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Recall that the argument principle states that given a meromorphic function $f$ and a compact region $K \subseteq \mathbb{C}$ whose boundary determines a simple contour and on which $f$ has no singularities, then:

$$ \frac{1}{2\pi i}\int_{\partial K} \frac{f'(w)}{f(w)}dw = (N(Z) - N(P))$$

where $N(Z)$ and $N(P)$ refer to the number of zeroes and poles $f$ has in the interior of the compact region.

I understand the statement and proof of the theorem, but I am a bit unclear as to why the contour integral of the logarithmic derivative is interpreted as the change in argument of $f$ along the simple contour $\partial K$.

I realize that $\frac{d}{dz} \log{f(z)} = \frac{f'(z)}{f(z)}$, so I can certainly understand why we would consider the quantity

$$\frac{1}{i} \int_{\partial K} \frac{f'(w)}{f(w)}dw$$

to measure the change in argument of $f$ along the contour if the contour is closed (since then the real part of the logarithm would vanish, so this quantity would indicate the change of argument).

However, too many times I have encountered teachers/websites/authors refer to

$$\int_{\gamma} \frac{f'(w)}{f(w)}dw$$

as the change of argument of $f$ even when $\gamma$ is not stated to be a closed contour. And this is precisely why I doubt my understanding of what's going on here.

Is there something obvious that I'm overlooking?

combinator
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  • Perhaps there's an implicit assumption that $\gamma$ be closed. And the imaginary part of the integral is of course the change of argument along $\gamma$ also for non-closed $\gamma$. – Daniel Fischer Feb 22 '14 at 21:04
  • I thought this might be a possibility, but what prompted me to ask this question today is that Stein & Shakarchi make it explicit that there is no such implicit assumption.

    They write that the integral of the logarithmic derivative along a curve is the change in argument of the function as it traverses the curve. In the next sentence, they say that if we assume that the curve is closed, then in that case the change in argument is determined entirely by the zeroes and poles of the function in the interior of the contour.

     – combinator Feb 22 '14 at 21:06
  • Do they perhaps refer to the imaginary part of the integral? Otherwise, can you quote the pertinent passage? (Hmm, maybe you just did. Anyway, the real part of the integral has nothing to do with the change of argument.) – Daniel Fischer Feb 22 '14 at 21:09
  • Sure, I'll quote the passage as it is written:

    "Note that in any case, the derivative of $\log{f(z)}$ is $\frac{f'(z)}{f(z)}$ which is single-valued, and the integral $\int_{\gamma} \frac{f'(z)}{f(z)}dz$ can be interpreted as the change in the argument of $f$ as $z$ traverses the curve $\gamma$. Moreover, assuming the curve is closed, this change of argument is determined entirely by the zeros and poles of $f$ inside $\gamma$."

     – combinator Feb 22 '14 at 21:12
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    I suppose you'll agree that if we add "the imaginary part of" at the right place, it makes good sense. And if we don't, I agree that it doesn't really. – Daniel Fischer Feb 22 '14 at 21:18
  • Yeah, sure. We're on the same page. I just had to ask because I've come across similar ambiguities from other sources, as well. – combinator Feb 22 '14 at 21:19

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