Complex analysis- show that the integral of this function exists

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Discussion Overview

The discussion revolves around a complex analysis problem involving the existence of an integral of the function f(z) - c/z, where f(z) is analytic and c is a constant. Participants explore the implications of Cauchy's integral theorem and the conditions under which the integral is defined, as well as the relationship between c and f(z).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes considering the function w(z) = zf(z) - c, asserting it is analytic and using Cauchy's integral theorem to evaluate the integral.
  • Another participant questions the validity of the integral if the contour crosses the origin, citing that the integral is not defined in that case.
  • A later reply acknowledges the concern about the contour and reiterates that the integral is defined only if the contour does not cross the origin.
  • Another participant challenges the approach by asking for clarification on the contour used for integration, suggesting that the evaluation may not yield new information if f(z) is analytic.
  • There is a mention of the residue theorem and its implications for the integral, indicating that the evaluation might lead to trivial results.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the contour choice and the validity of the integral. There is no consensus on whether the approach taken leads to a meaningful result.

Contextual Notes

Participants note the importance of the contour not crossing the origin for the integral to be defined, highlighting the dependence on the chosen path of integration.

jrp131191
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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 f(z) was analytic and to show that the integral of f(z)-\frac{c}{z} 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 w(z)=zf(z)-c, which is clearly analytic, then:

w(0)=\frac{1}{2\pi i }\oint \frac{zf(z)-c}{z}dz 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.
 
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jrp131191 said:
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 f(z) was analytic and to show that the integral of f(z)-\frac{c}{z} 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 w(z)=zf(z)-c, which is clearly analytic, then:

w(0)=\frac{1}{2\pi i }\oint \frac{zf(z)-c}{z}dz 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.


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

DonAntonio
 
DonAntonio said:
This question looks pretty weird and so does your answer: we have that if \,C:=\{z\in\Bbb C\;\;;\;\;|z|\leq c\,\,,\,\,c\in\Bbb R^+\}\, , 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.

DonAntonio

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.
 
jrp131191 said:
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 f(z) was analytic and to show that the integral of f(z)-\frac{c}{z} 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 w(z)=zf(z)-c, which is clearly analytic, then:

w(0)=\frac{1}{2\pi i }\oint \frac{zf(z)-c}{z}dz 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.

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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