Show that the given function maps the open unit disk into the upper half plane

In summary, the task is to prove that the function w=f(z)=i(\frac{1-z}{1+z}) maps the open unit disk in the complex plane {z \in C | z < 1} to the upper half-plane {w\in C|Im(w) >0}, and also maps the unit circle {z\in C||z|=1} to the real line. This can be shown by plugging in values less than one for z in w and getting an imaginary result, and plugging in one for z in w and getting a real result. To further prove this, one can choose a z with |z|<1 and show that Im(w)>0.
  • #1
Raziel2701
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Homework Statement


4. Let w = f(z) = [tex]i(\frac{1-z}{1+z})[/tex]. Show that f maps the open unit disk {z [tex]\in[/tex] C | z < 1} into the upper half-plane {w[tex]\in[/tex] C|Im(w) >0}, and maps the unit circle {z[tex]\in[/tex] C||z|=1} to the real line.


Homework Equations


I was given this hint:

"set w=[tex]i(\frac{1-z}{1+z})[/tex] and use the formula Im(w)= [tex]\frac{1}{2i}[/tex](w -[tex]\bar{w}[/tex])"


The Attempt at a Solution


This is cliche but, what does this mean in English? I've been trying to decipher some of this stuff, in order for me to know what to do, I must first understand what I'm being asked to do, so that would be my first request.

The second thought I have of this is that the hint given also doesn't mean much to me. So what exactly would be the topic I could read on to help me get more information on this concept I'm being tested on? I'm at a library right now, so if I were to pick up a book on complex variables, what topic more or less is this problem covering?

I need to be pushed on the right direction to solve this problem, right now I'm just more or less in the dark. The set notation is a bit cryptic for me. I get that z is an element of the set of complex numbers, but what exactly is it being said after the "|"?

Thanks!
And pardon the rough formatting.
 
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  • #2
Is it sufficient to show that since for the open disk z < 1, then if I plug a value less than one into w I get a result that is less than one and imaginary, and thus, it maps any point of the open disk into the upper half plane?

The same for the unit circle where |z| = 1. If I plug one I get zero for w, and thus I get the point (0,0) which is in the real line and therefore w maps any point on the unit circle to the real line?
 
  • #3
So you have shown that the map w, maps boundaries to boundaries, now choose a z with |z|<1 and show that Im(w)>0, so perhaps write [tex]z=re^{i\theta}[/tex] with r<1 and see if Im(w)>0.
 

1. What is the function being mapped?

The given function maps points in the open unit disk to points in the upper half plane. This means that for any point (x,y) in the unit disk, the function will output a point (u,v) in the upper half plane.

2. How do we know the function maps the open unit disk into the upper half plane?

In order for a function to map one set of points to another, it must satisfy two conditions: it must be one-to-one (each input has a unique output) and it must be onto (every point in the output set is covered by the function). These conditions can be proven by using mathematical techniques such as the Inverse Function Theorem.

3. What does "open unit disk" and "upper half plane" mean?

The open unit disk is a set of all points in the complex plane that have a distance from the origin less than 1. In other words, it is the interior of the unit circle. The upper half plane is the set of all complex numbers with an imaginary part (y-coordinate) greater than 0.

4. Can you provide an example of a function that maps the open unit disk into the upper half plane?

One example of such a function is f(z) = (z+2)/(z-1), where z is a complex number. This function satisfies the conditions of being one-to-one and onto, and can be proven to map the open unit disk into the upper half plane.

5. Why is it important to show that a function maps the open unit disk into the upper half plane?

This type of mapping is commonly used in complex analysis and other mathematical fields. It is important to show that a function maps one set of points into another because it allows for the use of different techniques and theorems that are specific to that type of mapping. In this case, proving that the function maps the open unit disk into the upper half plane allows us to use techniques such as conformal mapping and the Cauchy-Riemann equations in further analysis of the function.

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