Complex analysis (conformal?) mapping question probably easy

In summary, the conversation discusses finding a bijective mapping from the open unit disk to a sector, with a focus on finding a function that maps the possible values of r in the domain to the range and how to map the possible values for the argument of z injectively onto the given range. The suggestion is to use analytic functions, particularly 1/(1-z) to map the disk into a half plane and another function to take the half plane into the desired wedge.
  • #1
AxiomOfChoice
533
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].
 
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  • #2
Hi AxiomOfChoice! :smile:
AxiomOfChoice said:
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
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.
 

FAQ: Complex analysis (conformal?) mapping question probably easy

1. What is complex analysis and how is it different from real analysis?

Complex analysis is a branch of mathematics that deals with the study of functions on the complex plane, which is an extension of the real number line to include imaginary numbers. It differs from real analysis in that complex numbers have both a real and imaginary component, allowing for a richer understanding of functions and their properties.

2. What is a conformal mapping?

A conformal mapping is a type of complex function that preserves angles between curves. This means that the mapping does not distort the shape of the curves, but rather only changes their orientation. In other words, the mapping preserves the local geometry of the curves.

3. How is complex analysis used in real-world applications?

Complex analysis has many applications in different fields, including physics, engineering, and economics. It is used to solve problems involving electric circuits, fluid flow, heat transfer, and more. It also has applications in computer graphics and image processing.

4. Can you give an example of a conformal mapping?

One example of a conformal mapping is the complex exponential function, f(z) = ez. This function maps the complex plane onto the complex plane, preserving angles between curves. It is also known as a "rotation" mapping because it rotates the complex plane around the origin.

5. Is complex analysis difficult to learn?

Complex analysis can be challenging for beginners due to the introduction of complex numbers and the use of new concepts such as analytic functions and conformal mappings. However, with practice and a solid understanding of calculus, it can be mastered. As with any branch of mathematics, it requires dedication and persistence to fully grasp the concepts.

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