that mean the integral of -6i/[3+exp(-2it)] from 0 to 2π is 4πi?

That's a shame. Professors love interesting and curious students, and you do seem like one.

Unfortunately, my professors like students who can do the practice problems. I have finals on Monday and Tuesday. I'm pretty screwed.

The problem you run into when you integrate method 3 is not only with your parametrization, which you need to be more careful about, but also the fact that you never specify a branch cut. Method 1 sort of implies that you've done a branch cut, by integrating over [0,2π). But you don't even know it.

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

Ugh, I have Method 3 redone as an example for you at home. When I get my internet back up there I'll post it. Blame comcast. Wikipedia isn't that much help for this concept, if you ask me.

._.

This looks like something I wish I never dabbled in... I </3 complex... perhaps I just need a good deal more practice.

Am off to study with my friends. Whipping Lashing them into studying. Those people just need someone to make them study.

And yeah, thanks man. I need my top grades to continue studying...

Hadriel