I would like to show, that the Logarithm gives a Riemann surface. Therefore I define the set $M:=\{(z,w)\in D \vert \ f(z,w)=0 \}$ with $D:=ℂ^*\times ℂ$ and $f(z,w):=e^w-z$.
Afterwards I will use the following Theorems (I don't prove them now), which I'm going to mention here:
Theorem 1: Let $U\subseteq ℂ²$ open, $f:D\rightarrow ℂ$ a holomorph function and $M:=\{(z,w)\in U \vert \ f(z,w)=0 \}$ the corresponding zero set. If $f$ is nonsingular and $M$ is connected, then $M$ is a Riemann surface.
Theorem 2: Let $X$ be connected, $F:X\rightarrow Y$ a continuous function. Then $F(X)$ is also connected.
Now I'm going to prove that $M$ is a Riemann surface. Obviously $f$ is holomorph and for all $p=(z,w)\in M$, because of $\frac{\partial f}{\partial z}(z,w)=-1\neq 0$ the function $f$ is nonsingular. We can write the set $M$ as $M=\{(z,w)\in D\vert \ e^w=z\}=\{(e^w,w)\vert \ w\in ℂ\}$. Now have a look at the function $F:ℂ \rightarrow D, \quad F(w):=(e^w,w)$. This function $F$ ist continuous, as each component of the function is continuous. As $ℂ$ is obviously connected, with Theorem 2 we get, that $F(ℂ)=M$ is also connected. All conditions for Theorem 1 are now fulfilled and that's why $M$ is a Riemann surface.
So my questions are:
- Is this proof true?
- Does this proof really show, that that $M$ is the Riemann surface of the logarithm?
- Does the Riemann surface look like this https://de.wikipedia.org/wiki/Riemannsche_Fläche#/media/File:Riemann_surface_log.svg or like this http://demonstrations.wolfram.com/RiemannSurfaceOfTheLogarithm/ and why?
Thank you very much! :)