TITLE: Is image of boundary a boundary of image?
QUESTION [5 upvotes]: I have a question which appers in problems concerning Möbius transformation, for example
Let $A=\{ z\in \mathbb{C} : \|z\| <1, \Re(z)>0\}$ and $f(z)=\frac{z+i}{z-1}$
Determine $f(A)$. 
Often it is quite simple to find $f( \partial A)$ (image of boundary of $A$). If I knew that $f(\partial A)=\partial (f(A))$, I would only have to check whether $f(A)$ is inside  or outside the $\partial f(A)$.
What is sufficient condition on $f$ for having $f(\partial A)=\partial (f(A))$ ?

REPLY [5 votes]: The Möbius transformation maps circles and lines to circles and lines, so if $f$ is Möbius, we can be sure that $f (\partial A) = \partial( f(A))$.  
However, the result is not general to all transformations.  Consider the upper unit semicircle $A = \{ z \in \mathbb{C} : \|z \| \leq 1, Im(z) \geq 0\}$ and $f(z) = z^2$.  The boundary segment of the semicircle along the Real axis gets mapped by $f$ to the segment from $0$ to $1$ along the Real axis, which is interior to $f(A)$ (the unit circle).
Hope this helps!

Addendum
To show that $f(\partial A) = \partial(f(A))$, I'll use another fact of Möbius transformations:  Every Möbius transformation $f$ can be expressed as a composition of translations, inversions, scaling (magnification, etc.), and rotations.  Each such mapping preserves the property "$x$ is a boundary point."