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In the proof of Argument theorem, it is stated the name comes from the fact that the integral of Logarithmic derivative is proportional to change of argument of function. ie.,

$$\int _{ \gamma} \frac{f'(z)}{f(z)}dz =\int _{ \gamma} \frac{d}{dz} \ln(r(z)e^{i\theta})) dz =\int _{ \gamma} \frac{d}{dz} \ln(r(z)) dz + i \int _{ \gamma} \frac{d}{dz} \theta (z) dz = i \int _{ \gamma} \frac{d}{dz} \theta (z) dz $$

My question is how is

$$\int _{ \gamma} \frac{d}{dz} \ln(r(z)) dz = \int _{ \gamma} \frac{1}{r(z)} d(r(z) = 0?$$

Clearly it has poles whenever $f(z)$ is zero inside $\gamma$ and can't be outright equated to $0$ as its not analytic.

In other words I am asking for the proof of

$\int _{ \gamma} \frac{f'(z)}{f(z)}dz = i \int _{ \gamma} \frac{d}{dz} \theta (z) dz $

Linking to this question.

Phalaksha C G
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  • "Specifically, if $f(z)$ is a meromorphic function inside and on some closed contour $C$, and $f$ has no zeros or poles on $C$, then" – FShrike Sep 22 '23 at 13:13
  • It has no poles "on C" but has poles and zeroes inside Function needs to be analytic everywhere inside for integral in closed contour to be 0 – Phalaksha C G Sep 22 '23 at 13:18
  • Does it help to note that $r = |f|$ is real-valued and positive, so $\ln r$ is a primitive of its derivative away from the zeros and poles of $f$? – Andrew D. Hwang Sep 22 '23 at 14:27
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    The expression $\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is just a shorthand for $\int_0^1\frac{f'(\gamma(t))}{f(\gamma(t))}\gamma'(t)dt$. The integration takes place not over $\mathbb C$, but over the parameter space $[0,1]$. If we pick a continuous branch $\theta$ for $\arg(f\circ\gamma)$, we obtain a continuous branch for the function $g=\ln\circ f\circ\gamma=\ln|f\circ\gamma|+i\theta$ and $g$ is an anti-derivative of $\frac{f'(\gamma(t))}{f(\gamma(t))}\gamma'(t)$. Therefore the contour integral is equal to $g(1)-g(0)$. ... – user1551 Sep 22 '23 at 14:55
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    ... Whether $f$ has any zero or pole inside the track of $\gamma$ is irrelevant. It’s just the fundamental theorem of calculus. Anyway, since $g=\ln\circ f\circ\gamma=\ln|f\circ\gamma|+i\theta$, we have $g(1)-g(0)=i\big(\theta(1)-\theta(0)\big)$. Hence the contour integral is equal to $i$ times the net change in the argument of $f(\gamma(t))$. – user1551 Sep 22 '23 at 14:55
  • So this should be applicable to the argument function too, besides according to Cauchy-Goursat, Fundamental theorem is applicable only if f is analytic everywhere in the region isn't it? ie., integral being independent of path and only end values? – Phalaksha C G Sep 23 '23 at 02:59
  • @AndrewD.Hwang its real valued defined on a complex plane right? and through Cuachy Riemann equations it would't be analytic except at origin, I would expect it to be consistent with all equations if so? – Phalaksha C G Sep 23 '23 at 03:01
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    There is no conflict. Suppose $f$ is analytic. If the integrand (namely $f'(z)/f(z)$) is analytic (i.e., if $f$ has no zero) inside the region bounded by $\gamma$, the integral along $\gamma$ must be zero. Hence $\theta(1)=\theta(0)$. If $f'/f$ has a pole inside the region bounded by the curve, we will have $\theta(1)\ne\theta(0)$. – user1551 Sep 23 '23 at 09:39
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    This does not hold for $\frac{1}{z}$ in a closed contour around origin where integral is not zero but $2\pi i$ – Phalaksha C G Sep 23 '23 at 11:34
  • The point is, if $\gamma$ is defined on $[0, 1]$ (say), then$$\int_\gamma\frac{r'(z),dz}{r(z)} = \ln r(z)\big|_{z=\gamma(0)}^{z=\gamma(1)}=0$$since $\gamma(0)=\gamma(1)$. We can't say the same about the integral of $f'/f$ because in general no continuous choice of "$\ln f$" exists on the image of $\gamma$. – Andrew D. Hwang Sep 23 '23 at 12:01
  • Sorry I don't understand what you mean by "gamma defined on [0,1]". If its not analytic then shouldn't it depend on the path and not just end values? – Phalaksha C G Sep 23 '23 at 12:17
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    @PhalakshaCG “This does not hold for $\frac1z$...” why not? Let $f(z)=z$. Then $\frac{f'(z)}{f(z)}=\frac1z$. When $\gamma$ travels around the unit circle once counter-clockwise, $f'/f$ has a pole inside the region bounded by the track of $\gamma$. Therefore $\theta(1)\ne\theta(0)$, as mentioned in my previous comment. In fact, we have $\theta(1)-\theta(0)=2\pi$ and $\int_\gamma\frac{f'(z)}{f(z)}dz=g(1)-g(0)=i\big(\theta(1)-\theta(0)\big)=2\pi i$, as expected. – user1551 Sep 23 '23 at 15:29
  • Makes sense, I can see how it is same as change in argument, but it would be nice if there was a proof that holds for all functions. it would help if you could state the same as an answer – Phalaksha C G Sep 24 '23 at 04:52

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