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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?

Fizikus
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    Hint: Prove that the complement to such a curve is simply connected. Then use the monodromy theorem. – Moishe Kohan Nov 12 '17 at 01:15
  • @MoisheCohen: Easier said than done! How to prove that the complement to such a curve is simply connected? And, how does this fact constrain the function $F(x,y)$? – Fizikus Nov 12 '17 at 18:11
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    It depends on how much you know (I have no idea of your background). Do you know about Schoenflies theorem? Homology groups and Alexander duality? (If you want to use a spiral, none of these is needed, elementary tools will suffice.) It is not a particularly good idea to represent planar curves by equations (implicit or not) for the purpose of constructing branch cuts. All you need is a topological curve with simply connected complement. – Moishe Kohan Nov 12 '17 at 20:08
  • @MoisheCohen: Thanks, you have given me enough keywords to start googling in the right direction. If you could expand your comment a bit, I will gladly accept it as an answer. – Fizikus Nov 13 '17 at 10:43

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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:

  1. 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.

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

  3. 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

Moishe Kohan
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