Example of analytic function that maps circle to self intersecting curve Is there an explicit example that an analytic function maps the unit circle to a self intersecting curve? As unit circle is not homogenous to self intersecting curve, I am considering finding an analytic function that is not homeomorphism to satisfy the requirement.
Thanks for any suggestion.
 A: The function $f(z) = (z+1/2)^2$ is an example. It first takes $D(0,1)$ to $D(1/2,1)$ and then applies the map $w\to w^2$ to it. The last map we know very well, and from it you can verify that $f$ maps the upper half of the unit circle bijectively to a curve that starts in the first quadrant, moves left to the fourth quadrant, comes down and intersects the negative real axis, continues downward and rightward under $0$ and then comes back up to end at $1/4.$ The bottom half of the circle gets mapped to the conjugate of this. The total curve intersects itself exactly once, on the negative real axis, and the curve is regular.
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A: Take any two points $\exp(i\alpha) \ne \exp(i\beta)$ on the unit circle, and a nonconstant analytic function $f$ on a neighbourhood of the circle such that $f(\exp(i\alpha)) = f(\exp(i\beta))$.  Then the curve $\gamma(t) = f(\exp(it))$, $0 \le t \le 2\pi$ intersects itself ($\gamma(\alpha) = \gamma(\beta)$).   If you want the curve to be regular, require the zeros of $f'$ to be off the unit circle.  
For example, you might try
$$f(z) = \left(z - e^{i\alpha}\right)\left(z - e^{i\beta}\right)$$
A: I think $$cos(x)+i sin(2x)$$ is a natural, simple example. ($x=\frac{lnz}{i})$. It has a pole in $z=0$, however (because of the log).
By plugging in z, you get the following representation, which is more straight forward but doesn't show the figure behind it:
$$\frac{3z+\frac{1}{z}}{2}$$
