Complex analysis (conformal?) mapping question probably easy

  • #1

Homework Statement


We're supposed to find a bijective mapping from the open unit disk [itex]\{z : |z| < 1\}[/itex] to the sector [itex]\{z: z = re^{i \theta}, r > 0, -\pi/4 < \theta < \pi/4 \}[/itex].


Homework Equations





The Attempt at a Solution


This is confusing me. I tried to find a function that would map [itex][0,1)[/itex], which is the set of possible values of [itex]r[/itex] in the domain, injectively onto [itex](0,\infty)[/itex], which is the set of possible values of [itex]r[/itex] in the range. The best thing I could come up with is [itex]f(r) = \dfrac{1}{r(1-r)} - 4[/itex], but this is clearly not one-to-one, and it hits zero. What's more, I'm not sure how to find a function that will map the possible values for [itex]\text{Arg }z[/itex], which are [itex]-\pi < \text{Arg }z \leq \pi[/itex], injectively onto [itex](-\pi/4, \pi/4)[/itex].
 

Answers and Replies

  • #2
tiny-tim
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Hi AxiomOfChoice! :smile:
I tried to find a function that would map [itex][0,1)[/itex], which is the set of possible values of [itex]r[/itex] in the domain, injectively onto [itex](0,\infty)[/itex], which is the set of possible values of [itex]r[/itex] in the range.
why?? :redface:

Hint: get the boundary right, and everything else should fit in. :wink:
 
  • #3
Dick
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Don't try and mess around individually with r and theta. Just think about analytic functions. For example 1/(1-z) maps the disk into a half plane, right? Now find another function that can take a half plane into a wedge. Put them together.
 

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