Complex analysis- show that the integral of this function exists

[jrp131191]

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.

 

[DonAntonio]

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

 

[jrp131191]

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.

 

[Mute]

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?

 

[blue_raver22]

Great job on your exam! Your approach to this problem is correct. By considering the function w(z)=zf(z)-c, which is analytic, you were able to use Cauchy's integral theorem to evaluate the integral. Splitting the integral into two parts, one in terms of f(z) and the other in terms of c/z, and using the winding number to rearrange for c, shows that the integral exists and can be expressed in terms of an integral of f(z). This is a valid and appropriate method for showing that the integral exists. Keep up the good work!