1
$\begingroup$

I am asked to show that for $z\in \mathbb{C} \setminus \{0,1\}$, there exists an analytic (single-valued) function, $F(z)$ on $\mathbb{C} \setminus \{0,1\}$, such that $F'=f$, where $$f(z) = \frac{(1-2z)\cos(2\pi z)}{z^2 (1-z)^2}$$ I know that if $$\int_{\gamma} f(z) dz =0$$ for all closed contours, $\gamma$, then $f$ has an antiderivative. Furthermore, in the case of the given function above, $f(z)$, I know that Res$(f,0)=$ Res$(f,1)=0$, so using the Residue Theorem I know that for any simple closed contour, $\gamma$, we have $$\int_{\gamma} f(z) dz =0$$

However, to ensure that $f$ has an antiderivative, I need to show that this is true for all closed $\gamma$, not just simple closed $\gamma$. How can I go about finishing this last step of the proof?

$\endgroup$
  • $\begingroup$ Are you allowed to use Taylor series expansion of cosine? If so, the result follows immediately by considering the uniform convergence of the series on proper domains. $\endgroup$ – Ken Hung Aug 4 at 0:17
  • $\begingroup$ The Residue theorem holds for all (rectifiable) closed curves. Which version are you referring to? $\endgroup$ – Martin R Aug 4 at 0:18
  • $\begingroup$ For general rectifiable closed curve $\gamma$ in $\mathbb{C}\setminus\{0,1\}$, we have $$\int_{\gamma}f(z)\,\mathrm{d}z=2\pi i \sum_{z_0\in\{0,1\}}\operatorname{wind}(f,z_0)\operatorname{Res}(f,z_0),$$ where $$\operatorname{wind}(f,z_0)=\frac{1}{2\pi i}\int_{\gamma}\frac{\mathrm{d}z}{z}\in\mathbb{Z}$$ is the winding number. It can be proved by deforming $\gamma$ into a formal linear combination of simple closed curves. $\endgroup$ – Sangchul Lee Aug 4 at 0:25
  • $\begingroup$ What I was trying to emphasize is that in showing that the antiderivative of a certain function exists, one may try to construct a series expansion of the function and try to show that it converges uniformly on some domain. So that you can integrate (here I mean the inverse process of differentiation) the series term by term to find the antiderivative explicitly. $\endgroup$ – Ken Hung Aug 4 at 0:30
  • $\begingroup$ In this case, Taylor series expansion of cosine might be helpful. $\endgroup$ – Ken Hung Aug 4 at 0:31
1
$\begingroup$

Here is an alternative solution. Write

$$ g(z) = f(z) - \left( \frac{1}{z^2} - \frac{1}{(z-1)^2} \right). $$

Then $g$ has removable singularities at both $z=0$ and $z=1$, and so, $g$ extends to a holomorphic function on $\mathbb{C}$. In particular, $g$ has an antiderivative, say $G(z)$. Then

