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

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

log(3e^(2iθ + 1)

 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

Noah Wrote:

Noah Wrote:

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

2147483647 Wrote:

 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

2147483647 Wrote: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.

Noah Wrote:

Noah Wrote: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.)

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

2147483647 Wrote: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:

2147483647 Wrote: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.

2147483647 Wrote: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.

2147483647 Wrote: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.)

Noah Wrote:

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!

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

modular Wrote: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.

modular Wrote: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.