Complex analysis (conformal?) mapping question probably easy

AxiomOfChoice
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Homework Statement


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

Homework Equations


The Attempt at a Solution


This is confusing me. I tried to find a function that would map [0,1), which is the set of possible values of r in the domain, injectively onto (0,\infty), which is the set of possible values of r in the range. The best thing I could come up with is f(r) = \dfrac{1}{r(1-r)} - 4, 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 \text{Arg }z, which are -\pi < \text{Arg }z \leq \pi, injectively onto (-\pi/4, \pi/4).
 
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Hi AxiomOfChoice! :smile:
AxiomOfChoice said:
I tried to find a function that would map [0,1), which is the set of possible values of r in the domain, injectively onto (0,\infty), which is the set of possible values of r in the range.

why?? :redface:

Hint: get the boundary right, and everything else should fit in. :wink:
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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