6
$\begingroup$

Let $f$ be holomorphic in an open $\Omega \subset \mathbb{C}$ and $\gamma$ a closed curve in $\text{int}(\Omega)$, along which $f$ is never zero.

Are these hypotheses enough to claim $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is an integer? If not, what are the necessary and sufficient conditions for that?

I can prove that's an integer for some particular cases, for example when $\Omega$ is convex and $\gamma$ is a circle. But I've seen people claming this in many other contexts, like when $\Omega$ is an annulus around $0$ and $\gamma $ is an arbitrary curve inside it. This isn't obvious to me at all.

Thanks!

$\endgroup$
  • $\begingroup$ Look up the Delves-Lyness algorithm. $\endgroup$ – J. M. is a poor mathematician May 11 '16 at 17:48
  • 2
    $\begingroup$ We must require that $\gamma$ doesn't pass through any zero of $f$ of course. $\endgroup$ – Daniel Fischer May 11 '16 at 17:49
  • 1
    $\begingroup$ More or less a duplicate of math.stackexchange.com/questions/1197670/…, since the integral is the winding number of $f \circ \gamma$ with respect to zero. $\endgroup$ – Martin R May 11 '16 at 18:25
6
$\begingroup$

One must require that $\gamma$ doesn't pass through any zeros of $f$. If $\gamma$ passes through a zero of $f$, then the integral doesn't exist as a Lebesgue integral, but under mild assumptions on the regularity of $\gamma$ one can still interpret it as a principal value integral. However, in that case, the principal value need not be an integer.

If $\gamma$ doesn't pass through any zero of $f$, then noting that $\frac{f'}{f}$ is the derivative of any local branch of $\log f$ one deduces the assertion. Let's suppose that $\gamma \colon [0,1] \to \Omega \setminus f^{-1}(0)$, and set $z_0 = \gamma(0)$. Using the local existence of branches of $\log f$ on $\Omega \setminus f^{-1}(0)$, one finds that

$$f(\gamma(t)) = f(z_0)\cdot \exp \biggl( \int_{\gamma\lvert_{[0,t]}} \frac{f'(z)}{f(z)}\,dz\biggl)$$

for all $t\in [0,1]$. Since $\gamma(1) = \gamma(0)$ it follows that

$$\exp\biggl( \int_{\gamma} \frac{f'(z)}{f(z)}\,dz\biggr) = 1,$$

which is equivalent to the assertion.

$\endgroup$
1
$\begingroup$

Yes, because that's the Logarithmic Derivative of $f$ and if it's meromorphic, it will have residue $\pm n$, either from a zero of order $n$, or from a pole of order $n$ inside your contour.

Choosing any small circle $w=\rho\exp(i\theta)$ around the enclosed singularity or zero and using the principal branch of the logarithm,

$$ \frac{1}{2\pi i}\int_\gamma \frac{dw}{w}=\frac{\pm n}{2\pi i}\int_{-\pi}^\pi \frac{i\rho\exp(i\theta)d\theta}{\rho\exp(i\theta)}=\frac{\pm n}{2\pi i}\int_{-\pi}^\pi i d\theta=\frac{\pm n\cdot 2\pi i}{2\pi i}=\pm n $$

(With my apologies of course to Daniel Fischer's much more complete answer)

And btw, that's the winding number of $f$ as Martin R says in his comment.

$\endgroup$
  • 1
    $\begingroup$ Unless I am mistaken, this argument works only if $f$ is holomorphic in a simply-connected domain with the exception of isolated singularities, but not – for example – for $f$ holomorphic in an annulus. $\endgroup$ – Martin R May 11 '16 at 18:48
  • $\begingroup$ That's how I remember it as well, but I seem to miss the annulus situation. What exactly seems to be the problem of the argument with such domains? One would think that $\rho$ might be the problem (since $\rho$ doesn't go to $0$ in an annulus), but it looks like it cancels above, so it doesn't look like $\rho$ matters, afterall. Can you please shed some light on this? $\endgroup$ – Yiannis Galidakis May 11 '16 at 19:09
  • $\begingroup$ Take $\Omega$ as an annulus around $0$, for example. A priori, we could take $f$ with no zeros and no poles at the annulus and $\gamma$ a circle around $0$. In that case, I think what you said about residues, poles and zeros wouldn't apply. $\endgroup$ – rmdmc89 May 11 '16 at 22:27
  • $\begingroup$ Agreed, but I am talking about a meromorphic $f$. Doesn't make sense to consider an annulus around 0 when I know that $f$ has at least one pole/zero at 0. On the other hand, if $f$ is holomorphic in the given region by definition it doesn't have poles or zeros there, so by Cauchy-Goursat the integral would be zero, which is still an integer. $\endgroup$ – Yiannis Galidakis May 12 '16 at 0:35
  • $\begingroup$ I think we can only use Cauchy's theorem when the interior of the circle is contained in $\Omega$. That is why the example of the annulus and the circle around zero could be problematic for an arbitrary function. $\endgroup$ – rmdmc89 May 12 '16 at 1:44

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.