Mapping Argand Plane to Upper Half Plane

In summary, the conversation discusses finding a linear fractional transformation from a set D in the complex plane, where D is defined as z such that |Arg z| < alpha and alpha is less than or equal to pi. There is confusion about what D represents, but it is determined that for alpha = pi/2, D would consist of z with real values greater than 0. However, for alpha = pi, D would cover the entire complex plane.
  • #1
d2j2003
58
0

Homework Statement


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


Homework Equations





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|≤[itex]\pi[/itex] right? so wouldn't that just be the entire complex plane? If we consider the Argument from 0 to [itex]\pi[/itex] and from 0 to -[itex]\pi[/itex] since |-[itex]\pi[/itex]|=[itex]\pi[/itex] Is this correct??
 
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  • #2
anyone??
 
  • #3
I believe what this is saying is that you first select [itex]\alpha \leq \pi[/itex] and then form [itex]D := \left\{ z : \left| \mathit{arg} \ z \right| < \alpha \right\}[/itex]. This would not be the entire complex plane. I believe for say [itex]\alpha = \frac{\pi}{2}[/itex] would look like [itex]D = \left\{ z : \mathit{real} \ z > 0 \right\}[/itex].
 
  • #4
right but then if [itex]\alpha[/itex]=[itex]\pi[/itex] wouldn't it just be the entire plane? since if it is [itex]\pi[/itex]/2 then it is half of the plane..
 

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