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.


  • $\begingroup$ Look up the Delves-Lyness algorithm. $\endgroup$ – J. M. is a poor mathematician May 11 '16 at 17:48
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    $\begingroup$ We must require that $\gamma$ doesn't pass through any zero of $f$ of course. $\endgroup$ – Daniel Fischer May 11 '16 at 17:49
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    $\begingroup$ 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

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.


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.

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    $\begingroup$ 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

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