Complex Analysis - Proving a bijection on a closed disk

In summary: You can start by writing phi_w and phi_w* explicitly using the given formula and then simplifying. In summary, we are asked to prove that the function phi_w is a bijection on the closed disk \bar{D} for w not in \bar{D}. To do this, we first need to show that both phi_w and its inverse map the circle \mathbb{T} into itself. This can be done by substituting zz* into the formula for phi_w and simplifying.
  • #1

Homework Statement



For each [tex] w \in \mathbb{C} [/tex] define the function [tex] \phi_w [/tex] on the open set [tex] \mathbb{C}\backslash \{\bar{w}^{-1}\} [/tex] by [tex] \phi_w (z) = \frac{w - z}{1 - \bar{w}z}[/tex], for [tex] z \in \mathbb{C}\backslash \{\bar{w}^{-1}\} \back[/tex].

Prove that [tex] \phi_w : \bar{D} \mapsto \bar{D} [/tex] is a bijection on the closed disk [tex] \bar{D} [/tex] for [tex] w \notin \bar{D} [/tex].

Hint: Compute the inverse of [tex] \phi_w[/tex] and prove first that both [tex] \phi_w[/tex] and [tex](\phi_w)^{-1} [/tex] map the circle [tex]\mathbb{T} [/tex] into itself.

Homework Equations





The Attempt at a Solution



So following from the hint, I can calculate the inverse of [tex] \phi_w [/tex]. But I before I can start, I'm getting lost even at how I would show mapping the function and it's inverse to a circle...(I'm even more confused by "into itself".)
 
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  • #2
You want to show that if |z|=1 then |phi_w(z)|=1. Use that |z|=1 is equivalent to zz*=1.
 
  • #3
So I would substitute zz* into |phi_w(z)| to see if it's 1?
 
  • #4
semithinking said:
So I would substitute zz* into |phi_w(z)| to see if it's 1?

No. You want to show (phi_w(z))(phi_w(z))*=1 for any z such that zz*=1.
 

What is complex analysis?

Complex analysis is a branch of mathematics that deals with functions of complex numbers. It involves the study of complex-valued functions, their derivatives, and their properties.

What is a bijection?

A bijection is a function that is both injective (one-to-one) and surjective (onto). This means that each element in the domain maps to a unique element in the range, and every element in the range has at least one preimage in the domain.

What is a closed disk?

A closed disk is a subset of the complex plane that includes all points within a given distance (radius) from a fixed point (center). It is denoted by the symbol D(z, r) where z is the center and r is the radius.

How do you prove a function is a bijection on a closed disk?

To prove a function is a bijection on a closed disk, you must first show that it is both injective and surjective. This can be done by using algebraic or geometric arguments, depending on the specific function. Additionally, you must show that the function is continuous on the closed disk and has an inverse function that is also continuous.

Why is proving a bijection on a closed disk important?

Proving a bijection on a closed disk is important because it allows us to establish a one-to-one correspondence between complex numbers within the disk and points on the complex plane. This can be useful in many applications, such as solving complex equations or analyzing complex functions.

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