Mapping Argand Plane to Upper Half Plane

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    Complex Mapping
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Homework Help Overview

The discussion revolves around finding a linear fractional transformation that maps a specified domain D, defined by the argument of complex numbers, to the upper half-plane. The subject area involves complex analysis and transformations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the definition of the domain D and its implications based on the value of α. There is uncertainty about whether D represents the entire complex plane when α equals π, as well as how to visualize the domain for different values of α.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the domain D based on the argument of z. Some guidance has been offered regarding specific cases, but no consensus has been reached on the overall characterization of D.

Contextual Notes

There is a focus on the implications of the value of α in defining the domain D, particularly the distinction between different values of α and their geometric interpretations in the complex plane.

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Homework Statement


find linear fractional transformation from D={z:|Arg z| < \alpha}, \alpha≤\pi to the upper half plane


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The Attempt at a Solution



The problem I am having here what exactly D is.. (visualizing it) D is just z such that |Arg z|≤\pi right? so wouldn't that just be the entire complex plane? If we consider the Argument from 0 to \pi and from 0 to -\pi since |-\pi|=\pi Is this correct??
 
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I believe what this is saying is that you first select \alpha \leq \pi and then form D := \left\{ z : \left| \mathit{arg} \ z \right| &lt; \alpha \right\}. This would not be the entire complex plane. I believe for say \alpha = \frac{\pi}{2} would look like D = \left\{ z : \mathit{real} \ z &gt; 0 \right\}.
 
right but then if \alpha=\pi wouldn't it just be the entire plane? since if it is \pi/2 then it is half of the plane..
 

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