Integration and log of a complex number

[Daniiel]

Hey,

I know the answer to this integral is 2ipi as it was given but I trying to find out how its 2ipi.

Here is my working

[PLAIN]http://img707.imageshack.us/img707/1681/unledrny.jpg

I've been looking at this for ages and I can't work out what I've done wrong


thanks,

 

[susskind_leon]

The complex logarithm is ambiguous (as you have already found out) and therefore not always suitable. I would check http://en.wikipedia.org/wiki/Complex_logarithm#Constructing_branches_via_integration very carefully, I'm not sure if you can actually use the logarithm here.
You can always substitute $u=e^{it}$.
What will you get as an expression then?
What will be the integration path?
Can you spot a singularity? Calculate its residue and you're done :)

 

[Daniiel]

If I change variables to u = e^(it) will i get,

du/dt = ie^(it), so dt=1/(ie^(it)

so int (2u/(2u-1))e^(-it) du, this is where I am going to stuff up a lot, i did it two ways, since u =e^(it) they cancel, so you get in int (2/(2u-1)) du then that integral ends up as the nasty log in it again so then i tried
leaving the e^(it)
and i got the nasty log thing + u - 1/2

I've been looking over my notes and the only stuff about logs he has given use is that log(z)=log|z| + i arg(z), i tried in mathematica and wolframalpha separately and they gave different answers haha, wolfram gave 2ipi thought

 

[susskind_leon]

Forget logarithm! I mean, I can't really tell you why, but maybe someone else can.

So, let's go through it
$$\int_0^{2\pi} \frac{2ie^{it}}{2e^{it}-1} dt$$
$$u=e^{it}$$
$$du = i e^{it} dt$$
So the integral becomes
$$\int_S \frac{2}{2u -1} du$$
where you integrate over the unit circle.
Now, the function has a simple pole at u=1/2. Now calculate the residue of this point and you will get that it's 1. Apply the residue theorem and you will get that the value of the integral is 2 pi i.

 

[Daniiel]

Oh right,

Sorry this is the first time I've every read about residue or the residue theorem.

This question was actually a question that's really easy to answer with cauchy's integral theorem, but they asked to do it without using it, I know the answer to the question is 6ipi using CIF and i end up with 3[log(ew)] between 0..2Pi

its just integrating (z^2+z+1)/(z-1) dz over a circle of radius 2, so like 1 step using the cauchy integral formula haha

What I've done is used long division, found most things cancel and end up with the 3log buisness