Rigorous statement on the possible shapes of branch cuts I have read on numerous occasions, e.g. in this answer that a branch cut of the complex logarithm can be any curve that connects origin and complex infinity and does not intersect itself. I am looking for a more rigorous version of that statement.
My question: Is it possible to define a branch cut as a curve in the complex plane given by an implicit equation $F(x,y)=0$? If so, what are the conditions on $F$ so that it represents a branch cut?
 A: A compact subset $A$ of $S^2$ is called a Jordan arc is $A$ is homeomorphic to the interval $[0,1]$. 
Theorem. The complement to any Jordan arc in $S^2$ is simply connected. 
Proof. One proof you can find in Achille Hui's answer to this question. It relies upon Schoenflies Theorem:
For every Jordan loop $C\subset S^2$ there exists a homeomorphism $S^2\to S^2$ which takes $C$ to a round circle. 
Let me give another argument:


*

*By the Alexander duality theorem, $H_1(S^2 -A)\cong H^0(A)=0$ and $\tilde{H}_0(S^2-A)\cong H^1(A)=0$ (homology and cohomology is taken with integer coefficients). Hence, $U:=S^2-A$ is connected (the second isomorphism) and has trivial first homology group. 

*By Hurewicz theorem, $H_1(U)$ is isomorphic to the abelianization $G^{ab}$ of $G=\pi_1(U)$. 

*The fundamental group $G$ of every noncompact connected surface, such as  $U$, is free, of (possibly infinite) rank $r$ (see here), hence, $G^{ab}$ is isomorphic to the free abelian group of rank $r$. The latter is trivial if and only if $r=0$, i.e. $G=\{1\}$. 
To conclude, $S^2-A$ is simply connected. qed
