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 .
- 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 $\frac{z-r}{1-rz}, 0<r<1$ 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)