Still quite unsure about the chapter on orientation. For instance the Mobius map indeed preserves orientation, and I can show this rigorously, however I am still having difficulty understanding what it means for a holomorphic function $f$ to preserve orientation over some region in the complex plane. Which is why I asked a question regarding this $2$ days ago, but did not receive an answer.

The problem I am still thinking about is, I cannot show or provide a counter argument for why there does or does not exist some holomorphic function $f$ on a $\{z:|z|\le1\}$ such that it sends the unit circle with the counter- clockwise orientation into the unit circle with the clockwise orientation?

From the comment from this question asked earlier, one did suggest the use of the Argument principle which states that:

Argument Principle: If $f$ is meromorphic on a simply connected region $D$ and $\Gamma$ is a simply closed curve in $D$ not passing through the roots $z_j$ nor the poles $p_k$ of $f,$ then $$\frac{1}{2\pi i}\oint_{\Gamma}\frac{f'(z)}{f(z)}dz=n-m$$ Where $n$ is the number of roots of $f$(including multiplicities) in $\Gamma$ and $m$ is the number of poles of $f$ (including multiplicities) inside $\Gamma.$

I am really curious to find out why we can apply this to try and provide an explanation for my problem. I would appreciate some help because I am quite confused with orientation at the moment, and I think this problem is good in the sense it will help me understand some important notions on holomorphic functions which send or does not send certain orientation on a region to a different orientation of the same region.

I would really appreciate some rigorous argument for this problem. Much help will be appreciated.

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    $\begingroup$ Minor nitpick: You're missing a '$2\pi i$' in the formula of the argument principle. $\endgroup$ – Jonatan B. Bastos Apr 21 '18 at 4:43

[Following Kevin's hint on your previous question]. Suppose $f$ holomorphic on the unit disk $\overline{\mathbb D}$ maps $\partial\mathbb D$ into itself reversing orientation. Consider the integral $$ \frac1{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}\,dz,\ \ \ \ (\star)$$ where the counterclockwise orientation is assumed. We know its value is the number of zeros of $f$ in $\mathbb D$ by the Argument Principle (notice $f$ can't have any poles in $\mathbb D$ because it is holomorphic there). In particular this integral has to be non-negative.

On the other hand, performing the change of variables $z=e^{it}$, $$ \int_{|z|=1}\frac{f'(z)}{f(z)}\,dz=\int_0^{2\pi}\frac{f'(e^{it})}{f(e^{it})}\,ie^{it}\,dt=\int_0^{2\pi}\frac{\frac{d}{dt}\big[\,f( e^{it})\big]}{f(e^{it})}\,dt.\ \ \ \ (\ast) $$ We know that the derivative of $t\mapsto f(e^{it})$ as a curve from $[0,2\pi]$ to $\mathbb R^2$ is tangent to the unit circle $\partial\mathbb D$. Moreover, since $f$ reverses orientation the direction of this tangent vector must agree with the clockwise orientation. I.e. we have the following situation: enter image description here

Therefore $\frac{d}{dt}\big[f(e^{it})\big]\big/f(e^{it})$ has constant argument $-\pi/2$. Call $M(t)$ to its modulus. From $(\ast)$, $$ \int_{|z|=1}\frac{f'(z)}{f(z)}\,dz=\int_0^{2\pi}e^{-i\pi/2}M(t)\,dt=-i\int_0^{2\pi}M(t)\,dt. $$ From here it follows that $(\star)$ equals $\frac{-1}{2\pi}\int_0^{2\pi}M(t)\,dt$. Now, this is a non-positive number so (since we already knew $(\star)$ is non-negative) it has to be $\int_0^{2\pi}M(t)\,dt=0$ and hence $M$ is (a.e. and therefore everywhere by continuity) the $0$ function. This means that $f'=0$ on $\partial\mathbb D$, so $f'=0$ on $\overline{\mathbb D}$ (by the identity theorem for holomorphic functions for example), so $f$ must be constant. (!)

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    $\begingroup$ This was as clear as of an explanation I could have hoped for. This was very helpful thank you very much ! $\endgroup$ – Aurora Borealis Apr 21 '18 at 7:44

Supplementing Jonatan's answer, the reason for the preservation of orientation can also be seen by algebraic means.

Again if we consider $$\frac{1}{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}dz$$ With the counter clockwise orientation.

We can take a parametrization of the circle, $\gamma :[a,b] \rightarrow\mathbb{C}$, and perform the following manipulation:

$$\frac{1}{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}dz=\frac{1}{2\pi i}\int_{a}^{b}\frac{f'(\gamma(t))}{f(\gamma(t))}\gamma'(t)dt=\frac{1}{2\pi i}\int_{a}^{b}\frac{\frac{d}{dt}(f(\gamma(t)))}{f(\gamma(t))}dt$$

Which is, in turn $$\frac{1}{2\pi i}\int_{f\circ\gamma}\frac{1}{w}dw$$

But this is simply the winding number of the curve $f\circ \gamma$ about $0$. If we assume that $f$ maps the CCW unit circle into the CW unit circle then $f\circ \gamma$ is exactly a clockwise oriented unit circle, which means our integral must evaluate to $-1$.

But that is impossible by the argument principle, since $f$ is holomorphic in the unit circle and thus has no poles to speak of.


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