$$ f(z) = g(z) + \frac{1}{z^2} - \frac{1}{(z-1)^2} $$

has an antideriviative

$$ G(z) - \frac{1}{z} + \frac{1}{z-1}. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the response. I'm trying to follow your reasoning where you conclude that $g$ has an antiderivative. You say $g$ extends to a holomorphic function and so, in particular, $g$ has an antiderivative. However, just because a function is holomorphic does not mean that it has an antiderivative. As Martin pointed to in a comment above, $1/z$ is holomorphic in $\mathbb{C} \setminus \{0\}$ yet does not have a holomorphic antiderivative. $\endgroup$ – French Toast Crunch Aug 4 at 3:36
  • $\begingroup$ @FrenchToastCrunch, The crucial difference here is that the domain of $g$ is all of $\mathbb{C}$, which is simply connected. In such case, a holomorphic function always admits an antiderivative. (That is why we bother to remove the poles in this solution!) Alternatively, since $g$ is analytic on all of $\mathbb{C}$, it admits Taylor series that converges to $g$ on all of $\mathbb{C}$, and so, its antiderivative can be represented explicitly via termwise integration. $\endgroup$ – Sangchul Lee Aug 4 at 3:40
  • $\begingroup$ I see, I'm following now. The last bit I need to put the pieces together is understanding why, if $g$ is a function with removable discontinuities, it extends to a holomorphic function on $\mathbb{C}$. My basic knowledge of analytic continuation is that if $f_1$ and $f_2$ are analytic on domains $D_1$ and $D_2$, respectively, and $f_1=f_2$ on $D_1 \cap D_2$, where $D_1 \cap D_2$ is non-empty, then $f_2$ is the analytic continuation of $f_1$ on $D_2$. I'm trying to use this to conclude what you did about the extension of $g$.If you can clear up this last step I'd be happy to accept your answer. $\endgroup$ – French Toast Crunch Aug 4 at 4:20
  • 1
    $\begingroup$ @FrenchToastCrunch, It seems that you have not learned Riemann's theorem on removable singularities yet. The proof is actually short, and is included in the link. $\endgroup$ – Sangchul Lee Aug 4 at 15:44
2
$\begingroup$

The Residue theorem states that for any (rectifiable) closed curve $\gamma$ in $\mathbb{C} \setminus \{0,1\}$: $$ \int _{\gamma }f(z)\,dz=2\pi i \bigl( \operatorname {I} (\gamma ,0)\operatorname {Res} (f,0) + \operatorname {I} (\gamma ,1)\operatorname {Res} (f,1) \bigr) $$ and that is zero for the given function $f$ because both residues are zero (as you already calculated).

Alternatively you could use that $$ f(z) = \frac{\cos(2 \pi z)}{z^2} - \frac{\cos(2 \pi (z-1))}{(z-1)^2} $$ and show that $\frac{\cos(2 \pi z)}{z^2}$ has an antiderivative in $\mathbb{C} \backslash \{0\}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What is $I(\gamma,0)$? Is that an indicator function specifying that the curve $\gamma$ encloses the point $0$? $\endgroup$ – mjw Aug 4 at 1:12
  • $\begingroup$ @mjw: It is the winding number of $\gamma$ with respect to the point $0$. $\endgroup$ – Martin R Aug 4 at 1:16
  • $\begingroup$ Okay, thank you. $\endgroup$ – mjw Aug 4 at 1:19
  • $\begingroup$ @MartinR : Thank you for your response, in the link you provided, the claim is that the Residue Theorem holds for any rectifiable closed curve in $U$, where $U$ is a simply connected open subset of $\mathbb{C}$. However, $\mathbb{C} \setminus \{0,1\}$ is not a simply connected set. I like your alternate suggestion but ultimately I'd be faced with the same problem I was running into. Following along with your suggestion, how do you show $\cos(2 \pi z)/z^2$ has an antiderivative in $\mathbb{C} \setminus \{0\}$? $\endgroup$ – French Toast Crunch Aug 4 at 3:28
  • 1
    $\begingroup$ @FrenchToastCrunch: The curve integral is only defined for rectifiable curves. $\endgroup$ – Martin R Aug 4 at 4:19
0
$\begingroup$

The function $f$ is holomorphic on $\mathbb{C} \backslash \{0,1\}$ because $f$ is the quotient of holomorphic functions. Since $f$ is holomorphic in this set, it has a primitive $F$ there so that $F^\prime=f$. Or is the question to find $F$?

UPDATE

This answer is not complete. Leaving it here for a while because the comment from @Martin R was helpful.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A holomorphic function does not necessarily have a (holomorphic) antiderivative. The standard counterexample is $1/z$ in $\mathbb{C} \backslash \{0\}$. – Your argument would only work in a simply-connected domain. $\endgroup$ – Martin R Aug 4 at 0:59
  • $\begingroup$ Okay, so (in general) there is not necessarily an $F$ that works for the whole set, if not for every closed curve $\gamma$, $\oint_\gamma f(z) \,dz = 0$. I am glad I attempted an answer. I learned something. Thank you! $\endgroup$ – mjw Aug 4 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.