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Complex analysis- show that the integral of this function exists

  1. Nov 4, 2012 #1
    Hi, I just had my exam on complex analysis and would just like to know if I did this question correctly.

    It said that the function [itex]f(z)[/itex] was analytic and to show that the integral of [itex]f(z)-\frac{c}{z}[/itex] existed for some constant c, then to find a formula for c in term of an integral of f(z).

    I said consider the function [itex]w(z)=zf(z)-c[/itex], which is clearly analytic, then:

    [itex]w(0)=\frac{1}{2\pi i }\oint \frac{zf(z)-c}{z}dz[/itex] by cauchys integral theorem

    so the integral of the given function is simply equal to the function i defined by w evaluated at z=0

    then from there I just split the integral into 2 integrals, one in terms of f(z) and the other in terms of c/z in which the constant can come out, and then the integrand would be equal to 2pik with k= winding number, given the contour didnt cross the origin and rearranged for c.
  2. jcsd
  3. Nov 4, 2012 #2

    This question looks pretty weird and so does your answer: we have that if [itex]\,C:=\{z\in\Bbb C\;\;;\;\;|z|\leq c\,\,,\,\,c\in\Bbb R^+\}\,[/itex] , then:

    $$\oint_{\partial C}\,\frac{dz}{z}=\left\{\begin{array}{} 0&\,,\,\text{ if }\,\,0\notin C\\{}\\2\pi i&\,,\,\text{ if }\,\,0\in \stackrel\circ C\end{array}\right. $$

    If zero is on the integration path then the integral, as far as I can see, isn't defined.

  4. Nov 4, 2012 #3
    Yes sorry I forgot to mention that, I did write " given that C does not cross the origin".

    Edit: Ok I did mention it in the post lol. But I just want to know how I went with it. I was very scared that I wouldn't be able to do any of the proofs but I felt pretty confident about this one.
  5. Nov 4, 2012 #4


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    What contour are you integrating over? Just one that circles the origin some number of times? In the case of say, a simple circular contour around the origin, I don't see how what you've done really gives you anything non-trivial. By definition of w(z), w(0) = c, and if you evaluate the integral expression, ##\oint dz zf(z)/z = 0## by the residue theorem because f(z) is analytic, so all you get out in the end is c = c. Is something else missing from the statement of the question?
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