Find a+b in Regular Hexagon Complex Plane Problem

Click For Summary
SUMMARY

The problem involves calculating the area of the set S = {1/z | z ∈ R}, where R is the region outside a regular hexagon centered at the origin in the complex plane. The hexagon has opposite pairs of sides one unit apart, with one pair parallel to the imaginary axis. The area of S can be expressed as aπ + √b, where a and b are positive integers. The solution requires understanding Mobius transformations and potentially using calculus to determine the area accurately.

PREREQUISITES
  • Understanding of complex numbers and the complex plane
  • Familiarity with Mobius transformations
  • Basic calculus concepts for area calculation
  • Knowledge of geometric properties of regular hexagons
NEXT STEPS
  • Study Mobius transformations in detail
  • Learn about complex analysis techniques for area calculations
  • Explore geometric properties of regular polygons, specifically hexagons
  • Review inversion in the complex plane and its applications
USEFUL FOR

Students studying complex analysis, high school mathematics educators, and anyone interested in advanced geometry and transformation techniques.

ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let R be the region outside the hexagon, and let S = [itex]\{ 1/z |x \in R}[/itex]. Then the area of S has the form a [itex]\pi +\sqrt b[/itex], where a and b are positive integers. Find a+b.

Homework Equations


The Attempt at a Solution


This should only require high school math although there is probably a solution using the fact that 1/z is a Mobius transformation or something else in complex analysis.

The hexagon is contained in the closed of radius 1/sqrt(3) center at the origin, which means that S is contained inside of the closed disk of radius \sqrt3 centered at the origin. So basically we need to figure out what to subtract off of 3 \pi. Anyone know how to do that?

Please just give a hint.
 
Physics news on Phys.org
What you are trying to do is a good idea, but there doesn't seem to be an easy way to find the cutoffs, as you've noticed.

I caved and peeked at what others did. To solve it without calculus, you need to know what an inversion is. That said, there is a fairly simple solution with calculus that a lot of high school students are probably more familiar with than with inversions.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
24
Views
4K