Let us consider a function $f(z)$ holomorphic along and inside a contour $\Gamma$ not surrounding the origin. With reference to the following contour integrals: $$ \oint\limits_{\Gamma} \frac{f(z)}{z}\,dz \,\,\,\,\,\,\,\,\, , \,\,\,\,\,\,\,\,\, \oint\limits_{\Gamma} \frac{f(z)}{\bar{z}}\,dz \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; z = x + iy$$ the one at the left vanishes, while the same cannot in general be stated for the one at the right (as $f(z)/\bar{z}$ is not holomorphic). However, could there exist specific conditions sufficient to guarantee that the integral at the right would strictly result $\neq 0$ ?
The reason of my investigation on this subject stems from the realization that this sort of strict integral inequalities are hard to come by in text books. Moreover, it was inspired by this discussion Topological properties of complex valued Riemann sum limit curve and a particular integral inequality, while attempting to generalize the same "visual" approach to the closed contour integrals above.
As the integral at the right would in general depend also on the particular contour $\Gamma$, for a start I would thus restrict the investigation to simple rectangular contours of the following kind, and whereby $Y$ may later be allowed to $\rightarrow \infty$, while further restricting to those $f(z)$ such that the integral of $f(z)/z$ along the horizontal sides of the rectangle $\Gamma$ vanishes at infinity.
As a visual support to the description, the simple example $f(z) = (0.75 + 4z + z^2)^{-3}$ was selected, among other randomly tried ones, just because resulting in comfortably readable curves. In the plot below, the complex integrals of $f(z)/z$ and $f(z)/\bar{z}$ along the contour $\Gamma$ would then be approximated by the vectors joining start and end point of the two curves $\gamma 1$ and $\gamma 2$, respectively. A more thorough description of this approach can be found at Section III of Chapter 8 (at least in the 2012 edition I have) of Prof. Needham's excellent book "Visual Complex Analysis" http://usf.usfca.edu/vca/.
Segmenting the contour $\Gamma$ in many small vectors $\Delta_i$ of equal length $\Delta$, each curve is then obtained by drawing "tip to tail" the "infinitesimal" vectors $f(z_n)/z_n \; \Delta_n$ for $\gamma 1$, and $f(z_n)/\bar{z}_{n} \; \Delta_n$ for $\gamma 2$ (where $z_n = x_n + iy_n$ ). The vectors joining the start and end points of the respective curves would then correspond to complex Riemann sums numerically approximating said two integrals. Reassuringly, $\gamma 1$ is a closed curve (i.e. the corresponding contour integral is zero, as indeed so it shall be). But let us walk along it (for this numerical approximation example: $a=1$, $b=2$, $Y=0.9$, $\Delta = 0.001$):
- the red segment corresponds to the above mentioned sum computed along $\Gamma$ from $b+i0$ to $b+iY$
- the orange segment to the sum computed along $\Gamma$ from $b+iY$ to $a+iY$
- the blue segment to the sum computed along $\Gamma$ from $a+iY$ to $a-iY$
- the light blue segment to the sum computed along $\Gamma$ from $a-iY$ to $b-iY$
- the black segment to the sum computed along $\Gamma$ from $b-iY$ to $b+i0$
$\gamma 2$ is instead an open curve (i.e. the corresponding contour integral is $\neq 0$, as indeed so it could be), noting that arcs of the same color correspond to the same segment on the contour $\Gamma$.
A perhaps more natural representation aiding in intuitively understanding such result is provided by the so called Whewell equation (see the corresponding entry in Wikipedia), whereby a curve is described in terms of its tangent vector angle $\varphi$ considered as a function of arc length $s$, i.e. $\varphi = g(s)$.
For the numerical approximation curves $\gamma 1$ and $\gamma 2$ above, $\varphi (s)$ would hence be approximated by the phase (or argument) of the respective "infinitesimal" vectors $f(z_n)/z_n \; \Delta_n \,$ and $\,f(z_n)/\bar{z}_{n} \; \Delta_n \, = \, e^{\,i \, \theta_n} \,f(z_n)/z_n \; \Delta_n $, where $\theta_n \, \, = 2 \arctan \frac {y_n}{x_n}$. Therefore, the curve $\gamma 2$ can be obtained from the curve $\gamma 1$ by applying the transformation: $$(1) \;\;\;\;\;\;\; \varphi (s) \, \rightarrow \, \varphi (s) + \theta(s) \; , \;\;\;\;\; \theta(s)=2 \arctan \frac {y(s)}{x(s)} \;\; , \;\;\; on \; \Gamma \; \Longrightarrow \; 0 \leq |\theta(s)| < \pi$$ An heuristic explanation of why $\gamma 2$ ought instead to be an open curve may hence observe that the transformation $(1)$ will "open up" $\gamma 1$ along some segments (compare for example the two red ones), while "closing it back" along some others, so that if $\gamma 1$ closes back on itself the condition $0 \leq |\theta(s)| < \pi$ makes it impossible for the same to occur also for $\gamma 2$ (just meaning it as a preliminary conjecture). For the rectangular contour $\Gamma$ above we can observe the following:
- Along the first half of the contour (i.e.: $b+i0 \rightarrow b+iY \rightarrow a+i0$) $\theta(s)$ starts from $0$,
- progressively increases to asymptotic $\pi$ (in case $Y \rightarrow \infty$), but never crossing it over,
- to then progressively decrease back to $0$ at $a+i0$.
- Similarly so for the other half of $\Gamma$, though now with the opposite sign.
- Moreover, at the same "traveled distance" along $\Gamma$ corresponds the same value of arc length $s$ for both $\gamma 1$ and $\gamma 2$ (it suffices to observe how the lengths of the "infinitesimal" vectors $f(z_n)/z_n \; \Delta_n$ and $f(z_n)/\bar{z}_{n} \; \Delta_n$ share the same value), and thus also total length ought to be the same.
Expressing it in the language of Differential Geometry: for the reference rectangular contour $\Gamma$, the total curvature of $\gamma 2$ is conserved identical to the total curvature of $\gamma 1$ (as the rotation $\theta(s)$ always folds back to $0$ without ever having crossed over $\pi$). Moreover, also the total length is conserved.
Finally, I am now able to formulate my question:
Can a transformation of the kind $(1)$ above transform a closed curve into another closed curve, while conserving total curvature and length ?
If it could be proven that it cannot, then it would also follow (at least for the above described integrability conditions and rectangular contour $\Gamma$): $$ \oint\limits_{\Gamma} \frac{f(z)}{\bar{z}}\,dz \; \neq \; 0$$ By means of the complex form of Green's theorem, such a result could then be extended to the integral over the region $R$ surrounded by the contour $\Gamma$: $$ \oint\limits_{\Gamma} F(z,\bar{z}) \,dz \; = \; 2 i \iint\limits_{R} \, \frac{\partial F}{\partial \bar{z}} \, dA \;\;\;\;\;\;\;\;\;\;\; \Longrightarrow \;\;\;\;\;\;\;\;\;\;\; \iint\limits_{R} \, \frac{f(z)}{\bar{z\;}^2} \, dA \; \neq 0$$
Many thanks for any suggestion about how to best approach such problem.
