How to determine the orientation of the image of a curve under analytic map? Let $D = \{z\in \mathbb{C} : |z| \ < 1\}$ be the unit disk in the complex plane. 
Suppose I have an analytic map $f$ from $\mathbb{C} - D$ to $\overline{D}$ (the closed unit disk) which takes the unit circle to itself.
Then I have been told that if $\gamma$ is a curve traversing the unit circle clockwise then the image $f(\gamma)$ will traverse the unit circle anticlockwise. 
I can see this in examples but I wonder how to prove this in general, and also how to see in other examples which way the image curve will be traversed. Thanks!
 A: A function is analytic in a region if it has a derivative at every point in that region. Near $z_0$, $f(z)$ will look like
$$
f(z)=f(z_0)+f'(z_0)(z-z_0)+o(z-z_0)
$$
where $\lim\limits_{z\to z_0}\left|\,\frac{o(z-z_0)}{z-z_0}\,\right|=0$.
The way that complex multiplication works is that the magnitudes of the numbers are multiplied and the arguments are added to get the result. It is this property which guarantees the preservation of orientation of an analytic map. That is, if $z$ circles $z_0$ clockwise/counterclockwise) ($z-z_0$ circles the origin clockwise/counterclockwise), then $f(z)$ circles $f(z_0)$ clockwise/counterclockwise ($f(z)-f(z_0)=f'(z_0)(z-z_0)+o(z-z_0)$ circles the origin clockwise/counterclockwise).
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To see that this applies to larger shapes, not just small circles around a point, we can add a mesh 
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When the mesh is small enough, we can use the argument above on each cell of the mesh. The union of the mesh cells will then cycle clockwise/counterclockwise.

Note that if there are singularities in the region circled (which means $f$ does not have a derivative at every point of the region), this is not true. For example, under $f(z)=\frac1z$, $e^{it}$ circles the origin counterclockwise, but $e^{-it}$ circles clockwise.
