1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex integral over a circle

  1. Apr 3, 2012 #1
    1. let C be the circle |z| = 2 traveled once in the positive sense. Computer the following integrals...

    a.∫c zez/(2z-3) dz

    2. Relevant equations

    I am confused as to a step in my solution, but i believe a relevant equation is if i am integrating over a circle and the function is analytic at a point, i can remove it by integrating and removing the denominator and replacing it with 2pi*i in the numerator. However i do not know why. This seems to be a product of the cauchy reimann equations used to calculate the integral.

    3. The attempt at a solution

    So i need to get the equation in the form x/z-z0 so i divide everything by 2.

    I am left with .5*zez/(z-(2/3)) In this form i can see it is analytic at z=3/2 (i believe the word analytic to mean that it goes to infinity)

    so i remove the denominator, and replace it with a substitution of 2pi*i...

    2pi*i*.5*z*ez evaluated at z=3/2
    is equal to 3/2pi*i*e3/2

    Correct? i Know that sometimes the factor 2pi*i can be negative but i dont understandw hy..
  2. jcsd
  3. Apr 3, 2012 #2
    Okay so i have looked abck and tried to understand the derivation again, and have realized it comes as a consquence of the equation integral of f(z)/(z-z0)^m dz is equal to 2pi*i f^(m-1) (z0) / (m-1)!

    (if anyone finds my notation to be confusing or sloppy, i can rewrite it. also is there a link to a website that i can use to link equations to those neat pictures? it seems alot easier to understand than what im doing)

    so to apply that equation to the above, my function is .5*ze^z, my Z0 is 3/2, and i rewrite my equation

    where i know 0!=1, and anything raised to the 0 equals 1, so

    my integral evaluates to 2pi*i.
  4. Apr 3, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper

    You were closer the first time. It's the Cauchy Integral Theorem you want to look at. And 'analytic' doesn't mean 'goes to infinity', it means rather the opposite. I think you might want to read up on this before you tackle the problem. Your concepts are very confused.
  5. Apr 3, 2012 #4
    Okay so third times a charm, i figured out that it is not f^(m-1), but f with derivatives equal to m-1.

    So since my m=1 for this, i do not have to take a derivative but only evaluate it at 3/2 over !0.

    so my answer is pi * i * 3/2 * e^(3/2)
  6. Apr 3, 2012 #5
    I know, i am struggling in this class and im trying to keep up with it. I am going to re-read the chapter in order to understand the cauchy integral theorem but to be honest im pretty poor at learning math out of books. I am best at visualizing problems which is particularly difficult in complex equations since its a 4 dimensional problem.
  7. Apr 3, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper

    There's less detailed visualization required here than you might think. Your answer in the last post is correct. But I'd feel better if you were clearer on why that is.
  8. Apr 3, 2012 #7
    Okay so an integral is analytic if it satisfied the cauchy reimann equations. Basically what i know is if an equation satisfies the cauchy reimann equations, all is good and it can be worked with. I need to solve the second order cauchy reimann equation in order to prove a function is harmonic. Proving a function is harmonic is important (im a physics major) but i havn't had to do more than prove this.

    Anyhoo a function is analytic if the power series converges. (i see what you mean, analytic is good, not bad) So where the denominator "blows up" (when z=z0) is where we evaluate the integral around because we are really evaluating the power series there?

    If a function has a z0 outside the integral would the function then equal 0?
  9. Apr 3, 2012 #8


    User Avatar
    Science Advisor
    Homework Helper

    You are getting there. Your function has pole at z=3/2 and that's inside of the circle |z|=2. Just apply the Cauchy Integral Theorem.
  10. Apr 4, 2012 #9
    To be honest i can do the problems but the cauchy integral theorem still perplexes me. I understand the integral fails if there is a "hole" within the integral, and equals 0 everywhere else. So i guess what im really trying to integrate is the pole? So the entire point is to find the integral of the pole?

    I found this extremely helpful:


    But what confused me was that when winding multiple times over the integral the equation becomes 1/2pi*i, which seems counter intuitive to me since if the first integral is equal to 2pi*i, if i integrate twice it should simply be 4pi*i.

    The "derivative of f" was expressed as 1/(2pi*i)*(f(zo)/(z-zo)^2, but i thought i could just take the derivative of a complex number as i did a "regular" number...

    I also have another problem related to this, Should i create another thread? (i am going to sleep for now so ill post it here, as i suspect the answer is i shouldnt)

    1. let C be the square with vertices z= +-2 and +-2i traversed once in the positive sense.

    etc etc etc.

    i suspect i evaluate these just as i would the first problem. As from how i understand the integral as long as the pole is within the "area" im just going to be integrating it anyway, so i can effectively ignore the area that im supposedly integrating and just treat it as though im integrating the pole. The area as far as i know is just to stipulate if the pole is within it or not.
  11. Apr 4, 2012 #10


    User Avatar
    Science Advisor
    Homework Helper

    You aren't 'integrating the pole' you are integrating the function along the contour. In some cases you can do that directly. You should work out integral of f(z)dz around the contour |z|=1 for some simple functions, like f(z)=1/z, f(z)=1, f(z)=z. You can do this with the change of variable z=exp(it). Notice that the only integral of these that is nonzero has a pole at z=0. Finding contour integrals if the function or contours are more complicated would be extremely hard. But luckily you have these theorems that relate the hard job of finding the integral to the much easier job of finding and summing the residues of the poles. That's the general idea.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook