# Open mapping Theorem and Rouches Theorem

My question is related to the proof here: https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)

Consider the closed Ball $$B$$ with radius $$d$$ around $$z_0$$ and a holomorphic function $$g(z)$$ which vanishes at $$z_0$$. Then the closed ball is mapped to $$g(B)$$ which should be also closed. Let $$e={\rm min}_{z \in \partial B} |g(z)|$$ be the minimal value of $$|g(z)|$$ on the boundary of $$B$$, so that the open disk $$D$$ at $$0$$ with radius $$e$$ is fully contained in $$g(B)$$, which is my assumption, since otherwise $$|g(z)|$$ would have a maximum inside $$B$$.

Now in the above link it seems to require Rouches theorem to claim that for any $$w\in D$$ there is a solution to $$g(z)=w$$ in $$B$$, but I feel like this statement is somehow trivial which it apparently is not. If $$D$$ is contained in $$g(B)$$ shouldn‘t this be automatically the case? What am I overlooking here?

So is the following argument valid?

Using the notation above: The boundary of $$D$$ - which has modulus $$e$$ - maps onto the boundary of $$g^{-1}(D)$$. This is because $$g$$ is holomorphic and by the maximum modulus principle attains its maximum modulus at the boundary of $$g^{-1}(D)$$, which however by construction is $$e$$ and therefore the boundary of $$D$$. Thus any value inside $$g^{-1}(D)$$ maps onto a value $$g(z)$$ with modulus $$ since $$g$$ is not constant.

The same argument is true for $$B$$ i.e. the maximum modulus is attained at the boundary of $$B$$. Since $$|g(z_0)|=0$$ and $$|g(z)|$$ does not have a local maximum inside $$B$$, but $$|g(\partial B)|\geq e$$ we must have $$g^{-1}(D) \subseteq B$$ or $$D\subseteq g(B)$$. (For any $$\phi \in [0,2\pi]$$ and $$r_1 we have $$\left|g\left(z_0 + r_1\, {\rm e}^{i\phi}\right)\right|<\left|g\left(z_0 + r_2\, {\rm e}^{i\phi}\right)\right|<\left|g\left(z_0 + d\, {\rm e}^{i\phi}\right)\right|$$)

I'd be really thankful if somebody could look over this argument!

• You don't automatically have the inclusion $D\subseteq g(B)$, not without proving it. – quasi Nov 2 '18 at 1:08
• So would it be also possible to proof this part with maximum modulus principle for $|g(z)|$ which is then attained at the boundary of $g^{-1}(D)$. – Diger Nov 2 '18 at 11:41
• I don't see how that would immediately yield $D\subseteq g(B)$. But feel free to post what you think is a proof, either by editing your question, or by posting an answer. That way someone (maybe not me) will be able to review your proposed argument. – quasi Nov 2 '18 at 12:19
• I actually have issues to formulate it, but if the maximum is obtained at the boundary (which by definition is $e$), how can any $z$ inside $g^{-1}(D)$ attain a value $g(z)$ larger than $e$ in modulus. – Diger Nov 2 '18 at 14:39
• I added a picture above. The point is that I think since B is simply connected so is g(B) and the minimal radius e defines D which then is fully contained in g(B). Also I think a boundary of B becomes a boundary of g(B). – Diger Nov 2 '18 at 18:15