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Contours, integration and connected sets

  1. Sep 12, 2009 #1
    1. I am having trouble grasping cauchys theorm. To my understanding the integral for any closed contour in which f(z) is path connected and analytic =0.

    2. Relevant equations
    Example 1:
    Now the integral around the unit circle for 1/z. In the unit circle 1/z is anlaytic for all z, but not path connected because when z=0 1/z is not defined. Therefore the integral does not equal zero and is equal to 2πi. I am confused about the following.

    1) is my reasoning correct?

    2) the book says that this integral is equal to 2πi. However when i integrate using γ(t)=e^it i get the right answer. When i apply the fundamental theorm and use the antiderivative of 1/z = Lnz and evaluate from 0 to 2π. I get Ln(1)+2πi+Kπi-Ln(1)-0πi+πKi. Now since o and 2π give the same argument doesnt this equal zero?

    3) if indeed the integral of any analytic function in a path connected set is zero. Applying this reasoning to f(z) =1/z^2. With the unit circle. F(z) is analytic in the unit circle but not path conneceted because it is not defined at z=0. To my understanding it should thus not equal zero. But it Does equal zero when i compute it. These apparent contradictions are doing my head in so if any one understands this i would like some help.

    My concern is : I can solve these things no problem. I know that for 1/z^2+1 i can factorise the denominator and integrate in two regions where z+i, and z-i are analytic, but the above two examples seem to contradict my reasoning. So now i am confused and i am worried because if i am confused now i will be totally confused in an exam!

    3. The attempt at a solution
  2. jcsd
  3. Sep 12, 2009 #2


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    And therefore is NOT "analytic for all z in the unit disk" (inside the unit circle). What you mean is that 1/z is analytic in the "punctured" disk: [itex]\{(x,y)|0< x^2+ y^2\le 1\}[/itex]. Also both the disk and punctured disk are path connected since it is possible to draw a continuous curve from any point inside the set to any other point inside the set without leaving the set. The punctured disk is not simply connected since there exist a closed path (any circle about the origin) that cannot be contracted to a point inside the set.

    They "give the same argument" but are NOT the same number so, no, this is not 0. A better way to do this integral is:
    On the unit circle, [itex]z= e^{i\theta}[/itex] so [itex]dz= ie^{i\theta}d\theta}[/itex] and
    [tex]\int \frac{dz}{z}= \int_{\theta= 0}^{2\pi}\frac{i e^{i\theta}d\theta}{e^{i\theta}}[/tex][tex]= \int_{\theta= 0}^{2\pi} i d\theta= 2\pi i[/tex]

    "If A then B" does NOT imply "if not A then not B". "If F(z) is analytic in the unit disk then the integral around the unit circle is 0" does NOT imply "If F(z) is NOT analytic in the unit disk then the integral around the unit circle is NOT 0". If F(z) is NOT analytic then you do not know whether the integral is 0 or not without actually doing the integral.

    You need to be more careful about your terms:
    1: Distinguish between "circle" and "disk".
    2: Distinguish between "disk" and "punctured disk"
    3: The punctured disk is "path connected"; it is not "simply connected".
  4. Sep 12, 2009 #3
    Lets say the function f(z) =1/z. It is entirely analytic in the disk y(t)=2+e^it.So it is zero. What if i am asked to calculate it and show that it is zero. Then i would need to use Inz.

    Now this function cannot be evaluated with the method you showed me. so i would need to use a substitution where u=2+e^it. and then i would get {Ln(2+e^i2π+2kπi)-Ln(2+0π+2kπ)

    I am just trying to figure out cases that are not the "norm" because i know from experience thats the type of question you get in an exam.The books i am using are advanced engineering mathematics, and complex analysis< shaum series> they seem to just go through the most common type examples.
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