# Proving two domains not to be conformally equivalent and property of holomorphic map

I am trying questions in complex analysis of an institute in which I dont study . My course was marred by a non serious professor and hence I have to study from resources online and hence I could not have good problem solving skills in complex analysis .

Questions :1. Prove that {z:0<|z|<1} and {z:r<|z|<R} are not conformally equivalent if r>0 .

I have studied defination of conformally equivalent and related theorems from Ponnusamy Silvermann but I am unable to think about which result I should use .

1. Show that if a holomorphic map f maps U(Unit disc with centre z=0) into itself it need not have a fixed point in U even if it extends to a continuous map of the closure of U onto itself .

For this question I dont have any clue on how it should be solved.

It is my humble request to you to help as I cant ask for help in questions anywhere else . I admit that I for now is not able to provide reasonable attempt but I have studied from book (Ponnusamy Silvermann) in detail. Thanks!!

For 1: the result is easy since if $$f:\mathbb D^* \to A(r,R)$$ is a holomorphic map, $$f$$ is bounded hence $$0$$ is a removable singularity, which means that $$f$$ is a map from the unit disc to the annulus; if $$f$$ is $$1-1$$ on the punctured disc, it is $$1-1$$ on the full disc ($$f(0)=f(w)$$ means that some neighborhood of $$0$$ and $$w$$ are mapped to the same open set around the common value) and it obviously cannot be onto since $$f(\partial D)$$ is connected

For 2: take $$\frac{z-r}{1-rz}, 0 which is an automorphism of the closed unit disc and has its two fixed points at $$\pm 1$$; the result is sharp since the Brouwer fixed point theorem shows that any continuous map of the closed unit disc to itself has fixed points

(here the result is obvious but in general, for any Mobius transform, the equation $$f(z)=z$$ becomes a quadratic after eliminating denominators so it cannot have more than two solutions)

• can you please tell how does in answer of 1 , f being bounded implies 0 is a removable singularity ?
– user775699
Sep 1, 2020 at 17:10
• This is one of the first results in complex analysis - isolated singularities are removable, poles or essential, and they are removable precisely when the function is bounded on a small neighborhood of them - proof: take $g=z^2f$; it is obvious that $g$ is differentiable in the full disc with $g'(0)=0$ since $g(0)=0, g(z)/z \to 0$ by the boundness of $f$; hence (this is a very strong hence encompassing the power of complex analysis) $g$ is analytic; $g(0)=g'(0)=0$, so $g(z)=z^2h(z)$ for some analytic $h$; so extending $f(0)=h(0)$ gives the required anlaytic extension of $f$ to zero Sep 1, 2020 at 17:26
• I am really sorry man but I am not able to understand in 1. how it can't be onto as f($\delta D$) is connected . Can you please help me I want to understand the solution.
– user775699
Jan 21, 2021 at 10:24
• @Cornad In 2 . we are asked to gve an example of a question which doesn't have fixed points and you gave example of an automophism having fixed points. and you are using So, It question 2. wrong as if I use Brower fixed point theorem then I am getting the existence of fixed point.
– user775699
Jan 21, 2021 at 10:42
• for 2, there is no fixed point in the open unit disc, which is what the problem asks; as noted there is always a fixed point in the closed unit disc by classical theorems; for 1, the connectedness of the boundary of the unit disc and the fact that $f$ is assumed to be injective inside, means that the image of the unit disc has a connected boundary and the annulus plainly has not such, so it cannot be the image; that is true for even a continuous injective map by topological considerations only Jan 21, 2021 at 12:58